2264 lines
74 KiB
Plaintext
2264 lines
74 KiB
Plaintext
==========图论模板 LCA=========
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//LCA (Least Common Ancestors, 最近公共祖先)
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//在线dfs + ST O(nlogn + q)
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int dfso[N << 1], cnt; //欧拉序列, 即dfs序, 长度为2n-1, 下标从1开始
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int pos[N]; //pos[i]表示点i在欧拉序列中第一次出现的位置
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int dep[N << 1]; //欧拉序列对应的深度
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int p[N << 1] = { -1}, dp[N << 1][20];
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void initRMQ(int n) {
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for (int i = 1; i <= n; i++) { p[i] = p[i - 1] + !(i & (i - 1)); dp[i][0] = i; }
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for (int j = 1; j <= p[n]; j++) {
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for (int i = 1; i + (1 << j) - 1 <= n; i++) {
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dp[i][j] = dep[dp[i][j - 1]] < dep[dp[i + (1 << (j - 1))][j - 1]] ? dp[i][j - 1] : dp[i + (1 << (j - 1))][j - 1];
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}
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}
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}
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inline int queryRMQ(int l, int r) {
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if (l > r) { swap(l, r); }
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int k = p[r - l + 1];
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return dep[dp[l][k]] <= dep[dp[r - (1 << k) + 1][k]] ? dp[l][k] : dp[r - (1 << k) + 1][k];
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}
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void dfs(int u, int p, int d) {
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dfso[++cnt] = u; dep[cnt] = d; pos[u] = cnt;
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for (int i = head[u]; ~i; i = nxt[i]) {
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int v = to[i];
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if (v != p) { dfs(v, u, d + 1); dfso[++cnt] = u; dep[cnt] = d; }
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}
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}
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void initLCA(int rt, int n) {
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cnt = 0; dfs(rt, rt, 0); initRMQ(2 * n - 1);
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}
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inline int queryLCA(int u, int v) { //查询u和v的LCA编号
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return dfso[queryRMQ(pos[u], pos[v])];
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}
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//倍增法 O(nlogn + qlogn)
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const int DEP = 20;
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int fa[N][DEP]; //fa[i][j]表示结点i的第2^j个祖先
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int dep[N]; //深度
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void bfs(int rt) {
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queue<int> que; que.push(rt);
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dep[rt] = 0; fa[rt][0] = rt;
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while (!que.empty()) {
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int u = que.front(); que.pop();
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for (int i = 1; i < DEP; i++) { fa[u][i] = fa[fa[u][i - 1]][i - 1]; }
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for (int i = head[u]; ~i; i = nxt[i]) {
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int v = to[i];
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if (v != fa[u][0]) { dep[v] = dep[u] + 1; fa[v][0] = u; que.push(v); }
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}
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}
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}
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int queryLCA(int u, int v) {
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if (dep[u] > dep[v]) { swap(u, v); }
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for (int d = dep[v] - dep[u], i = 0; d; d >>= 1, i++) {
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if (d & 1) { v = fa[v][i]; }
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}
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if (u == v) { return u; }
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for (int i = DEP - 1; i >= 0; i--) {
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if (fa[u][i] != fa[v][i]) { u = fa[u][i]; v = fa[v][i]; }
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}
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return fa[u][0];
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}
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//离线Tarjan O(n + q)
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int n, fa[N], ancestor[N], ans[N];
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bool vis[N];
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int qhead[N], qto[M], qnxt[M], qtot;
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void init() {
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qtot = 0; memset(qhead, -1, sizeof(qhead));
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memset(ancestor, 0, sizeof(ancestor)); memset(vis, 0, sizeof(vis));
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for (int i = 0; i <= n; i++) { fa[i] = i; }
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}
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void qaddedge(int x, int y) {
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qto[qtot] = y; qnxt[qtot] = qhead[x]; qhead[x] = qtot++;
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qto[qtot] = x; qnxt[qtot] = qhead[y]; qhead[y] = qtot++;
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}
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int findfa(int n) { return n == fa[n] ? n : fa[n] = findfa(fa[n]); }
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inline void unite(int x, int y) {
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x = findfa(x); y = findfa(y);
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if (x != y) { fa[y] = x; }
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}
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void Tarjan_LCA(int u) {
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ancestor[u] = u; vis[u] = true;
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for (int i = head[u]; ~i; i = nxt[i]) {
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int v = to[i];
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if (!vis[v]) { Tarjan_LCA(v); unite(u, v); ancestor[findfa(u)] = u; }
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}
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for (int i = qhead[u]; ~i; i = qnxt[i]) {
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int v = qto[i];
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if (vis[v]) { ans[i >> 1] = ancestor[findfa(v)]; }
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}
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}
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==================图论-连通性==================
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//图的割点、桥与双连通分量
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//[点连通度与边连通度]
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//在一个无向连通图中, 如果有一个顶点集合, 删除这个顶点集合,
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//以及这个集合中所有顶点相关联的边以后, 原图变成多个连通块, 就称这个点集为割点集合
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//一个图的点连通度的定义为, 最小割点集合中的顶点数
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//类似的, 如果有一个边集合, 删除这个边集合以后, 原图变成多个连通块, 就称这个点集为割边集合
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//一个图的边连通度的定义为, 最小割边集合中的边数
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//[双连通图、割点与桥]
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//如果一个无向连通图的点连通度大于1, 则称该图是点双连通的(point biconnected), 简称双连通或重连通
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//一个图有割点, 当且仅当这个图的点连通度为1, 则割点集合的唯一元素被称为割点(cut point), 又叫关节点(articulation point)
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//如果一个无向连通图的边连通度大于1, 则称该图是边双连通的(edge biconnected), 简称双连通或重连通
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//一个图有桥, 当且仅当这个图的边连通度为1, 则割边集合的唯一元素被称为桥(bridge), 又叫关节边(articulation edge)
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//可以看出, 点双连通与边双连通都可以简称为双连通, 它们之间是有着某种联系的, 下文中提到的双连通, 均既可指点双连通, 又可指边双连通
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//[双连通分量]
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//在图G的所有子图G'中, 如果G'是双连通的, 则称G'为双连通子图
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//如果一个双连通子图G'它不是任何一个双连通子图的真子集, 则G'为极大双连通子图
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//双连通分量(biconnected component), 或重连通分量, 就是图的极大双连通子图。特殊的, 点双连通分量又叫做块
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//[求割点与桥]
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//对图深度优先搜索, 定义DFS(u)为u在搜索树(以下简称为树)中被遍历到的次序号
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//定义Low(u)为u或u的子树中能通过非父子边追溯到的最早的节点, 即DFS序号最小的节点
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//根据定义, 则有: Low(u) = Min {DFS(u) DFS(v) (u,v)为后向边(返祖边) 等价于DFS(v) < DFS(u)且v不为u的父亲节点Low(v) (u,v)为树枝边(父子边)}
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//一个顶点u是割点, 当且仅当满足(1)或(2)
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//(1)u为树根, 且u有多于一个子树
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//(2)u不为树根, 且满足存在(u,v)为树枝边(或称父子边, 即u为v在搜索树中的父亲), 使得DFS(u)<=Low(v)
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//一条无向边(u,v)是桥, 当且仅当(u,v)为树枝边, 且满足DFS(u)<Low(v)
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//[求双连通分量]
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//对于点双连通分量, 实际上在求割点的过程中就能顺便把每个点双连通分量求出
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//建立一个栈, 存储当前双连通分量, 在搜索图时, 每找到一条树枝边或后向边(非横叉边), 就把这条边加入栈中
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//如果遇到某时满足DFS(u)<=Low(v), 说明u是一个割点, 同时把边从栈顶一个个取出,
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//直到遇到了边(u,v), 取出的这些边与其关联的点, 组成一个点双连通分量
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//割点可以属于多个点双连通分量, 其余点和每条边只属于且属于一个点双连通分量
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//对于边双连通分量, 只需在求出所有的桥以后, 把桥边删除, 原图变成了多个连通块, 则每个连通块就是一个边双连通分量
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//桥不属于任何一个边双连通分量, 其余的边和每个顶点都属于且只属于一个边双连通分量
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//[构造双连通图]
