1009 lines
26 KiB
Plaintext
1009 lines
26 KiB
Plaintext
/** SPFA 单源最短路径算法 不支持负环*/
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namespace SPFA
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{
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const int MAXN = 1005;
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int d[MAXN];/// distance [ From S to ... ]
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int v[MAXN];/// visit
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int q[MAXN];/// 基于数组的队列(也可用queue等...)
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int mp[MAXN][MAXN]; /// mp[i][j] i --> j is connected.
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int n;/// n is the number of max Point .
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void spfa(int StartPoint) /// d[i] is the min distance from StartPoint to i ( Both >=1 )
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{
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memset(d,0x3f,sizeof(d));
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memset(v,0,sizeof(v));
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/*
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for(int i=1;i<MAXN;i++)
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d[i]=INF,v[i]=0;*/
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int cnt=0;
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q[cnt++]=StartPoint;
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v[StartPoint]=1;
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d[StartPoint]=0;
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while(cnt>0)
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{
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int c=q[--cnt];
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v[c]=0;
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for(int i=1;i<=n;i++)
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{
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/// Here : if your mp[i][j] use INF as infinite, then use mp[c][i]!=INF.
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/// Or you may use mp[i][j]!=-1 && d[i] > d[c] + mp[c][i]
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if( mp[c][i]!=INF && d[i]>d[c]+mp[c][i] )
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{
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d[i]=d[c]+mp[c][i];
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if(!v[i]) v[i]=1,q[cnt++]=i;
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}
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}
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}
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}
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}/// End of NameSpace SPFA
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线段树模板(Powered By HC TECH - Kiritow)
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include
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/// General includes
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#include <cstdio>
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#include <cstdlib>
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#include <cstring>
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#include <algorithm>
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using namespace std;
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最基础的线段树
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/// 最基础的线段树: 单点更新,区间运算(求和)
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namespace SegmentTree
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{
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const int MAXN = 1000100;
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const int MAXTREENODE = MAXN<<2;
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struct node
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{
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int lt,rt;
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int val;
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};
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node tree[MAXTREENODE];
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/// _internal_v is a indexer of SegmentTree. It guides the procedure to the right node.
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void build(int L,int R,int _internal_v=1) /// Build a tree, _internal_v is 1 by default.
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{
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tree[_internal_v].lt=L;
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tree[_internal_v].rt=R;
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if(L==R)
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{
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scanf("%d",&tree[_internal_v].val);
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/// Or: tree[_internal].val = VAL_BY_DEFAULT
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return;
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}
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int mid=(L+R)>>1;
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build(L,mid,_internal_v<<1);
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build(mid+1,R,_internal_v<<1|1);/// x<<1 == x*2; x<<1|1 == x*2+1; (faster == slower)
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/// SegmentTree Main Algorithm
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tree[_internal_v].val=tree[_internal_v<<1].val+tree[_internal_v<<1|1].val;
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}
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void update(int Pos,int Val,int _internal_v=1)/// Update a position, _internal_v is 1 by default.
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{
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if(tree[_internal_v].lt==tree[_internal_v].rt)
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{
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tree[_internal_v].val=Val;
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return;
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}
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/// Update Deep-Loop
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if(Pos <= tree[_internal_v<<1].rt) update(Pos,Val,_internal_v<<1);
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if(Pos >= tree[_internal_v<<1|1].lt) update(Pos,Val,_internal_v<<1|1);
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/// SegmentTree Main Algorithm
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tree[_internal_v].val = tree[_internal_v<<1].val+tree[_internal_v<<1|1].val;
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}
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int _internal_ans;
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inline void _internal_clear_ans()
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{
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_internal_ans=0;
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}
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inline int _internal_get_ans()
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{
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return _internal_ans;
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}
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void basic_query(int L,int R,int _internal_v=1)/// Query A Segment [L,R] , _internal_v is 1 by default.
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{
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if(tree[_internal_v].lt >= L && tree[_internal_v].rt <= R)
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{
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_internal_ans+=tree[_internal_v].val;
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return;
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}
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if(L <= tree[_internal_v<<1].rt) basic_query(L,R,_internal_v<<1);
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if(R >= tree[_internal_v<<1|1].lt) basic_query(L,R,_internal_v<<1|1);
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}
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int query(int L,int R)
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{
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_internal_clear_ans();
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basic_query(L,R);
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return _internal_get_ans();
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}
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}/// End of namespace SegmentTree
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属性线段树
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/// 延迟更新: 区间赋值更新, 区间运算(求和)
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namespace AttributeSegmentTree
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{
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const int MAXN = 100100;
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const int MAXTREENODE = MAXN << 2;
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const int ATTR_BY_DEFAULT=1;///默认初始化属性
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struct node
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{
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int lt,rt;
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int attr;
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};
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node tree[MAXTREENODE];
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void build(int L,int R,int _indexer=1)
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{
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tree[_indexer].lt=L;
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tree[_indexer].rt=R;
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tree[_indexer].attr=ATTR_BY_DEFAULT;
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if(L!=R)
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{
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int mid=(L+R)>>1;
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build(L,mid,_indexer<<1);
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build(mid+1,R,_indexer<<1|1);
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}
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}
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void update(int L,int R,int NewAttr,int _indexer=1)
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{
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if(tree[_indexer].attr==NewAttr) return; /// Same Attribute. Don't Need Change.