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//方法为首先求出所有的桥, 然后删除这些桥边, 剩下的每个连通块都是一个双连通子图
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//把每个双连通子图收缩为一个顶点, 再把桥边加回来, 最后的这个图一定是一棵树, 边连通度为1
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//统计出树中度为1的节点的个数, 即为叶节点的个数, 记为leaf
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//则至少在树上添加(leaf + 1) / 2条边, 就能使树达到边二连通, 所以至少添加的边数就是(leaf + 1) / 2
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//具体方法为, 首先把两个最近公共祖先最远的两个叶节点之间连接一条边, 这样可以把这两个点到祖先的路径上所有点收缩到一起,
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//因为一个形成的环一定是双连通的。然后再找两个最近公共祖先最远的两个叶节点, 这样一对一对找完, 恰好是(leaf + 1) / 2次, 把所有点收缩到了一起
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//有向图的强连通分量
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//Kosaraju算法 O(V + E)
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vector<int> e[N], re[N], pos;
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bool vis[N];
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int belong[N], num[N], scc, cnt; //所属分量, 分量点数, 分量个数
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void dfs(int u) {
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vis[u] = true;
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for (int i = 0; i < (int)e[u].size(); i++) {
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if (!vis[e[u][i]]) { dfs(e[u][i]); }
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}
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pos.push_back(u);
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}
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void rdfs(int u, int k) {
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vis[u] = true; belong[u] = k; cnt++;
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for (int i = 0; i < (int)re[u].size(); i++) {
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if (!vis[re[u][i]]) { rdfs(re[u][i], k); }
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}
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}
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void SCC(int n) {
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memset(vis, 0, sizeof(vis)); pos.clear();
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for (int i = 0; i < n; i++) { if (!vis[i]) { dfs(i); } }
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memset(vis, 0, sizeof(vis)); memset(num, 0, sizeof(num)); scc = 0;
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for (int i = (int)pos.size() - 1; i >= 0; i--) {
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if (!vis[pos[i]]) { cnt = 0; rdfs(pos[i], ++scc); num[scc] = cnt; }
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}
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}
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//Tarjan算法 O(V + E)
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vector<int> e[N];
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int low[N], dfn[N], index, belong[N], num[N], scc; //所属分量, 分量点数, 分量个数
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stack<int> stk;
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bool instack[N];
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void Tarjan(int u) {
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int v; low[u] = dfn[u] = ++index; stk.push(u); instack[u] = true;
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for (int i = 0; i < (int)e[u].size(); i++) {
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int v = e[u][i];
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if (!dfn[v]) { Tarjan(v); low[u] = min(low[u], low[v]); }
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else if (instack[v] && low[u] > dfn[v]) { low[u] = dfn[v]; }
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}
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if (low[u] == dfn[u]) {
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scc++;
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do {
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v = stk.top(); stk.pop();
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instack[v] = false; belong[v] = scc; num[scc]++;
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} while (v != u);
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}
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}
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void SCC(int n) {
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memset(dfn, 0, sizeof(dfn)); memset(num, 0, sizeof(num));
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memset(instack, 0, sizeof(instack)); index = scc = 0;
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while (!stk.empty()) { stk.pop(); }
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for (int i = 0; i < n; i++) { if (!dfn[i]) { Tarjan(i); } }
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}
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//无向图的割点和桥
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//Tarjan算法 O(V + E)
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int low[N], dfn[N], index, addblock[N], bridge;
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bool instack[N], cut[N], ecut[M];
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stack<int> stk;
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void Tarjan(int u, int p) {
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int son = 0; low[u] = dfn[u] = ++index;
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stk.push(u); instack[u] = true;
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for (int i = head[u]; ~i; i = nxt[i]) {
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int v = to[i];
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if (v == p) { continue; }
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if (!dfn[v]) {
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Tarjan(v, u); son++;
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low[u] = min(low[u], low[v]);
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if (u != p && low[v] >= dfn[u]) { //割点
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cut[u] = true; addblock[u]++;
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}
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if (low[v] > dfn[u]) { //割边
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bridge++; ecut[i] = ecut[i ^ 1] = true;
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}
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} else if (low[u] > dfn[v]) { low[u] = dfn[v]; }
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}
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if (u == p && son > 1) { cut[u] = true; }
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if (u == p) { addblock[u] = son - 1; }
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instack[u] = false; stk.pop();
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}
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void CUT(int n) {
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memset(dfn, 0, sizeof(dfn)); memset(addblock, 0, sizeof(addblock));
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memset(instack, 0, sizeof(instack)); memset(cut, 0, sizeof(cut)); memset(ecut, 0, sizeof(ecut));
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while (!stk.empty()) { stk.pop(); } index = bridge = 0;
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for (int i = 0; i < n; i++) { if (!dfn[i]) { Tarjan(i, i); } }
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}
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//无向图的边双连通分量
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//求出所有的桥以后, 把桥边删除, 原图变成了多个连通块, 则每个连通块就是一个边双连通分量
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//桥不属于任何一个边双连通分量, 其余的边和每个顶点都属于且只属于一个边双连通分量
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//Tarjan算法 O(V + E)
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int low[N], dfn[N], index, belong[N], bridge, block;
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bool instack[N], ecut[M];
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stack<int> stk;
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void Tarjan(int u, int p) {
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low[u] = dfn[u] = ++index; stk.push(u); instack[u] = true;
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for (int i = head[u]; ~i; i = nxt[i]) {
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int v = to[i];
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if (v == p) { continue; }
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if (!dfn[v]) {
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Tarjan(v, u);
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low[u] = min(low[u], low[v]);
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if (low[v] > dfn[u]) { //割边
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bridge++; ecut[i] = ecut[i ^ 1] = true;
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}
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} else if (low[u] > dfn[v]) { low[u] = dfn[v]; }
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}
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if (low[u] == dfn[u]) {
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int v; block++;
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do {
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v = stk.top(); stk.pop(); instack[v] = false; belong[v] = block;
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} while (u != v);
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}
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}
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void EBCC(int n) {
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memset(dfn, 0, sizeof(dfn)); memset(instack, 0, sizeof(instack)); memset(cut, 0, sizeof(cut));
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while (!stk.empty()) { stk.pop(); } index = bridge = block = 0;
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for (int i = 0; i < n; i++) { if (!dfn[i]) { Tarjan(i, -1); } }
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}
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//无向图的点双连通分量
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//对于点双连通分量, 实际上在求割点的过程中就能顺便把每个点双连通分量求出
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//建立一个栈, 存储当前双连通分量, 在搜索图时, 每找到一条树枝边或后向边(非横叉边), 就把这条边加入栈中
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//如果遇到某时满足DFS(u)<=Low(v), 说明u是一个割点, 同时把边从栈顶一个个取出,
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//直到遇到了边(u,v), 取出的这些边与其关联的点, 组成一个点双连通分量
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//割点可以属于多个点双连通分量, 其余点和每条边只属于且属于一个点双连通分量
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//Tarjan算法 O(V + E)
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int low[N], dfn[N], index, belong[N], block; //所属分量, 分量点数, 分量个数
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stack<int> stk;
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bool instack[N];
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void Tarjan(int u, int p) {
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low[u] = dfn[u] = ++index;
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stk.push(u); instack[u] = true;
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for (int i = head[u]; ~i; i = nxt[i]) {
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int v = to[i];
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if (v == p) { continue; }
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if (!dfn[v]) {
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Tarjan(v, u);
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low[u] = min(low[u], low[v]);
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if (low[v] >= dfn[u]) { //u为割点
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int vv; block++;
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do {
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vv = stk.top(); stk.pop(); instack[vv] = false;
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belong[vv] = block;
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} while (vv != v);
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}
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} else if (instack[v] && low[u] > dfn[v]) { low[u] = dfn[v]; }
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}
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}
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void BCC(int n) {
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memset(dfn, 0, sizeof(dfn)); memset(instack, 0, sizeof(instack));
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while (!stk.empty()) { stk.pop(); } index = block = 0;
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for (int i = 0; i < n; i++) { if (!dfn[i]) { Tarjan(i, -1); } }
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}
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//构造双连通图
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//方法为首先求出所有的桥, 然后删除这些桥边, 剩下的每个连通块都是一个双连通子图
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//把每个双连通子图收缩为一个顶点, 再把桥边加回来, 最后的这个图一定是一棵树, 边连通度为1
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//统计出树中度为1的节点的个数, 即为叶节点的个数, 记为leaf
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//则至少在树上添加(leaf + 1) / 2条边, 就能使树达到边二连通, 所以至少添加的边数就是(leaf + 1) / 2
|
||
//具体方法为, 首先把两个最近公共祖先最远的两个叶节点之间连接一条边, 这样可以把这两个点到祖先的路径上所有点收缩到一起,
|
||
//因为一个形成的环一定是双连通的。然后再找两个最近公共祖先最远的两个叶节点, 这样一对一对找完, 恰好是(leaf + 1) / 2次, 把所有点收缩到了一起
|
||
|
||
|
||
=============================图论-匹配============================
|
||
//图论点、边集和二分图的相关概念和性质
|
||
//点覆盖: 点覆盖集即一个点集, 使得所有边至少有一个端点在集合里
|
||
//边覆盖: 边覆盖集即一个边集, 使得所有点都与集合里的边邻接
|
||
//独立集: 独立集即一个点集, 集合中任两个结点不相邻, 则称V为独立集
|
||
//团: 团即一个点集, 集合中任两个结点相邻
|
||
//边独立集: 边独立集即一个边集, 满足边集中的任两边不邻接
|
||
//支配集: 支配集即一个点集, 使得所有其他点至少有一个相邻点在集合里
|
||
//边支配集: 边支配集即一个边集, 使得所有边至少有一条邻接边在集合里
|
||
//最小路径覆盖: 用尽量少的不相交简单路径覆盖有向无环图G的所有顶点, 即每个顶点严格属于一条路径, 路径的长度可能为0(单个点)
|
||
//匹配: 匹配是一个边集, 满足边集中的边两两不邻接
|
||
//在匹配中的点称为匹配点或饱和点；反之, 称为未匹配点或未饱和点
|
||
//交错轨(alternating path): 图的一条简单路径, 满足任意相邻的两条边, 一条在匹配内, 一条不在匹配内
|
||
//增广轨(augmenting path): 是一个始点与终点都为未匹配点的交错轨
|
||
//最大匹配(maximum matching): 具有最多边的匹配
|
||
//匹配数(matching number): 最大匹配的大小
|
||
//完美匹配(perfect matching): 匹配了所有点的匹配
|
||
//完备匹配(complete matching): 匹配了二分图较小集合（二分图X, Y中小的那个）的所有点的匹配
|
||
//增广轨定理: 一个匹配是最大匹配当且仅当没有增广轨
|
||
//所有匹配算法都是基于增广轨定理: 一个匹配是最大匹配当且仅当没有增广轨
|
||
//最大匹配数 = 最小覆盖数 = 左边匹配点 + 右边未匹配点
|
||
//最小边覆盖 = 图中点的个数 - 最大匹配数 = 独立数
|
||
|
||
//二分图最大匹配
|
||
//Hungary + dfs
|
||
//邻接矩阵 O(V^2)
|
||
int uN, vN, match[N]; //左右点的数目
|
||
bool g[N][N], check[N];
|
||
bool dfs(int u) {
|
||
for (int v = 0; v < vN; v++) {
|
||
if (g[u][v] && !check[v]) {
|
||
check[v] = true;
|
||
if (match[v] == -1 || dfs(match[v])) {
|
||
match[v] = u; match[u] = v; return true;
|
||
}
|
||
}
|
||
}
|
||
return false;
|
||
}
|
||
//邻接表 O(V*E)
|
||
int head[N], to[M], nxt[M], tot, uN, match[N];
|
||
bool check[N];
|
||
void init() { tot = 0; memset(head, -1, sizeof(head)); }
|
||
void addedge(int x, int y) { to[tot] = y; nxt[tot] = head[x]; head[x] = tot++; }
|
||
bool dfs(int u) {
|
||
for (int i = head[u]; ~i; i = nxt[i]) {
|
||
int v = to[i];
|
||
if (!check[v]) {
|
||
check[v] = true;
|
||
if (match[v] == -1 || dfs(match[v])) {
|
||
match[v] = u; match[u] = v; return true;
|
||
}
|
||
}
|
||
}
|
||
return false;
|
||
}
|
||
int Hungary() {
|
||
memset(match, -1, sizeof(match));
|
||
int res = 0;
|
||
for (int u = 0; u < uN; u++) {
|
||
if (match[u] == -1) {
|
||
memset(check, 0, sizeof(check));
|
||
if (dfs(u)) { res++; }
|
||
}
|
||
}
|
||
return res;
|
||
}
|
||
//Hungary + bfs + 邻接矩阵 O(V*E)
|
||
bool g[N][N];
|
||
int uN, vN, match[N], check[N], pre[N];
|
||
int Hungary() {
|
||
memset(match, -1, sizeof(match));
|
||
memset(check, -1, sizeof(check));
|
||
queue<int> que;
|
||
int res = 0;
|
||
for (int i = 0; i < uN; i++) {
|
||
if (match[i] == -1) {
|
||
while (!que.empty()) { que.pop(); }
|
||
que.push(i); pre[i] = -1;
|
||
bool flag = false;
|
||
while (!que.empty() && !flag) {
|
||
int u = que.front(); que.pop();
|
||
for (int v = 0; v < vN && !flag; v++) {
|
||
if (g[u][v] && check[v] != i) {
|
||
check[v] = i; que.push(match[v]);
|
||
if (~match[v]) { pre[match[v]] = u; }
|
||
else {
|
||
flag = true;
|
||
for (int a = u, b = v; ~a;) {
|
||
int t = match[a]; match[a] = b; match[b] = a; a = pre[a]; b = t;
|
||
}
|
||
}
|
||
}
|
||
}
|
||
}
|
||
if (~match[i]) { res++; }
|
||
}
|
||
}
|
||
return res;
|
||
}
|
||
//Hungary + bfs + 邻接表 O(V*E)
|
||
int head[N], to[M], nxt[M], tot, uN, match[N], check[N], pre[N];
|
||
void init() { tot = 0; memset(head, -1, sizeof(head)); }
|
||
void addedge(int x, int y) { to[tot] = y; nxt[tot] = head[x]; head[x] = tot++; }
|
||
int Hungary() {
|
||
memset(match, -1, sizeof(match));
|
||
memset(check, -1, sizeof(check));
|
||
queue<int> que;
|
||
int res = 0;
|
||
for (int i = 0; i < uN; i++) {
|
||
if (match[i] == -1) {
|
||
while (!que.empty()) { que.pop(); }
|
||
que.push(i); pre[i] = -1;
|
||
bool flag = false;
|
||
while (!que.empty() && !flag) {
|
||
int u = que.front(); que.pop();
|
||
for (int i = head[u]; ~i && !flag; i = nxt[i]) {
|
||
int v = to[i];
|
||
if (check[v] != i) {
|
||
check[v] = i; que.push(match[v]);
|
||
if (~match[v]) { pre[match[v]] = u; }
|
||
else {
|
||
flag = true;
|
||
for (int a = u, b = v; ~a;) {
|
||
int t = match[a]; match[a] = b; match[b] = a; a = pre[a]; b = t;
|
||
}
|
||
}
|
||
}
|
||
}
|
||
}
|
||
if (~match[i]) { res++; }
|
||
}
|
||
}
|
||
return res;
|
||
}
|
||
//Hopcroft-Karp + vector + O(sqrt(V)*E)
|
||
const int INF = 0x3f3f3f3f;
|
||
vector<int> g[N];
|
||
int uN, matchx[N], matchy[N], dx[N], dy[N], dis;
|
||
bool check[N];
|
||
bool SearchP() {
|
||
memset(dx, -1, sizeof(dx));
|
||
memset(dy, -1, sizeof(dy));
|
||
queue<int> que; dis = INF;
|
||
for (int i = 0; i < uN; i++) { if (matchx[i] == -1) { dx[i] = 0; que.push(i); } }
|
||
while (!que.empty()) {
|
||
int u = que.front(); que.pop();
|
||
if (dx[u] > dis) { break; }
|
||
for (size_t i = 0; i < g[u].size(); i++) {
|
||
int v = g[u][i];
|
||
if (dy[v] == -1) {
|
||
dy[v] = dx[u] + 1;
|
||
if (matchy[v] == -1) { dis = dy[v]; }
|
||
else { dx[matchy[v]] = dy[v] + 1; que.push(matchy[v]); }
|
||
}
|
||
}
|
||
}
|
||
return dis != INF;
|
||
}
|
||
bool dfs(int u) {
|
||
for (size_t i = 0; i < g[u].size(); i++) {
|
||
int v = g[u][i];
|
||
if (!check[v] && dy[v] == dx[u] + 1) {
|
||
check[v] = true;
|
||
if (~matchy[v] && dy[v] == dis) { continue; }
|
||
if (matchy[v] == -1 || dfs(matchy[v])) {
|
||
matchy[v] = u; matchx[u] = v; return true;
|
||
}
|
||
}
|
||
}
|
||
return false;
|
||
}
|
||
int HK() {
|
||
memset(matchx, -1, sizeof(matchx));
|
||
memset(matchy, -1, sizeof(matchy));
|
||
int res = 0;
|
||
while (SearchP()) {
|
||
memset(check, 0, sizeof(check));
|
||
for (int i = 0; i < uN; i++) {
|
||
if (matchx[i] == -1 && dfs(i)) { res++; }
|
||
}
|
||
}
|
||
return res;
|
||
}
|
||
//二分图最佳匹配
|
||
//Kuhn-Munkers + 邻接矩阵 O(uN^2*vN)
|
||
//若求最小权匹配,可将权值取相反数,结果取相反数 点的编号从0开始
|
||
const int INF = 0x3f3f3f3f;
|
||
int uN, vN, g[N][N];
|
||
int matchy[N], lx[N], ly[N]; //y中各点匹配状态, 可行顶标
|
||
int slack[N]; //松弛数组
|
||
bool visx[N], visy[N];
|
||
bool dfs(int u) {
|
||
visx[u] = true;
|
||
for (int v = 0; v < vN; v++) {
|
||
if (visy[v]) { continue; }
|
||
int tmp = lx[u] + ly[v] - g[u][v];
|
||
if (tmp == 0) {
|
||
visy[v] = true;
|
||
if (matchy[v] == -1 || dfs(matchy[v])) { matchy[v] = u; return true; }
|
||
} else if (slack[v] > tmp) { slack[v] = tmp; }
|
||
}
|
||
return false;
|
||
}
|
||
int KM() {
|
||
memset(matchy, -1, sizeof(matchy));
|
||
memset(ly, 0, sizeof(ly));
|
||
for (int i = 0; i < uN; i++) { lx[i] = *max_element(g[i], g[i] + vN); }
|
||
for (int u = 0; u < uN; u++) {
|
||
memset(slack, 0x3f, sizeof(slack));
|
||
for (;;) {
|
||
memset(visx, 0, sizeof(visx));
|
||
memset(visy, 0, sizeof(visy));
|
||
if (dfs(u)) { break; }
|
||
int d = INF;
|
||
for (int i = 0; i < vN; i++) { if (!visy[i] && d > slack[i]) { d = slack[i]; } }
|
||
for (int i = 0; i < uN; i++) { if (visx[i]) { lx[i] -= d; } }
|
||
for (int i = 0; i < vN; i++) { if (visy[i]) { ly[i] += d; } else { slack[i] -= d; } }
|
||
}
|
||
}
|
||
int res = 0;
|
||
for (int i = 0; i < vN; i++) { if (matchy[i] != -1) { res += g[matchy[i]][i]; } }
|
||
return res;
|
||
}
|
||
//二分图多重匹配
|
||
int uN, vN, g[N][N], match[N][N]; //match[i][0]为个数
|
||
int num[N]; //右边最大的匹配数
|
||
bool check[N];
|
||
bool dfs(int u) {
|
||
for (int v = 0; v < vN; v++) {
|
||
if (g[u][v] && !check[v]) {
|
||
check[v] = true;
|
||
if (match[v][0] < num[v]) { match[v][++match[v][0]] = u; return true; }
|
||
for (int i = 1; i <= num[0]; i++) {
|
||
if (dfs(match[v][i])) { match[v][i] = u; return true; }
|
||
}
|
||
}
|
||
}
|
||
return false;
|
||
}
|
||
int Hungary() {
|
||
int res = 0;
|
||
for (int i = 0; i < vN; i++) { match[i][0] = 0; }
|
||
for (int u = 0; u < uN; u++) {
|
||
memset(check, false, sizeof(check));
|
||
res += dfs(u);
|
||
}
|
||
return res;
|
||
}
|
||
//一般图最大匹配
|
||
//邻接矩阵 O(V^3)
|
||
int n, match[N];
|
||
bool g[N][N], check[N];
|
||
bool aug(int now) {
|
||
bool ret = false; check[now] = true;
|
||
for (int i = 0; i < n; i++) {
|
||
if (!check[i] && g[now][i]) {
|
||
if (match[i] == -1) { match[now] = i; match[i] = now; ret = true; }
|
||
else {
|
||
check[i] = true;
|
||
if (aug(match[i])) { match[now] = i, match[i] = now, ret = true; }