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if(tree[_indexer].lt==L&&tree[_indexer].rt==R)
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{
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/// Right this segment. Update.
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tree[_indexer].attr=NewAttr;
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return;
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}
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/// This segment has only 1 attribute. New attribute is different.
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/// So change this segment's manager's attribute to -1 ( Different Attribute in this segment )
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if(tree[_indexer].attr!=-1)
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{
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tree[_indexer<<1].attr=tree[_indexer<<1|1].attr=tree[_indexer].attr;
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tree[_indexer].attr=-1;
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}
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/// If This segment has already had several attributes, operate its subtree by Deep-Loop.
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int mid=(tree[_indexer].lt+tree[_indexer].rt)>>1;
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if(L>mid)
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{
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update(L,R,NewAttr,_indexer<<1|1);
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}
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else if(R<=mid)
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{
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update(L,R,NewAttr,_indexer<<1);
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}
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else
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{
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update(L,mid,NewAttr,_indexer<<1);
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update(mid+1,R,NewAttr,_indexer<<1|1);
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}
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}
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#define ValueOfAttr(Attr) (Attr)
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int AttrSumUp(int _indexer=1)
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{
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if(tree[_indexer].attr!=-1)
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{
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return ValueOfAttr(tree[_indexer].attr)*(tree[_indexer].rt-tree[_indexer].lt+1);
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}
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else
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{
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return AttrSumUp(_indexer<<1)+AttrSumUp(_indexer<<1|1);
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}
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}
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}/// End of namespace AttributeSegmentTree
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成段更新的线段树(LAZY思想)
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/// 延迟更新: 区间运算更新(加法), 区间运算(求和)
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namespace LazySegmentTree
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{
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const int MAXN = 100100;
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const int MAXTREENODE = MAXN << 2;
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struct node
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{
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int lt,rt;
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int val;
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int add;
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};
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node tree[MAXTREENODE];
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void _internal_PushUp(int _indexer)
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{
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tree[_indexer].val=tree[_indexer<<1].val+tree[_indexer<<1|1].val;
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}
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void _internal_PushDown(int _indexer)
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{
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if(tree[_indexer].add!=0)
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{
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/// Broadcast this add value to Left and Right sub-tree node.
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tree[_indexer<<1].add+=tree[_indexer].add;
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tree[_indexer<<1|1].add+=tree[_indexer].add;
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/// Confirm this change by calculate and add changes to sub-trees.
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tree[_indexer<<1].val+=tree[_indexer].add * (tree[_indexer<<1].rt-tree[_indexer<<1].lt+1);
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tree[_indexer<<1|1].val+=tree[_indexer].add *(tree[_indexer<<1|1].rt-tree[_indexer<<1|1].lt+1);
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/// Now Clear this node's add value.
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tree[_indexer].add=0;
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}
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}
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void build(int L,int R,int _indexer=1)
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{
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tree[_indexer].lt=L;
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tree[_indexer].rt=R;
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tree[_indexer].add=0;/// This must be set to 0.
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if(L==R)
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{
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//scanf("%d",&tree[_indexer].val);
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tree[_indexer].val = 0;
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return;
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}
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int mid=(L+R)>>1;
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build(L,mid,_indexer<<1);
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build(mid+1,R,_indexer<<1|1);
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/// Update this val from down to up. (>.<)
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_internal_PushUp(_indexer);
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}
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void update(int L,int R,int ValToAdd,int _indexer=1)
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{
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/// Return when L or R exceeds range. So smart !
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if(R<tree[_indexer].lt||L>tree[_indexer].rt) return;
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if(L<=tree[_indexer].lt&&R>=tree[_indexer].rt)
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{
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/// This range is covered. So just add the 'add' value, which is called "LAZY"
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tree[_indexer].add+=ValToAdd;
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tree[_indexer].val+=ValToAdd*(tree[_indexer].rt-tree[_indexer].lt+1);
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return;
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}
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_internal_PushDown(_indexer);
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/// This ... Hum.. Seems not so clever...
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update(L,R,ValToAdd,_indexer<<1);
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update(L,R,ValToAdd,_indexer<<1|1);
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_internal_PushUp(_indexer);
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}
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int ans;
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void basic_query(int L,int R,int _indexer=1)
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{
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/// Data to find is not in this range.