|
||
check[i] = false;
|
||
}
|
||
if (ret) { break; }
|
||
}
|
||
}
|
||
check[now] = false; return ret;
|
||
}
|
||
//邻接表 O(V*E)
|
||
int head[N], to[M], nxt[M], tot, match[N];
|
||
bool check[N];
|
||
void init() { tot = 0; memset(head, -1, sizeof(head)); }
|
||
void addedge(int x, int y) { to[tot] = y; nxt[tot] = head[x]; head[x] = tot++; }
|
||
bool aug(int now) {
|
||
bool ret = false; check[now] = true;
|
||
for (int i = head[now]; ~i; i = nxt[i]) {
|
||
int v = to[i];
|
||
if (!check[v]) {
|
||
if (match[v] == -1) { match[now] = v; match[v] = now; ret = true; }
|
||
else {
|
||
check[v] = true;
|
||
if (aug(match[v])) { match[now] = v; match[v] = now; ret = true; }
|
||
check[v] = false;
|
||
}
|
||
if (ret) { break; }
|
||
}
|
||
}
|
||
check[now] = false; return ret;
|
||
}
|
||
int graphMatch() {
|
||
memset(match, -1, sizeof(match));
|
||
memset(check, 0, sizeof(check));
|
||
for (int i = 0, j = n; i < n && j >= 2; i++) {
|
||
if (match[i] == -1 && aug(i)) { i = -1, j -= 2; }
|
||
}
|
||
int ret = 0;
|
||
for (int i = 0; i < n; i++) { ret += (match[i] != -1); }
|
||
return ret >> 1;
|
||
}
|
||
//带花树 + vector O(V^3) 点的编号从0开始
|
||
int n, match[N], fa[N], nxt[N], mark[N], vis[N], t;
|
||
vector<int> e[N]; queue<int> que;
|
||
int findfa(int n) { return n == fa[n] ? n : fa[n] = findfa(fa[n]); }
|
||
void unite(int x, int y) { fa[findfa(x)] = findfa(y); }
|
||
int lca(int x, int y) {
|
||
for (t++;; swap(x, y)) {
|
||
if (x == -1) { continue; }
|
||
if (vis[x = findfa(x)] == t) { return x; }
|
||
vis[x] = t; x = match[x] == -1 ? -1 : nxt[match[x]];
|
||
}
|
||
}
|
||
void group(int a, int p) {
|
||
for (int b, c; a != p; unite(a, b), unite(b, c), a = c) {
|
||
b = match[a]; c = nxt[b];
|
||
if (findfa(c) != p) { nxt[c] = b; }
|
||
if (mark[b] == 2) { mark[b] = 1; que.push(b); }
|
||
if (mark[c] == 2) { mark[c] = 1; que.push(c); }
|
||
}
|
||
}
|
||
void aug(int st) {
|
||
for (int i = 0; i < n; i++) { fa[i] = i; }
|
||
memset(nxt, -1, sizeof(nxt)); memset(vis, -1, sizeof(vis)); memset(mark, 0, sizeof(mark));
|
||
while (!que.empty()) { que.pop(); } que.push(st); mark[st] = 1;
|
||
while (match[st] == -1 && !que.empty()) {
|
||
int u = que.front(); que.pop();
|
||
for (int i = 0; i < (int)e[u].size(); i++) {
|
||
int v = e[u][i];
|
||
if (v == match[u] || findfa(u) == findfa(v) || mark[v] == 2) { continue; }
|
||
if (mark[v] == 1) {
|
||
int p = lca(u, v);
|
||
if (findfa(u) != p) { nxt[u] = v; }
|
||
if (findfa(v) != p) { nxt[v] = u; }
|
||
group(u, p); group(v, p);
|
||
} else if (match[v] == -1) {
|
||
nxt[v] = u;
|
||
for (int j = v, k, l; ~j; j = l) { k = nxt[j]; l = match[k]; match[j] = k; match[k] = j; }
|
||
break;
|
||
} else { nxt[v] = u; que.push(match[v]); mark[match[v]] = 1; mark[v] = 2; }
|
||
}
|
||
}
|
||
}
|
||
int solve(int n) {
|
||
memset(match, -1, sizeof(match)); t = 0; int ret = 0;
|
||
for (int i = 0; i < n; i++) { if (match[i] == -1) { aug(i); } }
|
||
for (int i = 0; i < n; i++) { ret += (match[i] > i); }
|
||
return ret;
|
||
}
|
||
//ural 1099
|
||
int main() {
|
||
int u, v;
|
||
scanf("%d", &n);
|
||
while (~scanf("%d%d", &u, &v)) { e[--u].push_back(--v); e[v].push_back(u); }
|
||
int ans = solve(n);
|
||
printf("%d\n", ans * 2);
|
||
for (int u = 0; u < n; u++) {
|
||
if (u < match[u]) { printf("%d %d\n", u + 1, match[u] + 1); }
|
||
}
|
||
}
|
||
|
||
============图论-生成树==========
|
||
//最小生成树
|
||
//Prim + 邻接矩阵 O(V^2)
|
||
typedef int wtype;
|
||
const wtype INF = 0x3f3f3f3f;
|
||
int n; wtype mp[N][N], cost[N];
|
||
bool vis[N];
|
||
wtype Prim(int src) {
|
||
memset(vis, 0, sizeof(vis)); memcpy(cost, mp[src], sizeof(mp[src]));
|
||
wtype ret = 0; vis[src] = true;
|
||
for (int j = 1; j < n; j++) {
|
||
int u = -1; wtype mn = INF;
|
||
for (int i = 0; i < n; i++) {
|
||
if (!vis[i] && mn > cost[i]) { mn = cost[i]; u = i; }
|
||
}
|
||
if (mn == INF) { return -1; } //原图不连通
|
||
vis[u] = true; ret += mn;
|
||
for (int v = 0; v < n; v++) {
|
||
if (!vis[v] && cost[v] > mp[u][v]) { cost[v] = mp[u][v]; }
|
||
}
|
||
}
|
||
return ret;
|
||
}
|
||
//Prim + priority_queue + 邻接表 O(ElogV)
|
||
typedef int wtype;
|
||
int head[N], to[M], nxt[M], tot; wtype len[M], cost[N];
|
||
bool vis[N];
|
||
void init() { tot = 0; memset(head, -1, sizeof(head)); }
|
||
void addedge(int x, int y, wtype z) { to[tot] = y; len[tot] = z; nxt[tot] = head[x]; head[x] = tot++; }
|
||
struct Node {
|
||
int v; wtype w;
|
||
bool operator<(const Node &r)const { return w > r.w; }
|
||
};
|
||
wtype Prim(int src) {
|
||
memset(cost, 0x3f, sizeof(cost)); memset(vis, 0, sizeof(vis));
|
||
wtype ret = 0;
|
||
priority_queue<Node> que; que.push((Node) {src, cost[src] = 0});
|
||
while (!que.empty()) {
|
||
int u = que.top().v; que.pop();
|
||
if (vis[u]) { continue; }
|
||
vis[u] = true; ret += cost[u];
|
||
for (int i = head[u]; ~i; i = nxt[i]) {
|
||
int v = to[i];
|
||
if (!vis[v] && cost[v] > len[i]) { cost[v] = len[i]; que.push((Node) {v, cost[v]}); }
|
||
}
|
||
}
|
||
return ret;
|
||
}
|
||
//Kruskal + 邻接表 O(ElogE)
|
||
typedef int wtype;
|
||
struct Edge {
|
||
int u, v; wtype w;
|
||
bool operator<(const Edge &r)const { return w < r.w; }
|
||
} edge[M];
|
||
int n, fa[N], tot; //加边前赋值为0
|
||
void addedge(int u, int v, wtype w) { edge[tot].u = u; edge[tot].v = v; edge[tot++].w = w; }
|
||
int findfa(int x) { return fa[x] == -1 ? x : fa[x] = findfa(fa[x]); }
|
||
wtype Kruskal() {
|
||
memset(fa, -1, sizeof(fa));
|
||
sort(edge, edge + tot);
|
||
int cnt = 0; wtype ret = 0;
|
||
for (int i = 0; i < tot; i++) {
|
||
int u = edge[i].u, v = edge[i].v, t1 = findfa(u), t2 = findfa(v); wtype w = edge[i].w;
|
||
if (t1 != t2) { ret += w; fa[t1] = t2; cnt++; }
|
||
if (cnt == n - 1) { break; }
|
||
}
|
||
if (cnt < n - 1) { return -1; } //不连通
|
||
return ret;
|
||
}
|
||
//次小生成树
|
||
//Prim + 邻接矩阵 O(V^2 + E)
|
||
//求最小生成树时, 用数组Mx[i][j]来表示MST中i到j最大边权
|
||
//求完后, 直接枚举所有不在MST中的边, 替换掉最大边权的边, 更新答案
|
||
typedef int wtype;
|
||
const wtype INF = 0x3f3f3f3f;
|
||
int n, pre[N];
|
||
wtype mp[N][N], cost[N], Mx[N][N]; //Mx[i][j]表示在最小生成树中从i到j的路径中的最大边权
|
||
bool vis[N], used[N][N];
|
||
wtype Prim() {
|
||
memset(Mx, 0, sizeof(Mx));
|
||
memset(pre, 0, sizeof(pre));
|
||
memset(vis, 0, sizeof(vis));
|
||
memset(used, 0, sizeof(used));
|
||
wtype ret = 0; vis[0] = true; pre[0] = -1;
|
||
for (int i = 1; i < n; i++) { cost[i] = mp[0][i]; }
|
||
for (int j = 1; j < n; j++) {
|
||
int u = -1; wtype mn = INF;
|
||
for (int j = 0; j < n; j++)
|
||
if (!vis[j] && mn > cost[j]) { mn = cost[j]; u = j; }
|
||
if (mn == INF) { return -1; } //原图不连通
|
||
vis[u] = true; ret += mn;
|
||
used[u][pre[u]] = used[pre[u]][u] = true;
|
||
for (int v = 0; v < n; v++) {
|
||
if (vis[v]) { Mx[v][u] = Mx[u][v] = max(Mx[v][pre[u]], cost[u]); }
|
||
if (!vis[v] && cost[v] > mp[u][v]) { cost[v] = mp[u][v]; pre[v] = u; }
|
||
}
|
||
}
|
||
return ret;
|
||
}
|
||
//Kruskal + 邻接表 O(VElogE)
|
||
typedef int wtype;
|
||
const wtype INF = 0x3f3f3f3f;
|
||
struct Edge {
|
||
int u, v; wtype w;
|
||
bool operator<(const Edge &r)const { return w < r.w; }
|
||
} edge[M];
|
||
int n, fa[N], path[N], tot; //加边前赋值为0
|
||
void addedge(int u, int v, wtype w) { edge[tot].u = u; edge[tot].v = v; edge[tot++].w = w; }
|
||
int findfa(int x) { return fa[x] == -1 ? x : fa[x] = findfa(fa[x]); }
|
||
wtype Kruskal() {
|
||
memset(fa, -1, sizeof(fa));
|
||
sort(edge, edge + tot);
|
||
int cnt = 0; wtype ret = 0;
|
||
for (int i = 0; i < tot; i++) {
|
||
int u = edge[i].u, v = edge[i].v, t1 = findfa(u), t2 = findfa(v); wtype w = edge[i].w;
|
||
if (t1 != t2) { ret += w; fa[t1] = t2; path[cnt++] = i; }
|
||
if (cnt == n - 1) { break; }
|
||
}
|
||
if (cnt < n - 1) { return -1; } //不连通
|
||
return ret;
|
||
}
|
||
wtype KruskalSec() {
|
||
wtype ret = INF;
|
||
for (int x = 0; x < n - 1; x++) {
|
||
memset(fa, -1, sizeof(fa));
|
||
int cnt = 0; wtype tmp = 0;
|
||
for (int i = 0; i < tot; i++) {
|
||
if (i != path[x]) {
|
||
int u = edge[i].u, v = edge[i].v, t1 = findfa(u), t2 = findfa(v); wtype w = edge[i].w;
|
||
if (t1 != t2) { tmp += w; fa[t1] = t2; cnt++; }
|
||
if (cnt == n - 1) { if (tmp < ret) { ret = tmp; } break; }
|
||
}
|
||
}
|
||
}
|
||
if (ret == INF) { return -1; } //不存在
|
||
return ret;
|
||
}
|
||
//最小树形图
|
||
//朱刘算法 O(VE)
|
||
typedef int wtype;
|
||
const int N = 1005;
|
||
const int M = 40005;
|
||
const wtype INF = 0x3f3f3f3f;
|
||
struct Edge { int u, v; wtype w; } edge[M];
|
||
int n, m, pre[N], id[N], vis[N];
|
||
wtype g[N][N], in[N];
|
||
wtype Zhuliu(int root) {
|
||
wtype res = 0; int u, v;
|
||
for (;;) {
|
||
memset(in, 0x3f, sizeof(in));
|
||
memset(id, -1, sizeof(id));
|
||
memset(vis, -1, sizeof(vis));
|
||
for (int i = 0; i < m; i++) {
|
||
if (edge[i].u != edge[i].v && edge[i].w < in[edge[i].v]) {
|
||
pre[edge[i].v] = edge[i].u; in[edge[i].v] = edge[i].w;
|
||
}
|
||
}
|
||
for (int i = 0; i < n; i++) {
|
||
if (i != root && in[i] == INF) { return -1; } //不存在最小树形图
|
||
}
|
||
int tn = 0; in[root] = 0;
|
||
for (int i = 0; i < n; i++) {
|
||
res += in[i]; v = i;
|
||
while (vis[v] != i && id[v] == -1 && v != root) {
|
||
vis[v] = i; v = pre[v];
|
||
}
|
||
if (v != root && id[v] == -1) {
|
||
for (int u = pre[v]; u != v ; u = pre[u]) { id[u] = tn; }
|
||
id[v] = tn++;
|
||
}
|
||
}
|
||
if (tn == 0) { break; } //没有有向环
|
||
for (int i = 0; i < n; i++) {
|
||
if (id[i] == -1) { id[i] = tn++; }
|
||
}
|
||
for (int i = 0; i < m;) {
|
||
v = edge[i].v; edge[i].u = id[edge[i].u]; edge[i].v = id[edge[i].v];
|
||
if (edge[i].u != edge[i].v) { edge[i++].w -= in[v]; }
|
||
else { swap(edge[i], edge[--m]); }
|
||
}
|
||
n = tn; root = id[root];
|
||
}
|
||
return res;
|
||
}
|
||
//POJ 3164
|
||
int main() {
|
||
int C = 0, T, u, v, w;
|
||
scanf("%d", &T);
|
||
while (++C <= T) {
|
||
memset(g, 0x3f, sizeof(g));
|
||
scanf("%d%d", &n, &m);
|
||
while (m--) {
|
||
scanf("%d%d%d", &u, &v, &w);
|
||
if (u == v) { continue; }
|
||
g[u][v] = min(g[u][v], w);
|
||
}
|
||
for (int i = 0; i < n; i++) {
|
||
for (int j = 0; j < n; j++) {
|
||
if (g[i][j] < INF) { edge[m].u = i; edge[m].v = j; edge[m++].w = g[i][j]; }
|
||
}
|
||
}
|
||
wtype ans = Zhuliu(0);
|
||
printf("Case #%d: ", C);
|
||
if (ans == -1) { puts("Possums!"); }
|
||
else { printf("%d\n", ans); }
|
||
}
|
||
}
|
||
//曼哈顿距离最小生成树
|
||
//Kruskal O(VlogV)
|
||
const int INF = 0x3f3f3f3f;
|
||
struct Point {
|
||
int x, y, id;
|
||
bool operator<(const Point &r)const { return x < r.x || (x == r.x && y < r.y); }
|
||
} p[N];
|
||
struct Edge { //有效边
|
||
int u, v, w;
|
||
bool operator<(const Edge &r)const { return w < r.w; }
|
||
} edge[N << 2];
|
||
struct BIT { //树状数组, 找y-x大于当前的, 但是y+x最小的
|
||
int min_val, pos;
|
||
void init() { min_val = INF; pos = -1; }
|
||
} bit[N];
|
||
int n, tot, fa[N];
|
||
int a[N], b[N];
|
||
void addedge(int u, int v, int w) { edge[tot].u = u; edge[tot].v = v; edge[tot++].w = w; }
|
||
int findfa(int x) { return fa[x] == -1 ? x : fa[x] = findfa(fa[x]); }
|
||
inline int cost(Point a, Point b) { return abs(a.x - b.x) + abs(a.y - b.y); }
|
||
inline int lowbit(int x) { return x & (-x); }
|
||
void update(int i, int val, int pos) {
|
||
for (; i > 0; i -= lowbit(i)) {
|
||
if (val < bit[i].min_val) { bit[i].min_val = val; bit[i].pos = pos; }
|
||
}
|
||
}
|
||
int query(int i, int m) { //查询[i, m]的最小值位置
|
||
int min_val = INF, pos = -1;
|
||
for (; i <= m; i += lowbit(i)) {
|
||
if (bit[i].min_val < min_val) { min_val = bit[i].min_val; pos = bit[i].pos; }
|
||
}
|
||
return pos;
|
||
}
|
||
void MMST() {
|
||
tot = 0;
|
||
for (int d = 0; d < 4; d++) { //4种坐标变换
|
||
if (d == 1 || d == 3) { for (int i = 0; i < n; i++) { swap(p[i].x, p[i].y); } }
|
||
else if (d == 2) { for (int i = 0; i < n; i++) { p[i].x = -p[i].x; } }
|
||
sort(p, p + n);
|
||
for (int i = 0; i < n; i++) { a[i] = b[i] = p[i].y - p[i].x; }
|
||
sort(b, b + n);
|
||
int m = unique(b, b + n) - b;
|
||
for (int i = 1; i <= m; i++) { bit[i].init(); }
|
||
for (int i = n - 1 ; i >= 0; i--) {
|
||
int pos = lower_bound(b, b + m, a[i]) - b + 1, ans = query(pos, m);
|
||
if (ans != -1) { addedge(p[i].id, p[ans].id, cost(p[i], p[ans])); }
|
||
update(pos, p[i].x + p[i].y, i);
|
||
}
|
||
}
|
||
}
|
||
int Kruskal() {
|
||
MMST();
|
||
memset(fa, -1, sizeof(fa));
|
||
sort(edge, edge + tot);
|
||
int ret = 0;
|
||
for (int i = 0, k = 0; i < tot; i++) {
|
||
int u = edge[i].u, v = edge[i].v, t1 = findfa(u), t2 = findfa(v);
|
||
if (t1 != t2) {
|
||
fa[t1] = t2; ret += edge[i].w;
|
||
if (++k == n - 1) { return ret; }
|
||
}
|
||
}
|
||
}
|
||
//POJ3241 求曼哈顿最小生成树上第k大的边
|
||
int Kruskal(int k) {
|
||
MMST(n, p);
|
||
memset(fa, -1, sizeof(fa));
|
||
sort(edge, edge + tot);
|
||
for (int i = 0; i < tot; i++) {
|
||
int u = edge[i].u, v = edge[i].v, t1 = findfa(u), t2 = findfa(v);
|
||
if (t1 != t2) {
|
||
fa[t1] = t2;
|
||
if (--k == 0) { return edge[i].w; }
|
||
}
|
||
}
|
||
}
|
||
//生成树计数
|
||
//Matrix-Tree定理(Kirchhoff 矩阵-树定理)
|
||
//1、G的度数矩阵D[G]是一个n*n的矩阵, 并且满足: 当i ≠ j时,dij = 0; 当i = j时, dij等于vi的度数
|
||
//2、G的邻接矩阵A[G]也是一个n*n的矩阵, 并且满足: 如果vi vj之间有边直接相连, 则aij = 1, 否则为0
|
||
//我们定义G的Kirchhoff矩阵(也称为拉普拉斯算子)C[G]为C[G] = D[G] - A[G], 则Matrix-Tree定理可以
|
||
//描述为: G的所有不同的生成树的个数等于其Kirchhoff矩阵C[G]任何一个n - 1阶主子式的行列式的绝对值
|
||
//所谓n - 1阶主子式, 就是对于r(1 ≤ r ≤ n), 将C[G]的第r行、第r列同时去掉后得到的新矩阵, 用Cr[G]表示。
|
||
|
||
|
||
============图论-网络流=========
|
||
//最大流
|
||
//SAP + 邻接矩阵 O(V^2*E) 点的编号默认从0开始
|
||
typedef long long ftype;
|
||
int n, dis[N], pre[N], cur[N], gap[N];
|
||
ftype cap[N][N];
|
||
ftype SAP(int src, int sink, int n) {
|
||
memset(dis, 0, sizeof(dis));
|
||
memset(cur, 0, sizeof(cur));
|
||
memset(gap, 0, sizeof(gap));
|
||
int u = src; ftype mxflow = 0, aug = -1; pre[src] = src; gap[0] = n;
|
||
while (dis[src] < n) {
|
||
loop:
|
||
for (int v = cur[u]; v < n; ++v) {
|
||
if (cap[u][v] > 0 && dis[u] == dis[v] + 1) {
|
||
if (aug == -1 || aug > cap[u][v]) { aug = cap[u][v]; }
|
||
pre[v] = u; u = cur[u] = v;
|
||
if (v == sink) {
|
||
for (u = pre[u]; v != src; v = u, u = pre[u]) {
|
||
cap[u][v] -= aug; cap[v][u] += aug;
|
||
}
|
||
mxflow += aug; aug = -1;
|
||
}
|
||
goto loop;
|
||
}
|
||
}
|
||
int mndis = n - 1;
|
||
for (int v = 0; v < n; v++) {
|
||
if (cap[u][v] > 0 && mndis > dis[v]) { cur[u] = v; mndis = dis[v]; }
|
||
}
|
||
if (--gap[dis[u]] == 0) { break; }
|
||
dis[u] = mndis + 1; gap[dis[u]]++; u = pre[u];
|
||
}
|
||
return mxflow;
|
||
}
|
||
//ISAP + 邻接表 O(V^2*E)
|
||
typedef long long ftype;
|
||
const ftype INF = 0x3f3f3f3f;
|
||
int head[N], to[M], nxt[M], tot, dis[N], pre[N], cur[N], gap[N];
|
||
ftype cap[M];
|
||
inline void init() { tot = 0; memset(head, -1, sizeof(head)); }
|
||
inline void addedge(int x, int y, ftype w, ftype rw = 0) {
|
||
to[tot] = y; cap[tot] = w; nxt[tot] = head[x]; head[x] = tot++;
|
||
to[tot] = x; cap[tot] = rw; nxt[tot] = head[y]; head[y] = tot++;
|
||
}
|
||
ftype ISAP(int src, int sink, int n) {
|
||
memset(dis, 0, sizeof(dis));
|
||
memset(gap, 0, sizeof(gap));
|
||
memcpy(cur, head, sizeof(head));
|
||
int u = src, v; ftype mxflow = 0; pre[u] = -1; gap[0] = n;
|
||
while (dis[src] < n) {
|
||
if (u == sink) {
|
||
ftype mndis = INF;
|
||
for (int i = pre[u]; ~i; i = pre[to[i ^ 1]]) {
|
||
if (mndis > cap[i]) { mndis = cap[i]; }
|
||
}
|
||
for (int i = pre[u]; ~i; i = pre[to[i ^ 1]]) {
|
||
cap[i] -= mndis; cap[i ^ 1] += mndis;
|
||
}
|
||
u = src; mxflow += mndis;
|
||
continue;
|
||
}
|
||
bool flag = false;
|
||
for (int i = cur[u]; ~i; i = nxt[i]) {
|
||
v = to[i];
|
||
if (cap[i] > 0 && dis[v] + 1 == dis[u]) {
|
||
flag = true; cur[u] = pre[v] = i; break;
|
||
}
|
||
}
|
||
if (flag) { u = v; continue; }
|
||
int mndis = n;
|
||
for (int i = head[u]; ~i; i = nxt[i]) {
|
||
if (cap[i] > 0 && dis[to[i]] < mndis) { mndis = dis[to[i]]; cur[u] = i; }
|
||
}
|
||
if (--gap[dis[u]] == 0) { return mxflow; }
|
||
dis[u] = mndis + 1; gap[dis[u]]++;
|
||
if (u != src) { u = to[pre[u] ^ 1]; }
|
||
}
|
||
return mxflow;
|
||
}
|
||
//ISAP + bfs标号 + queue + 邻接表 O(V^2*E)
|
||
typedef long long ftype;
|
||
const ftype INF = 0x3f3f3f3f;
|
||
int n, head[N], to[M], nxt[M], tot, dis[N], pre[N], cur[N], gap[N];
|
||
ftype cap[M];
|
||
inline void init() { tot = 0; memset(head, -1, sizeof(head)); }
|
||
inline void addedge(int x, int y, ftype w, ftype rw = 0) {
|
||
to[tot] = y; cap[tot] = w; nxt[tot] = head[x]; head[x] = tot++;
|
||
to[tot] = x; cap[tot] = rw; nxt[tot] = head[y]; head[y] = tot++;
|
||
}
|
||
void bfs(int sink) {
|
||
memset(dis, -1, sizeof(dis));
|
||
memset(gap, 0, sizeof(gap));
|
||
dis[sink] = 0; gap[0] = 1;
|
||
queue<int> que; que.push(sink);
|
||
while (!que.empty()) {
|
||
int u = que.front(); que.pop();
|
||
for (int i = head[u], v; ~i; i = nxt[i]) {
|
||
v = to[i];
|
||
if (~dis[v]) { continue; }
|
||
dis[v] = dis[u] + 1; gap[dis[v]]++; que.push(v);
|
||
}
|
||
}
|
||
}
|
||
ftype ISAP(int src, int sink, int n) {
|
||
bfs(sink);
|
||
memcpy(cur, head, sizeof(head));
|
||
int u = pre[src] = src, v, i; ftype mxflow = 0;
|
||
while (dis[sink] < n) {
|
||
if (u == sink) {
|
||
ftype mndis = INF; int inser;
|
||
for (i = src; i != sink; i = to[cur[i]]) {
|
||
if (mndis > cap[cur[i]]) { mndis = cap[cur[i]]; inser = i; }
|
||
}
|
||
for (i = src; i != sink; i = to[cur[i]]) {
|
||
cap[cur[i]] -= mndis; cap[cur[i] ^ 1] += mndis;
|
||
}
|
||
mxflow += mndis; u = inser;
|
||
}
|
||
for (i = cur[u]; ~i; i = nxt[i]) {
|
||
v = to[i];
|
||
if (dis[v] + 1 == dis[u] && cap[i] > 0) { break; }
|
||
}
|
||
if (~i) { cur[u] = i; pre[to[i]] = u; u = to[i]; }
|
||
else {
|
||
if (--gap[dis[u]] == 0) { break; }
|
||
int mndis = n;
|
||
for (i = head[u]; ~i; i = nxt[i]) {
|
||
if (cap[i] > 0 && mndis > dis[to[i]]) { cur[u] = i; mndis = dis[to[i]]; }
|
||
}
|
||
dis[u] = mndis + 1; gap[dis[u]]++; u = pre[u];
|
||
}
|
||
}
|
||
return mxflow;
|
||
}
|
||
//Dinic O(V^2*E)
|
||
typedef long long ftype;
|
||
const ftype INF = 0x3f3f3f3f;
|
||
int head[N], to[M], nxt[M], tot, dis[N], cur[N], src, sink;
|
||
ftype cap[M];
|
||
inline void init() { tot = 0; memset(head, -1, sizeof(head)); }
|
||
inline void addedge(int x, int y, ftype w, ftype rw = 0) {
|
||
to[tot] = y; cap[tot] = w; nxt[tot] = head[x]; head[x] = tot++;
|
||
to[tot] = x; cap[tot] = rw; nxt[tot] = head[y]; head[y] = tot++;
|
||
}
|
||
bool bfs() {
|
||
memset(dis, 0, sizeof(dis)); dis[src] = 1;
|
||
queue<int> que; que.push(src);
|
||
while (!que.empty()) {
|
||
int u = que.front(); que.pop();
|
||
for (int i = head[u], v; ~i; i = nxt[i]) {
|
||
v = to[i];
|
||
if (cap[i] > 0 && !dis[v]) { dis[v] = dis[u] + 1; que.push(v); }
|
||
}
|
||
}
|
||
return dis[sink];
|
||
}
|
||
ftype dfs(int u, ftype delta) {
|
||
if (u == sink || delta == 0) { return delta; }
|
||
ftype ret = 0;
|
||
for (int &i = cur[u], v; delta && ~i; i = nxt[i]) {
|
||
v = to[i];
|
||
if (cap[i] > 0 && dis[v] == dis[u] + 1) {
|
||
ftype aug = dfs(v, min(cap[i], delta));
|
||
if (!aug) { continue; }
|
||
cap[i] -= aug; cap[i ^ 1] += aug; delta -= aug; ret += aug;
|
||
if (!delta) { break; }
|
||
}
|
||
}
|
||
return ret;
|
||
}
|
||
ftype Dinic() {
|
||
ftype ret = 0;
|
||
while (bfs()) {
|
||
memcpy(cur, head, sizeof(head));
|
||
ret += dfs(src, INF);
|
||
}
|
||
return ret;
|
||
}
|
||
//HLPP Highest Label Preflow Push O(V^3)
|
||
typedef long long ftype;
|
||
struct Edge { int v; ftype c; } edge[M];
|
||
vector<int> e[N];
|
||
int n, m, src, sink, c[N << 1], d[N], done[N];
|
||
ftype w[N];
|
||
bool vis[N];
|
||
void init(int _n, int _src, int _sink) {
|
||
n = _n; m = 0; src = _src; sink = _sink;
|
||
for (int i = 0; i <= n; ++i) { e[i].clear(); }
|
||
}
|
||
inline void addEdge(int x, int y, ftype w, ftype rw = 0) {
|
||
edge[m].v = y; edge[m].c = w; e[x].push_back(m++);
|
||
edge[m].v = x; edge[m].c = rw; e[y].push_back(m++);
|
||
}
|
||
void bfs() {
|
||
memset(c, 0, sizeof(c));
|
||
fill(d, d + n, n + 1);
|
||
queue<int> que; que.push(sink);
|
||
c[n + 1] = n - 1; d[src] = n; d[sink] = 0;
|
||
while (!que.empty()) {
|
||
int u = que.front(); que.pop();
|
||
c[n + 1]++; c[d[u]]++;
|
||
for (int i = 0, v; i < (int)e[u].size(); i++) {
|
||
v = edge[e[u][i]].v; ftype c = edge[e[u][i] ^ 1].c;
|
||
if (d[v] == n + 1 && c > 0) { d[v] = d[u] + 1; que.push(v); }
|
||
}
|
||
}
|
||
}
|
||
int HLPP(int n) {
|
||
memset(w, 0, sizeof(w));
|
||
memset(done, 0, sizeof(done));
|
||
memset(vis, 0, sizeof(vis));
|
||
bfs();
|
||
int todo = -1;
|
||
vector<queue<int>> que(n << 1);
|
||
vis[src] = vis[sink] = true;
|
||
for (int i = 0, v; i < (int)e[src].size(); i++) {
|
||
Edge &arc = edge[e[src][i]], &cra = edge[e[src][i] ^ 1]; v = arc.v;
|
||
w[v] += arc.c; cra.c += arc.c; arc.c = 0;
|
||
if (!vis[v]) { vis[v] = true; que[d[v]].push(v); todo = max(todo, d[v]); }
|
||
}
|
||
while (todo >= 0) {
|
||
if (que[todo].empty()) { todo--; continue; }
|
||
int u = que[todo].front(); que[todo].pop();
|
||
vis[u] = false;
|
||
while (done[u] < (int)e[u].size()) {
|
||
Edge &arc = edge[e[u][done[u]]]; int v = arc.v;
|
||
if (d[u] == d[v] + 1 && arc.c > 0) {
|
||
Edge &cra = edge[e[u][done[u]] ^ 1]; ftype f = min(w[u], arc.c);
|
||
w[u] -= f; w[v] += f; arc.c -= f; cra.c += f;
|
||
if (!vis[v]) { vis[v] = true; que[d[v]].push(v); }
|
||
if (w[u] == 0) { break; }
|
||
}
|
||
done[u]++;
|
||
}
|
||
if (w[u] > 0) {
|
||
int du = d[u];
|
||
c[d[u]]--; d[u] = n << 1;
|
||
for (int i = 0, v; i < (int)e[u].size(); i++) {
|
||
Edge &arc = edge[e[u][i]]; v = arc.v;
|
||
if (d[u] > d[v] + 1 && arc.c > 0) { d[u] = d[v] + 1; done[u] = i; }
|
||
}
|
||
c[d[u]]++;
|
||
if (c[du] == 0) {
|
||
for (int i = 0; i < n; ++i) {
|
||
if (d[i] > du && d[i] < n + 1) { c[d[i]]--; c[n + 1]++; d[i] = n + 1; }
|
||
}
|
||
}
|
||
vis[u] = true; que[d[u]].push(u); todo = d[u];
|
||
}
|
||
}
|
||
return w[sink];
|
||
}
|
||
//有上下界的网络流
|
||
//无源汇上下界最大流
|
||
//增加超级源汇点, 边的容量变为原弧的上界减去下界的差, 记录每个点的入流下界和-出流下界和
|
||
//当下界和大于0时连一条超级源点到这个点的边, 容量为这个下界和
|
||
//当下界和小于0时连一条这个点到超级汇点的边, 容量为下界和的绝对值
|
||
//求一遍最大流, 如果源的出度满流的话, 即存在最大流, 否则则不存在最大流
|
||
//有源汇上下界最大流
|
||
//从汇连一条容量INF的边到源, 将有源汇转化为无源汇
|
||
//判断是否满流后, 去掉超级源汇及其边(head[]删除即可, 调用最大流算法时, 点的数量要包括超级源汇)
|
||
//再跑一遍原源到原汇的最大流, 输出即可
|
||
//有源汇上下界最小流
|
||
//增加超级源汇点, 求一遍超级源到超级汇的最大流
|
||
//从原汇点连一条容量INF的边到原源点, 再求一遍超级源到超级汇的最大流
|
||
//当且仅当超级源的出度满流时有可行解, 解为原汇点到原源点的反向弧
|
||
////最小费用最大流 MCMF O(V*E*f)
|
||
//最小费用流: 若dis[sink]为负则继续增广
|
||
//最小费用最大流: 若dis[sink]不为INF则继续增广
|
||
//求最大费用只需取相反数, 结果取相反数即可
|
||
typedef long long ftype;
|
||
const ftype INF = 0x3f3f3f3f;
|
||
int head[N], to[M], nxt[M], tot, cur[N];
|
||
ftype cap[M], cost[M], flow[N], dis[N], mncost, mxflow;
|
||
bool vis[N];
|
||
inline void init() { tot = 0; memset(head, -1, sizeof(head)); }
|
||
inline void addedge(int x, int y, ftype w, ftype c) {
|
||
to[tot] = y; cap[tot] = w; cost[tot] = c; nxt[tot] = head[x]; head[x] = tot++;
|
||
to[tot] = x; cap[tot] = 0; cost[tot] = -c; nxt[tot] = head[y]; head[y] = tot++;
|
||
}
|
||
bool SPFA(int src, int sink) {
|
||
memset(dis, 0x3f, sizeof(dis));
|
||
memset(vis, 0, sizeof(vis));
|
||
dis[src] = 0; cur[src] = -1; flow[src] = INF;
|
||
queue<int> que; que.push(src);
|
||
while (!que.empty()) {
|
||
int u = que.front(); que.pop();
|
||
vis[u] = false;
|
||
for (int i = head[u], v; ~i; i = nxt[i]) {
|
||
v = to[i];
|
||
if (cap[i] > 0 && dis[v] > dis[u] + cost[i]) {
|
||
dis[v] = dis[u] + cost[i]; flow[v] = min(flow[u], cap[i]); cur[v] = i;
|
||
if (!vis[v]) { vis[v] = true; que.push(v); }
|
||
}
|
||
}
|
||
}
|
||
if (dis[sink] == INF) { return false; }
|
||
mxflow += flow[sink]; mncost += flow[sink] * dis[sink];
|
||
for (int i = cur[sink]; ~i; i = cur[to[i ^ 1]]) {
|
||
cap[i] -= flow[sink]; cap[i ^ 1] += flow[sink];
|
||
}
|
||
return true;
|
||
}
|
||
ftype MCMF(int src, int sink) {
|
||
mxflow = mncost = 0;
|
||
while (SPFA(src, sink));
|
||
return mncost;
|
||
}
|
||
|
||
===========图论-应用==========
|
||
//拓扑排序 O(V^2)
|
||
//queue->普通判断用, priority_queue->字典序
|
||
//邻接矩阵
|
||
int n, mp[N][N], in[N], ret[N];
|
||
int topoSort() {
|
||
queue<int> que; int k = 0; bool flag = false;
|
||
for (int i = 0; i < n; i++) { if (in[i] == 0) { que.push(i); } }
|
||
while (!que.empty()) {
|
||
if (que.size() > 1) { flag = true; }
|
||
int u = que.front(); que.pop(); ret[k++] = u;
|
||
for (int v = 0; v < n; v++) {
|
||
if (mp[u][v] && --in[v] == 0) { que.push(v); }
|
||
}
|
||
}
|
||
return k < n ? -1 : flag ? 0 : 1; //有环, 不唯一, 唯一
|
||
}
|
||
//邻接表
|
||
int head[N], to[M], nxt[M], tot, n, in[N], ret[N];
|
||
void init() { tot = 0; memset(head, -1, sizeof(head)); }
|
||
void addedge(int x, int y) { to[tot] = y; nxt[tot] = head[x]; head[x] = tot++; }
|
||
bool topoSort() {
|
||
priority_queue<int> que; int k = 0;
|
||
for (int i = 0; i < n; i++) { if (in[i] == 0) { que.push(i); } }
|
||
while (!que.empty()) {
|
||
int u = que.top(); que.pop(); ret[k++] = u;
|
||
for (int i = head[u]; ~i; i = nxt[i]) {
|
||
if (--in[to[i]] == 0) { que.push(to[i]); }
|
||
}
|
||
}
|
||
return k == n;
|
||
}
|
||
//染色判断二分图
|
||
bool col[N];
|
||
bool Color(int u) {
|
||
for (int i = head[u]; ~i; i = nxt[i]) {
|
||
int v = to[i];
|
||
if (!col[v]) {
|
||
col[v] = !col[u];
|
||
if (!Color(v)) { return false; }
|
||
} else if (col[v] == col[u]) { return false; }
|
||
}
|
||
return true;
|
||
}
|
||
//欧拉回路: 每条边只经过一次, 要求回到起点
|
||
//欧拉路径: 每条边只经过一次, 不要求回到起点
|
||
//欧拉路径的判断:
|
||
//无向图: 连通(不考虑度为0的点), 每个顶点度数都为偶数或者有且仅有两个点的度数不为偶数
|
||
//有向图: 基图连通(把边当成无向边, 同样不考虑度为0的点), 每个顶点出度等于入度
|
||
//或者有且仅有一个点的出度比入度多1, 有且仅有一个点的出度比入度少1, 其余出度等于入度
|
||
//混合图: 如果存在欧拉回路, 一点存在欧拉路径了。否则如果有且仅有两个点的(出度 - 入度)是奇数,
|
||
//那么给这个两个点加边, 判断是否存在欧拉回路
|
||
//欧拉回路判断:
|
||
//无向图: 连通(不考虑度为0的点), 每个顶点度数都为偶数
|
||
//有向图: 基图连通(同样不考虑度为0的点), 每个顶点出度等于入度
|
||
//混合图: 基图连通(不考虑度为0的点), 然后借助网络流判定
|
||
//首先给原图中的每条无向边随便指定一个方向, 将原图改为有向图G'
|
||
//设D[i]为G'中(点i的出度 - 点i的入度), 若存在D[i]为奇数, 或者图不连通, 则无解
|
||
//若初始D值都是偶数, 则将G'改装成网络: 设立源点S和汇点T
|
||
//对于每个D[i] > 0的点i, 连边<S, i>, 容量为D[i] / 2; 对于每个D[j] < 0的点j, 连边<j, T>, 容量为-D[j] / 2
|
||
//G'中的每条边在网络中仍保留, 容量为1, 求这个网络的最大流, 若S引出的所有边均满流, 则原混合图是欧拉图
|
||
//将网络中所有流量为1的不与S或T关联的边在G'中改变方向,形成的新图G''一定是有向欧拉图
|
||
//欧拉回路 + 邻接矩阵 O(N^2)
|
||
//求欧拉路径/回路经过的点 支持自环和重边
|
||
int n, mp[N][N], path[N], cnt;
|
||
void dfsu(int u) {
|
||
for (int v = n - 1; v >= 0; v--) {
|
||
while (mp[u][v]) { mp[u][v]--; mp[v][u]--; dfsu(v); }
|
||
}
|
||
path[cnt++] = u;
|
||
}
|
||
void dfsd(int u) {
|
||
for (int v = n - 1; v >= 0; v--) {
|
||
while (mp[u][v]) { mp[u][v]--; dfsd(v); }
|
||
}
|
||
path[cnt++] = u;
|
||
}
|
||
//无向图 SGU101
|
||
int head[N], to[M], nxt[M], tot, deg[N], path[M], cnt;
|
||
bool vis[M];
|
||
void init() {
|
||
tot = 0; cnt = 0;
|
||
memset(head, -1, sizeof(head)); memset(vis, 0, sizeof(vis)); memset(deg, 0, sizeof(deg));
|
||
}
|
||
void addedge(int x, int y) {
|
||
to[tot] = y; nxt[tot] = head[x]; head[x] = tot++;
|
||
to[tot] = x; nxt[tot] = head[y]; head[y] = tot++;
|
||
}
|
||
void dfs(int u) {
|
||
for (int &i = head[u]; ~i;) {
|
||
if (!vis[i]) { vis[i] = vis[i ^ 1] = true; int t = i; dfs(to[i]); path[cnt++] = t; }
|
||
else { i = nxt[i]; }
|
||
}
|
||
}
|
||
int main() {
|
||
int n, u, v;
|
||
while (~scanf("%d", &n)) {
|
||
init();
|
||
for (int i = 0; i < n; i++) {
|
||
scanf("%d%d", &u, &v); addedge(u, v); deg[u]++; deg[v]++;
|
||
}
|
||
int s = -1, cnto = 0;
|
||
for (int i = 0; i <= 6; i++) {
|
||
if (s == -1 && deg[i] > 0) { s = i; }
|
||
if (deg[i] & 1) { cnto++; s = i; }
|
||
}
|
||
if (cnto != 0 && cnto != 2) { puts("No solution"); continue; }
|
||
dfs(s);
|
||
if (cnt != n) { puts("No solution"); continue; }
|
||
for (int i = 0; i < cnt; i++) {
|
||
printf("%d %c\n", (path[i] >> 1) + 1, path[i] & 1 ? '+' : '-');
|
||
}
|
||
}
|
||
}
|
||
//有向图 POJ2337
|
||
//给出n个小写字母组成的单词, 要求将n个单词连接起来, 输出字典序最小的解
|
||
int head[N], to[M], nxt[M], tot, in[N], out[N], path[N], cnt;
|
||
bool vis[M];
|
||
string str[N];
|
||
void init() { tot = 0; memset(head, -1, sizeof(head)); }
|
||
void addedge(int x, int y) {
|
||
to[tot] = y; nxt[tot] = head[x]; head[x] = tot++;
|
||
}
|
||
void dfs(int u) {
|
||
for (int &i = head[u]; ~i;) {
|
||
if (!vis[i]) { vis[i] = true; int t = i; dfs(to[i]); path[cnt++] = n - t - 1; }
|
||
else { i = nxt[i]; }
|
||
}
|
||
}
|
||
int main() {
|
||
int T, n;
|
||
char s[25];
|
||
scanf("%d", &T);
|
||
while (T--) {
|
||
init(); cnt = 0;
|
||
memset(vis, 0, sizeof(vis));
|
||
memset(in, 0, sizeof(in));
|
||
memset(out, 0, sizeof(out));
|
||
scanf("%d", &n);
|
||
for (int i = 0; i < n; i++) { scanf("%s", s); str[i] = s; }
|
||
sort(str, str + n);
|
||
int s = 26;
|
||
for (int i = n - 1; i >= 0; i--) { //字典序大的先加入
|
||
int u = str[i][0] - 'a', v = str[i][str[i].size() - 1] - 'a';
|
||
addedge(u, v); out[u]++; in[v]++;
|
||
s = min(s, min(u, v));
|
||
}
|
||
int cnt1 = 0, cnt2 = 0;
|
||
for (int i = 0; i < 26; i++) {
|
||
if (out[i] - in[i] == 1) { cnt1++; s = i; } //如果有一个出度比入度大1的点, 就从这个点出发, 否则从最小的点出发
|
||
else if (out[i] - in[i] == -1) { cnt2++; }
|
||
else if (out[i] - in[i] != 0) { cnt1 = 3; }
|
||
}
|
||
if (!((cnt1 == 0 && cnt2 == 0) || (cnt1 == 1 && cnt2 == 1))) { puts("***"); continue; }
|
||
dfs(s);
|
||
if (cnt != n) { puts("***"); continue; }
|
||
for (int i = cnt - 1; i >= 0; i--) {
|
||
printf("%s%c", str[path[i]].c_str(), i > 0 ? '.' : '\n');
|
||
}
|
||
}
|
||
}
|
||
//有向图 并查集 HDU1116
|
||
//判断所有单词能不能连成一串, 如果有多个重复的单词, 也必须满足这样的条件
|
||
int n, fa[N], in[N], out[N], p[N];
|
||
bool vis[N];
|
||
char s[M];
|
||
void init() {
|
||
memset(vis, 0, sizeof(vis)); memset(in, 0, sizeof(in)); memset(out, 0, sizeof(out));
|
||
for (int i = 0; i < N; i++) { fa[i] = i; }
|
||
}
|
||
int findfa(int n) {
|
||
return n == fa[n] ? n : fa[n] = findfa(fa[n]);
|
||
}
|
||
inline void unite(int x, int y) {
|
||
x = findfa(x); y = findfa(y);
|
||
if (x != y) { fa[y] = x; }
|
||
}
|
||
int main() {
|
||
int T, n;
|
||
scanf("%d", &T);
|
||
while (T--) {
|
||
init();
|
||
scanf("%d", &n);
|
||
while (n--) {
|
||
scanf("%s", s); int u = s[0] - 'a', v = s[strlen(s) - 1] - 'a';
|
||
unite(u, v); out[u]++; in[v]++; vis[u] = vis[v] = true;
|
||
}
|
||
int cnt = 0, k = 0;
|
||
for (int i = 0; i < N; i++) {
|
||
fa[i] = findfa(i);
|
||
if (vis[i] && fa[i] == i) { cnt++; }
|
||
}
|
||
if (cnt > 1) { puts("The door cannot be opened."); continue; } //不连通
|
||
for (int i = 0; i < N; i++) {
|
||
if (in[i] != out[i]) { p[k++] = i; if (k > 2) { break; } }
|
||
}
|
||
if (k == 0 || (k == 2 && ((out[p[0]] - in[p[0]] == 1 && in[p[1]] - out[p[1]] == 1) ||
|
||
(out[p[1]] - in[p[1]] == 1 && in[p[0]] - out[p[0]] == 1)))) {
|
||
puts("Ordering is possible.");
|
||
} else { puts("The door cannot be opened."); }
|
||
}
|
||
}
|
||
//混合图 POJ1637 (本题保证连通)
|
||
//判断欧拉回路 需ISAP + 邻接表 O(V^2*E)
|
||
int in[N], out[N];
|
||
int main() {
|
||
int T, u, v, w;
|
||
scanf("%d", &T);
|
||
while (T--) {
|
||
init();
|
||
memset(in, 0, sizeof(in)); memset(out, 0, sizeof(out));
|
||
scanf("%d%d", &n, &m);
|
||
while (m--) {
|
||
scanf("%d%d%d", &u, &v, &w); out[u]++; in[v]++;
|
||
if (w == 0) { addedge(u, v, 1); } //双向
|
||
}
|
||
bool flag = true;
|
||
for (int i = 1; i <= n; i++) {
|
||
if (out[i] - in[i] > 0) { addedge(0, i, (out[i] - in[i]) >> 1); }
|
||
else if (in[i] - out[i] > 0) { addedge(i, n + 1, (in[i] - out[i]) >> 1); }
|
||
if (out[i] - in[i] & 1) { flag = false; break; }
|
||
}
|
||
if (!flag) { puts("impossible"); continue; }
|
||
ISAP(0, n + 1, n + 2);
|
||
for (int i = head[0]; ~i; i = nxt[i]) {
|
||
if (cap[i] > 0 && cap[i] > flow[i]) { flag = false; break; }
|
||
}
|
||
puts(flag ? "possible" : "impossible");
|
||
}
|
||
}
|
||
//2-SAT
|
||
//染色法
|
||
const int N = 20005;
|
||
const int M = 100005;
|
||
int head[N], to[M], nxt[M], tot;
|
||
bool vis[N]; //染色标记
|
||
int S[N], top; //栈
|
||
void init() { tot = 0; memset(head, -1, sizeof(head)); }
|
||
void addedge(int x, int y) { to[tot] = y; nxt[tot] = head[x]; head[x] = tot++; }
|
||
bool dfs(int u) {
|
||
if (vis[u ^ 1]) { return false; }
|
||
if (vis[u]) { return true; }
|
||
vis[u] = true; S[top++] = u;
|
||
for (int i = head[u]; ~i; i = nxt[i]) {
|
||
if (!dfs(to[i])) { return false; }
|
||
}
|
||
return true;
|
||
}
|
||
bool twoSAT(int n) {
|
||
memset(vis, 0, sizeof(vis));
|
||
for (int i = 0; i < n; i += 2) {
|
||
if (vis[i] || vis[i ^ 1]) { continue; }
|
||
top = 0;
|
||
if (!dfs(i)) {
|
||
while (top) { vis[S[--top]] = false; }
|
||
if (!dfs(i ^ 1)) { return false; }
|
||
}
|
||
}
|
||
return true;
|
||
}
|
||
//HDU 1814
|
||
int main() {
|
||
int n, m, u, v;
|
||
while (~scanf("%d%d", &n, &m)) {
|
||
init();
|
||
while (m--) {
|
||
scanf("%d%d", &u, &v); u--; v--;
|
||
addedge(u, v ^ 1); addedge(v, u ^ 1);
|
||
}
|
||
if (twoSAT(n << 1)) {
|
||
for (int i = 0; i < n << 1; i++) {
|
||
if (vis[i]) { printf("%d\n", i + 1); }
|
||
}
|
||
} else { printf("NIE\n"); }
|
||
}
|
||
}
|
||
//Tarjan强连通缩点
|
||
const int N = 1005;
|
||
const int M = 100005;
|
||
int head[N], to[M], nxt[M], tot;
|
||
int num[N], Low[N], DFN[N], S[N], Belong[N], idx, top, scc; //Belong数组的值1~scc
|
||
bool instack[N];
|
||
void init() { tot = 0; memset(head, -1, sizeof(head)); }
|
||
void addedge(int x, int y) { to[tot] = y; nxt[tot] = head[x]; head[x] = tot++; }
|
||
void Tarjan(int u) {
|
||
Low[u] = DFN[u] = ++idx; S[top++] = u; instack[u] = true;
|
||
for (int i = head[u]; ~i; i = nxt[i]) {
|
||
int v = to[i];
|
||
if (!DFN[v]) { Tarjan(v); Low[u] = min(Low[u], Low[v]); }
|
||
else if (instack[v] && Low[u] > DFN[v]) { Low[u] = DFN[v]; }
|
||
}
|
||
if (Low[u] == DFN[u]) {
|
||
scc++;
|
||
do { v = S[--top]; instack[v] = false; Belong[v] = scc; num[scc]++; } while (v != u);
|
||
}
|
||
}
|
||
bool solvable(int n) { //n是总个数, 需要选择一半
|
||
memset(DFN, 0, sizeof(DFN));
|
||
memset(instack, 0, sizeof(instack));
|
||
memset(num, 0, sizeof(num));
|
||
idx = scc = top = 0;
|
||
for (int i = 0; i < n; i++) { if (!DFN[i]) { Tarjan(i); } }
|
||
for (int i = 0; i < n; i += 2) { if (Belong[i] == Belong[i ^ 1]) { return false; } }
|
||
return true;
|
||
}
|
||
//拓扑排序求任意一组解部分
|
||
queue<int> q1, q2;
|
||
vector<vector<int>> dag; //缩点后的逆向DAG图
|
||
char color[N]; //染色, 为'R'是选择的
|
||
int cf[N], indeg[N]; //入度
|
||
void solve(int n) {
|
||
dag.assign(scc + 1, vector<int>());
|
||
memset(indeg, 0, sizeof(indeg));
|
||
memset(color, 0, sizeof(color));
|
||
for (int u = 0; u < n; u++) {
|
||
for (int i = head[u]; ~i; i = nxt[i]) {
|
||
int v = to[i];
|
||
if (Belong[u] != Belong[v]) { dag[Belong[v]].push_back(Belong[u]); indeg[Belong[u]]++; }
|
||
}
|
||
}
|
||
for (int i = 0; i < n; i += 2) {
|
||
cf[Belong[i]] = Belong[i ^ 1]; cf[Belong[i ^ 1]] = Belong[i];
|
||
}
|
||
while (!q1.empty()) { q1.pop(); }
|
||
while (!q2.empty()) { q2.pop(); }
|
||
for (int i = 1; i <= scc; i++) { if (indeg[i] == 0) { q1.push(i); } }
|
||
while (!q1.empty()) {
|
||
int u = q1.front(); q1.pop();
|
||
if (color[u] == 0) { color[u] = 'R'; color[cf[u]] = 'B'; }
|
||
for (int i = 0; i < (int)dag[u].size(); i++) {
|
||
if (--indeg[dag[u][i]] == 0) { q1.push(dag[u][i]); }
|
||
}
|
||
}
|
||
}
|
||
int change(char s[]) {
|
||
int ret = 0, i = 0;
|
||
while (s[i] >= '0' && s[i] <= '9') { ret *= 10; ret += s[i++] - '0'; }
|
||
return (ret << 1) + (s[i] != 'w');
|
||
}
|
||
//POJ3648
|
||
int main() {
|
||
int n, m, u, v;
|
||
char s1[10], s2[10];
|
||
while (scanf("%d%d", &n, &m), (n || m)) {
|
||
init();
|
||
while (m--) {
|
||
scanf("%s%s", s1, s2); u = change(s1); v = change(s2);
|
||
addedge(u ^ 1, v); addedge(v ^ 1, u);
|
||
}
|
||
addedge(1, 0);
|
||
if (solvable(n << 1)) {
|
||
solve(n << 1);
|
||
for (int i = 1; i < n; i++) {
|
||
//注意这一定是判断color[Belong[
|
||
if (color[Belong[i << 1]] == 'R') { printf("%dw", i); }
|
||
else { printf("%dh", i); }
|
||
putchar(i != n - 1 ? ' ' : '\n');
|
||
}
|
||
} else { printf("bad luck\n"); }
|
||
}
|
||
}
|
||
//最大团
|
||
//搜索 O(n*2^n)
|
||
int mp[N][N], stk[N][N], dp[N], ans;
|
||
bool dfs(int crt, int tot) {
|
||
if (!crt) {
|
||
if (tot > ans) { ans = tot; return true; }
|
||
return false;
|
||
}
|
||
for (int i = 0, u, nxt; i < crt; i++) {
|
||
u = stk[tot][i]; nxt = 0;
|
||
if (crt - i + tot <= ans) { return false; }
|
||
if (dp[u] + tot <= ans) { return false; }
|
||
for (int j = i + 1; j < crt; j++) {
|
||
int v = stk[tot][j];
|
||
if (g[u][v]) { stk[tot + 1][nxt++] = v; }
|
||
}
|
||
if (dfs(nxt, tot + 1)) { return true; }
|
||
}
|
||
return false;
|
||
}
|
||
int maxClique(int n) {
|
||
ans = 0;
|
||
for (int i = n - 1, j, k; i >= 0; i--) {
|
||
for (j = i + 1, k = 0; j < n; j++) {
|
||
if (g[i][j]) { stk[1][k++] = j; }
|
||
}
|
||
dfs(k, 1); dp[i] = ans;
|
||
}
|
||
return ans;
|
||
}
|
||
//随机贪心 O(T*n^2)
|
||
const int T = 1000;
|
||
int mp[N][N], id[N], ansn, ans[N]; bool del[N];
|
||
void solve(int n) {
|
||
memset(del, 0, sizeof(del)); int k = 0;
|
||
for (int i = 0, j; i < n; i++) {
|
||
if (del[i]) { continue; }
|
||
for (j = i + 1, k++; j < n; j++) { if (!mp[id[i]][id[j]]) { del[j] = true; } }
|
||
}
|
||
if (k > ansn) {
|
||
ansn = k;
|
||
for (int i = k = 0; i < n; i++) { if (!del[i]) { ans[k++] = id[i]; } }
|
||
}
|
||
}
|
||
void maxClique(int n) {
|
||
for (int i = 0; i < n; i++) { id[i] = i; }
|
||
for (int t = 0; t < T; t++) {
|
||
for (int i = 0; i < n; i++) { swap(id[i], id[rand() % n]); } solve();
|
||
}
|
||
}
|
||
//最大独立集
|
||
//随机算法 O(T*(V+E))
|
||
const int T = 1000;
|
||
int q[N], pos[N]; bool del[N];
|
||
int solve(int n) {
|
||
int ans = 0;
|
||
for (int t = 0; t < T; t++) {
|
||
int ret = 0, top = n; memset(del, 0, sizeof(del));
|
||
for (int i = 1; i <= n; i++) { q[i] = pos[i] = i; }
|
||
while (top) {
|
||
int x = rand() % top + 1, u = q[x]; q[x] = q[top--]; pos[q[x]] = x; ret++;
|
||
for (int i = head[u]; ~i; i = nxt[i]) {
|
||
int v = to[i];
|
||
if (!del[v]) { del[v] = true; x = pos[v]; q[x] = q[top--]; pos[q[x]] = x; }
|
||
}
|
||
}
|
||
ans = max(ans, ret);
|
||
}
|
||
return ans;
|
||
}
|
||
|
||
==========图论-最短路=========
|
||
//邻接表
|
||
typedef int wtype;
|
||
const wtype INF = 0x3f3f3f3f;
|
||
int head[N], to[M], nxt[M], tot; wtype len[M];
|
||
void init() { tot = 0; memset(head, -1, sizeof(head)); }
|
||
void addedge(int x, int y, wtype z) { to[tot] = y; len[tot] = z; nxt[tot] = head[x]; head[x] = tot++; }
|
||
//Dijkstra + priority_queue + 邻接矩阵 O(V^2)
|
||
typedef int wtype;
|
||
struct Node {
|
||
int v; wtype w;
|
||
bool operator<(const Node &r)const { return w > r.w; }
|
||
};
|
||
int n, pre[N]; wtype mp[N][N], dist[N]; bool vis[N]; //邻接矩阵初始化为INF
|
||
void Dijkstra(int src) {
|
||
memset(dist, 0x3f, sizeof(dist));
|
||
memset(vis, 0, sizeof(vis)); memset(pre, -1, sizeof(pre));
|
||
priority_queue<Node> que; que.push((Node) {src, dist[src] = 0});
|
||
while (!que.empty()) {
|
||
int u = que.top().v; que.pop();
|
||
if (vis[u]) { continue; }
|
||
vis[u] = true;
|
||
for (int v = 0; v < n; v++) {
|
||
if (!vis[v] && dist[v] > dist[u] + mp[u][v]) {
|
||
dist[v] = dist[u] + mp[u][v]; pre[v] = u; que.push((Node) {v, dist[v]});
|
||
}
|
||
}
|
||
}
|
||
}
|
||
//Dijkstra + priority_queue + vector存边 O((V+E)logV)
|
||
typedef int wtype;
|
||
struct Node {
|
||
int v; wtype w;
|
||
bool operator<(const Node &r)const { return w > r.w; }
|
||
};
|
||
vector<Node> e[N]; wtype dist[N]; bool vis[N];
|
||
void Dijkstra(int src) {
|
||
memset(dist, 0x3f, sizeof(dist));
|
||
memset(vis, 0, sizeof(vis));
|
||
priority_queue<Node> que; que.push((Node) {src, dist[src] = 0});
|
||
while (!que.empty()) {
|
||
int u = que.top().v; que.pop();
|
||
if (vis[u]) { continue; }
|
||
vis[u] = true;
|
||
for (int i = 0; i < (int)e[u].size(); i++) {
|
||
int v = e[u][i].v; wtype w = e[u][i].w;
|
||
if (!vis[v] && dist[v] > dist[u] + w) {
|
||
dist[v] = dist[u] + w; que.push((Node) {v, dist[v]});
|
||
}
|
||
}
|
||
}
|
||
}
|
||
//Dijkstra + priority_queue + 邻接表 O((V+E)logV)
|
||
typedef int wtype;
|
||
struct Node {
|
||
int v; wtype w;
|
||
bool operator<(const Node &r)const { return w > r.w; }
|
||
};
|
||
wtype dist[N];
|
||
void Dijkstra(int src) {
|
||
memset(dist, 0x3f, sizeof(dist));
|
||
priority_queue<Node> que; que.push((Node) {src, dist[src] = 0});
|
||
while (!que.empty()) {
|
||
int u = que.top().v; wtype w = que.top().w; que.pop();
|
||
if (w > dist[u]) { continue; }
|
||
for (int i = head[u]; ~i; i = nxt[i]) {
|
||
int v = to[i];
|
||
if (dist[v] > len[i] + w) {
|
||
dist[v] = w + len[i]; que.push((Node) {v, dist[v]});
|
||
}
|
||
}
|
||
}
|
||
}
|
||
//SPFA + queue/stack + vector存边 O(kE)
|
||
typedef int wtype;
|
||
struct Edge { int v; wtype w; };
|
||
vector<Edge> e[N]; int cnt[N], pre[N]; wtype dist[N]; bool vis[N];
|
||
bool SPFA(int src) {
|
||
memset(dist, 0x3f, sizeof(dist)); dist[src] = 0;
|
||
memset(cnt, 0, sizeof(cnt)); cnt[src] = 1;
|
||
memset(vis, 0, sizeof(vis)); vis[src] = true;
|
||
memset(pre, -1, sizeof(pre));
|
||
queue<int> que; que.push(src); //stack
|
||
while (!que.empty()) {
|
||
int u = que.front(); que.pop();
|
||
vis[u] = false;
|
||
for (int i = 0; i < (int)e[u].size(); i++) {
|
||
int v = e[u][i].v; wtype w = e[u][i].w;
|
||
if (dist[v] > dist[u] + w) {
|
||
dist[v] = dist[u] + w; pre[v] = u;
|
||
if (!vis[v]) {
|
||
vis[v] = true; que.push(v);
|
||
if (++cnt[v] > n) { return false; } //有负环回路
|
||
}
|
||
}
|
||
}
|
||
}
|
||
return true; //没有负环回路
|
||
}
|
||
//SPFA + SLF + LLL + 邻接表 O(kE)
|
||
typedef int wtype;
|
||
wtype dist[N];
|
||
bool vis[M];
|
||
void SPFA(int src) {
|
||
memset(dist, 0x3f, sizeof(dist)); dist[src] = 0;
|
||
memset(vis, 0, sizeof(vis)); vis[src] = true;
|
||
deque<int> que; que.push_back(src);
|
||
ll sum = 0;
|
||
while (!que.empty()) {
|
||
int u = que.front(); que.pop_front();
|
||
if (!que.empty() && (ll)dist[u] * que.size() > sum) { que.push_back(u); continue; }
|
||
sum -= dist[u]; vis[u] = false;
|
||
for (int i = head[u]; ~i; i = nxt[i]) {
|
||
int v = to[i]; wtype d = dist[u] + len[i];
|
||
if (d < dist[v]) {
|
||
if (vis[v]) { sum += d - dist[v]; }
|
||
dist[v] = d;
|
||
if (!vis[v]) {
|
||
if (!que.empty() && dist[que.front()] > dist[v]) { que.push_front(v); }
|
||
else { que.push_back(v); }
|
||
sum += dist[v]; vis[v] = true;
|
||
}
|
||
}
|
||
}
|
||
}
|
||
}
|
||
//差分约束系统
|
||
//a向b连一条权值为c的有向边表示b ? a <= c, 用SPFA判断是否存在负环, 存在即无解
|
||
//Bellman-Ford + vector O(VE)
|
||
//可以处理负边权图, 可以判断是否存在负环回路, 当且仅当图中不包含从源点可达的负权回路时返回true
|
||
typedef int wtype;
|
||
struct Edge { int u, v; wtype w; };
|
||
vector<Edge> e;
|
||
wtype dist[N];
|
||
bool BellmanFord(int src) {
|
||
memset(dist, 0x3f, sizeof(dist)); dist[src] = 0;
|
||
for (int i = 1; i < n; i++) {
|
||
bool flag = false;
|
||
for (int j = 0; j < (int)e.size(); j++) {
|
||
int u = e[j].u, v = e[j].v; wtype w = e[j].w;
|
||
if (dist[v] > dist[u] + w) { dist[v] = dist[u] + w; flag = true; }
|
||
}
|
||
if (!flag) { return true; } //没有负环回路
|
||
}
|
||
for (int j = 0; j < (int)e.size(); j++) {
|
||
int u = e[j].u, v = e[j].v; wtype w = e[j].w;
|
||
if (dist[v] > dist[u] + w) { return false; } //有负环回路
|
||
}
|
||
return true; //没有负环回路
|
||
}
|
||
//Floyd 带路径记录 O(V^3)
|
||
typedef int wtype;
|
||
int n, pre[N][N]; wtype mp[N][N]; //初始化为INF
|
||
void Floyd() {
|
||
for (int i = 1; i <= n; i++) {
|
||
for (int j = 1; j <= n; j++) { pre[i][j] = (i == j) ? -1 : i; }
|
||
}
|
||
for (int k = 1; k <= n; k++) {
|
||
for (int i = 1; i <= n; i++) {
|
||
for (int j = 1; j <= n; j++) {
|
||
if (mp[i][k] + mp[k][j] < mp[i][j]) {
|
||
mp[i][j] = mp[i][k] + mp[k][j]; pre[i][j] = pre[k][j];
|
||
}
|
||
}
|
||
}
|
||
}
|
||
}
|
||
//无负权图的最小环
|
||
//有向图: mp[i][i] = INF, 然后跑floyd, ans = min(d[i][i])
|
||
//求无向图中经过至少3个点的最小环:
|
||
typedef int wtype;
|
||
const wtype INF = 0x3f3f3f3f;
|
||
int n; wtype mp[N][N], d[N][N]; //初始化为INF
|
||
wtype cycFloyd() {
|
||
memcpy(d, mp, sizeof(mp)); wtype ret = INF;
|
||
for (int k = 1; k <= n; k++) {
|
||
for (int i = 1; i < k; i++) {
|
||
for (int j = i + 1; j < k; j++) {
|
||
ans = min(ans, d[i][j] + mp[i][k] + mp[k][j]);
|
||
}
|
||
}
|
||
for (int i = 1; i <= n; i++) {
|
||
for (int j = 1; j <= n; j++) {
|
||
d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
|
||
}
|
||
}
|
||
}
|
||
}
|
||
//Astar求k短路
|
||
typedef int wtype;
|
||
const wtype INF = 0x3f3f3f3f;
|
||
int head[N], to[M], nxt[M], tot; wtype len[M];
|
||
int rhead[N], rto[M], rnxt[M], rtot; wtype rlen[M];
|
||
void init() { tot = rtot = 0; memset(head, -1, sizeof(head)); memset(rhead, -1, sizeof(rhead)); }
|
||
void addedge(int x, int y, wtype z) {
|
||
to[tot] = y; len[tot] = z; nxt[tot] = head[x]; head[x] = tot++;
|
||
rto[rtot] = x; rlen[rtot] = z; rnxt[rtot] = rhead[y]; rhead[y] = rtot++;
|
||
}
|
||
struct Node {
|
||
int v; wtype w;
|
||
bool operator<(const Node &r)const { return w > r.w; }
|
||
};
|
||
int vis[N]; wtype dist[N], ans[N]; //前k短路
|
||
void Astar(int src, int des, int k) {
|
||
memset(dist, 0x3f, sizeof(dist)); memset(ans, 0x3f, sizeof(ans)); memset(vis, 0, sizeof(vis));
|
||
priority_queue<Node> que; que.push((Node) {des, dist[des] = 0});
|
||
while (!que.empty()) {
|
||
int u = que.top().v; wtype w = que.top().w; que.pop();
|
||
if (w > dist[u]) { continue; }
|
||
for (int i = rhead[u]; ~i; i = rnxt[i]) {
|
||
int v = rto[i];
|
||
if (dist[v] > rlen[i] + w) {
|
||
dist[v] = w + rlen[i]; que.push((Node) {v, dist[v]});
|
||
}
|
||
}
|
||
}
|
||
if (dist[src] < INF) { que.push((Node) {src, dis[src]}); }
|
||
while (!que.empty()) {
|
||
int u = que.top().v; wtype w = que.top().w; que.pop();
|
||
if (u == des && vis[des] <= k) { ans[vis[des]] = w; }
|
||
if (vis[u] > k) { continue; }
|
||
for (int i = head[u]; ~i; i = nxt[i]) {
|
||
int v = to[i]; que.push((Node) {v, w - dist[u] + dist[v] + len[i]});
|
||
}
|
||
}
|
||
}
|
||
|
||
|
||
===========Tarjan 算法求一个无向图的桥
|
||
int dfs[MAXN], low[MAXN], tim = 0;
|
||
inline void Dfs(int u) {
|
||
dfs[u] = low[u] = ++tim;
|
||
for (Edge *p = a[u]; p; p = p->next) if (p->flag) {
|
||
if (!dfs[p->y]) {
|
||
p->opt->flag = false;
|
||
Dfs(p->y);
|
||
low[u] = min(low[u], low[p->y]);
|
||
} else low[u] = min(low[u], dfs[p->y]);
|
||
}
|
||
for (Edge *p = a[u]; p; p = p->next) if (p->opt->flag == false && low[p->y] > dfs[u]) {
|
||
p->bridge = p->opt->bridge = true;
|
||
}
|
||
}
|
||
|
||
=========Hopcroft 算法，其中二分图左边点的编号为1-n，右边点编号为n+1 - N，match[i]表示左边的i匹配上的右边的点，0无匹配，link[i]表示右边的i匹配上的左边的点，0无匹配
|
||
|
||
const int MAXN = 10000;
|
||
int match[MAXN];
|
||
int levelx[MAXN], levely[MAXN], link[MAXN];
|
||
int d[MAXN];
|
||
inline bool Bfs(void) {
|
||
int head = 1, tail = 0;
|
||
memset(levely, 0, sizeof levely);
|
||
for (int i = 1; i <= n; i++) {
|
||
if (!match[i]) d[++tail] = i;
|
||
levelx[i] = 0;
|
||
}
|
||
bool ret = false;
|
||
while (head <= tail) {
|
||
int now = d[head++];
|
||
for (Edge *p = a[now]; p; p = p->next) if (levely[p->y] == 0) {
|
||
levely[p->y] = levelx[now] + 1;
|
||
if (link[p->y] == 0) ret = true;
|
||
else levelx[link[p->y]] = levely[p->y] + 1, d[++tail] = link[p->y];
|
||
}
|
||
}
|
||
return ret;
|
||
}
|
||
bool Find(int u) {
|
||
for (Edge *p = a[u]; p; p = p->next) if (levely[p->y] == levelx[u] + 1) {
|
||
levely[p->y] = 0;
|
||
if (link[p->y] == 0 || Find(link[p->y])) {
|
||
match[u] = p->y; link[p->y] = u;
|
||
return true;
|
||
}
|
||
}
|
||
return false;
|
||
}
|
||
inline void Match(void) {
|
||
while (Bfs())
|
||
for (int i = 1; i <= n; i++)
|
||
if (!match[i]) Find(i);
|
||
}
|
||
inline void clear(void) {
|
||
memset(match, 0, sizeof match);
|
||
memset(link, 0, sizeof link);
|
||
memset(levelx, 0, sizeof levelx);
|
||
memset(levely, 0, sizeof levely);
|
||
}
|
||
|
||
|
||
=============KM算法求最小权完美匹配（注意不是全局最大权匹配），如果两边点数不相等或者非完美二分图，则可以添加虚拟点和权值为inf的边
|
||
|
||
若求最大权匹配，则将边权取反即可
|
||
const int MAXN = 210;
|
||
int nx, ny; // 左边的点标号从1到nx，右边点标号从1到ny
|
||
long long inf, cost[MAXN][MAXN], fx[MAXN], fy[MAXN], dist[MAXN]; //权值若为long long的话，只需改动此行即可
|
||
int used[MAXN], maty[MAXN], which[MAXN];
|
||
inline void AddEdge(int x, int y, int z)
|
||
{
|
||
cost[x][y] = z;
|
||
}
|
||
pair<int, long long> KM(void)
|
||
{
|
||
for (int x = 1; x <= nx; x++)
|
||
{
|
||
int y0 = 0;
|
||
maty[0] = x;
|
||
for (int y = 0; y <= ny; y++)
|
||
{
|
||
dist[y] = inf + 1;
|
||
used[y] = false;
|
||
}
|
||
do
|
||
{
|
||
used[y0] = true;
|
||
int x0 = maty[y0], y1;
|
||
long long delta = inf + 1;
|
||
for (int y = 1; y <= ny; y++) if (!used[y])
|
||
{
|
||
long long curdist = cost[x0][y] - fx[x0] - fy[y];
|
||
if (curdist < dist[y])
|
||
{
|
||
dist[y] = curdist;
|
||
which[y] = y0;
|
||
}
|
||
if (dist[y] < delta)
|
||
{
|
||
delta = dist[y];
|
||
y1 = y;
|
||
}
|
||
}
|
||
for (int y = 0; y <= ny; y++) if (used[y])
|
||
{
|
||
fx[maty[y]] += delta;
|
||
fy[y] -= delta;
|
||
}
|
||
else dist[y] -= delta;
|
||
y0 = y1;
|
||
}
|
||
while (maty[y0] != 0);
|
||
do
|
||
{
|
||
int y1 = which[y0];
|
||
maty[y0] = maty[y1];
|
||
y0 = y1;
|
||
}
|
||
while (y0);
|
||
}
|
||
long long ret = 0;
|
||
int npair = 0;
|
||
for (int y = 1; y <= ny; y++)
|
||
{
|
||
int x = maty[y];
|
||
if (cost[x][y] < inf)
|
||
{
|
||
ret += cost[x][y];
|
||
npair++;
|
||
}
|
||
}
|
||
return make_pair(npair, ret);
|
||
}
|
||
inline void clear(void)
|
||
{
|
||
memset(fx, 0x9f, sizeof fx);
|
||
memset(fy, 0x9f, sizeof fy);
|
||
memset(cost, 0x3f, sizeof cost);
|
||
memset(maty, 0, sizeof maty);
|
||
inf = cost[0][0];
|
||
}
|
||
|
||
==================Dinic求解最大流，有当前弧优化
|
||
struct Edge
|
||
{
|
||
int y, f;
|
||
Edge *next, *opt;
|
||
}*a[MAXN], DATA[MAXM << 1], *data = DATA;
|
||
inline void Add(int x, int y, int c)
|
||
{
|
||
Edge *tmp = data++;
|
||
tmp->y = y;
|
||
tmp->f = c, tmp->next = a[x];
|
||
a[x] = tmp;
|
||
tmp = data++;
|
||
tmp->y = x;
|
||
tmp->f = 0;
|
||
tmp->next = a[y];
|
||
a[y] = tmp;
|
||
a[x]->opt = a[y];
|
||
a[y]->opt = a[x];
|
||
}
|
||
int n, m, vs, vt, L;
|
||
int level[MAXN], d[MAXN];
|
||
inline bool Bfs(void)
|
||
{
|
||
memset(level, -1, sizeof level);
|
||
d[1] = vs;
|
||
level[vs] = 0;
|
||
int head = 1, tail = 1;
|
||
while (head <= tail)
|
||
{
|
||
int now = d[head++];
|
||
e[now] = a[now];
|
||
for (Edge *p = a[now]; p; p = p->next) if (p->f > 0&& level[p->y] == -1)
|
||
level[d[++tail] = p->y] = level[now] + 1;
|
||
}
|
||
return level[vt] != -1;
|
||
}
|
||
inline int Extend(int u, int sum)
|
||
{
|
||
if (u == vt) return sum;
|
||
int r = 0, t;
|
||
for (Edge *p = e[u]; p && r < sum; p = p->next) if (level[p->y] == level[u] + 1 && p->f > 0)
|
||
{
|
||
t = std::min(sum - r, p->f);
|
||
t = Extend(p->y, t);
|
||
p->f -= t, p->opt->f += t, r += t;
|
||
e[u] = p;
|
||
}
|
||
if (!r) level[u] = -1;
|
||
return r;
|
||
}
|
||
inline int Dinic(void)
|
||
{
|
||
int r = 0, t;
|
||
while (Bfs())
|
||
{
|
||
while ((t = Extend(vs, inf))) r += t;
|
||
}
|
||
return r;
|
||
}
|
||
|
||
|
||
============spfa求解最小费用最大流
|
||
struct Edge
|
||
{
|
||
int y, f, c;
|
||
Edge *next, *opt;
|
||
Edge(int y, int f, int c, Edge *next):y(y), f(f), c(c), next(next) {}
|
||
}*a[MAXN];
|
||
inline void AddEdge(int x, int y, int f, int c)
|
||
{
|
||
a[x] = new Edge(y, f, c, a[x]);
|
||
a[y] = new Edge(x, 0, -c, a[y]);
|
||
a[x]->opt = a[y];
|
||
a[y]->opt = a[x];
|
||
}
|
||
int d[MAXN], vis[MAXN], dis[MAXN];
|
||
Edge *path[MAXN];
|
||
inline pair<int, int> Spfa(void)
|
||
{
|
||
int Flow = 0, Cost = 0;
|
||
while (true)
|
||
{
|
||
memset(vis, 0, sizeof vis);
|
||
memset(dis, 0x7f, sizeof dis);
|
||
memset(path, 0, sizeof path);
|
||
int head = 1, tail = 1, sum = 1;
|
||
d[1] = vs;
|
||
vis[vs] = true;
|
||
dis[vs] = 0;
|
||
while (sum)
|
||
{
|
||
int now = d[head++];
|
||
if (head == MAXN) head = 1;
|
||
sum--;
|
||
for (Edge *p = a[now]; p; p = p->next) if (p->f > 0 && dis[p->y] > dis[now] + p->c)
|
||
{
|
||
dis[p->y] = dis[now] + p->c;
|
||
path[p->y] = p;
|
||
if (!vis[p->y])
|
||
{
|
||
++tail;
|
||
if (tail == MAXN) tail = 1;
|
||
sum++;
|
||
d[tail] = p->y;
|
||
vis[p->y] = true;
|
||
}
|
||
}
|
||
vis[now] = false;
|
||
}
|
||
if (dis[vt] == dis[0]) return make_pair(Flow, Cost);
|
||
int tmp = vt, Min = ~0U>>1;
|
||
while (path[tmp])
|
||
{
|
||
Min = min(Min, path[tmp]->f);
|
||
tmp = path[tmp]->opt->y;
|
||
}
|
||
Flow += Min;
|
||
tmp = vt;
|
||
while (path[tmp])
|
||
{
|
||
path[tmp]->f -= Min;
|
||
path[tmp]->opt->f += Min;
|
||
Cost += Min * path[tmp]->c;
|
||
tmp = path[tmp]->opt->y;
|
||
}
|
||
}
|
||
return make_pair(Flow, Cost);
|
||
}
|
||
|
||
================最小树形图算法
|
||
使用邻接矩阵存图，返回-1表示无最小树形图。否则，tmp[i].second表示i的入边
|
||
map[i][j] = 1 表示表示存在一条从i到j的有向边
|
||
dis[i][j] 表示从i到j边的长度
|
||
|
||
typedef pair<int, int> PII;
|
||
PII tmp[MAXN];
|
||
int dis[MAXN][MAXN];
|
||
int map[MAXN][MAXN];
|
||
int vis[MAXN], q[MAXN], inp[MAXN];
|
||
int Dfs(int u)
|
||
{
|
||
vis[u] = true;
|
||
int ret = 1;
|
||
for (int i = 1; i <= n; i++) if (map[u][i] && !vis[i]) ret += Dfs(i);
|
||
return ret;
|
||
}
|
||
int Zhuliu(void)
|
||
{
|
||
memset(map, 0, sizeof map);
|
||
memset(vis, 0, sizeof vis);
|
||
if (Dfs(1) != n) return -1;
|
||
int done = 0;
|
||
while (true)
|
||
{
|
||
memset(vis, 0, sizeof vis);
|
||
memset(inp, 0, sizeof inp);
|
||
int Ans = 0;
|
||
for (int i = 1; i <= n; i++) vis[i] = i;
|
||
for (int i = 2; i <= n; i++)
|
||
{
|
||
tmp[i] = PII(1000000000, 0);
|
||
for (int j = 1; j <= n; j++) if (map[j][i] == 1) tmp[i] = min(tmp[i], PII(dis[j][i], j));
|
||
inp[tmp[i].second]++;
|
||
if (tmp[i].second) Ans += tmp[i].first;
|
||
}
|
||
int head = 1, tail = 0;
|
||
for (int i = 1; i <= n; i++) if (!inp[i]) q[++tail] = i;
|
||
while (head <= tail)
|
||
{
|
||
int now = q[head++];
|
||
if (!--inp[tmp[now].second]) q[++tail] = tmp[now].second;
|
||
}
|
||
bool ok = true;
|
||
for (int i = 1, t; i <= n; i++) if (inp[i] > 0)
|
||
{
|
||
t = i;
|
||
ok = false;
|
||
do
|
||
{
|
||
inp[t] = -i;
|
||
t = tmp[t].second;
|
||
vis[t] = i;
|
||
}
|
||
while (t != i);
|
||
}
|
||
if (ok) return Ans + done;
|
||
for (int i = 1; i <= n; i++) if (inp[i] < 0)
|
||
{
|
||
done += tmp[i].first;
|
||
for (int j = 1; j <= n; j++) if (vis[j] != vis[i])
|
||
{
|
||
if (map[i][j])
|
||
{
|
||
map[vis[i]][vis[j]] = 1;
|
||
dis[vis[i]][vis[j]] = min(dis[vis[i]][vis[j]], dis[i][j]);
|
||
}
|
||
if (map[j][i])
|
||
{
|
||
dis[vis[j]][vis[i]] = min(dis[vis[j]][vis[i]], dis[j][i] - tmp[i].first);
|
||
map[vis[j]][vis[i]] = 1;
|
||
}
|
||
}
|
||
if (vis[i] != i) for (int j = 1; j <= n; j++) map[i][j] = map[j][i] = 0;
|
||
}
|
||
}
|
||
}
|
||
|