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if(R<tree[_indexer].lt||L>tree[_indexer].rt) return;
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/// Data to find is right in this range , or covers this range.
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if(L<=tree[_indexer].lt&&R>=tree[_indexer].rt)
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{
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ans+=tree[_indexer].val;
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return ;
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}
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_internal_PushDown(_indexer);
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int mid=(tree[_indexer].lt+tree[_indexer].rt)>>1;
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if(R<=mid)
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basic_query(L,R,_indexer<<1);
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else if(L>mid)
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basic_query(L,R,_indexer<<1|1);
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else
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{
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basic_query(L,mid,_indexer<<1);
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basic_query(mid+1,R,_indexer<<1|1);
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}
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}
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int query(int L,int R)
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{
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ans=0;
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basic_query(L,R);
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return ans;
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}
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}/// End of namespace LazySegmentTree
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成段更新，区间合并的线段树
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/// 区间赋值更新, 区间合并, 查找左端
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namespace AttributeMergeSegmentTree
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{
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const int MAXN = 100100;
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const int MAXTREENODE = MAXN << 3;
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const int ATTR_BY_DEFAULT=-1;///默认初始化属性 -1 无需操作 0 子树有住户离开 1 子树有住户进入 (来自POJ 3667)
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struct node
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{
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int lt,rt;
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int lsum,rsum,sum;
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int attr;
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};
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node tree[MAXTREENODE];
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void build(int L,int R,int _indexer=1)
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{
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tree[_indexer].lt=L;
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tree[_indexer].rt=R;
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tree[_indexer].attr=ATTR_BY_DEFAULT;
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tree[_indexer].lsum=tree[_indexer].rsum=tree[_indexer].sum=R-L+1;
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if(L!=R)
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{
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int mid=(L+R)>>1;
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build(L,mid,_indexer<<1);
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build(mid+1,R,_indexer<<1|1);
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}
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}
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void update(int L,int R,int NewAttr,int _indexer=1)
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{
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if(L==tree[_indexer].lt&&R==tree[_indexer].rt)
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{
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tree[_indexer].lsum=tree[_indexer].rsum=tree[_indexer].sum=
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NewAttr==1 ? 0 : tree[_indexer].rt-tree[_indexer].lt+1 ; /// Same as R-L+1
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tree[_indexer].attr=NewAttr;
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return;
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}
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/// Push Down (updated)
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if(tree[_indexer].attr!=-1)
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{
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/// Sync the Attribute to sub-tree
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tree[_indexer<<1].attr=tree[_indexer<<1|1].attr=tree[_indexer].attr;
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tree[_indexer].attr=-1;
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tree[_indexer<<1].rsum=tree[_indexer<<1].lsum=tree[_indexer<<1].sum= tree[_indexer<<1].attr==1 ? 0 : tree[_indexer<<1].rt-tree[_indexer<<1].lt+1;
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tree[_indexer<<1|1].rsum=tree[_indexer<<1|1].lsum=tree[_indexer<<1|1].sum= tree[_indexer<<1|1].attr==1 ? 0 : tree[_indexer<<1|1].rt-tree[_indexer<<1|1].lt+1;
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}
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int mid=(tree[_indexer].lt+tree[_indexer].rt)>>1;
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if(R<=mid)
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{
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update(L,R,NewAttr,_indexer<<1);
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}
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else if(L>mid)
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{
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update(L,R,NewAttr,_indexer<<1|1);
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}
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else
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{
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update(L,mid,NewAttr,_indexer<<1);
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update(mid+1,R,NewAttr,_indexer<<1|1);
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}
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/// Push Up (updated)
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tree[_indexer].lsum=tree[_indexer<<1].lsum; /// left & left
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tree[_indexer].rsum=tree[_indexer<<1|1].rsum; /// right & right
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if(tree[_indexer<<1].lsum == tree[_indexer<<1].rt-tree[_indexer].lt+1)
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{
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/// Father.LeftSum == RightSon.LeftSum + LeftSon.LeftSum
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tree[_indexer].lsum+=tree[_indexer<<1|1].lsum;
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}
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/// Why tree[_indexer].rsum but not tree[_indexer<<1|1].rsum ???
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if(tree[_indexer].rsum==tree[_indexer<<1|1].rt-tree[_indexer<<1|1].lt+1)
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{
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/// Father.RightSum == LeftSon.RightSum + RightSon.RightSum
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tree[_indexer].rsum+=tree[_indexer<<1].rsum;
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}
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tree[_indexer].sum=max(max(tree[_indexer<<1].sum,tree[_indexer<<1|1].sum),tree[_indexer<<1].rsum+tree[_indexer<<1|1].lsum);
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}
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int query(int Val,int _indexer=1)
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{
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if(tree[_indexer].lt==tree[_indexer].rt)
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{
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return tree[_indexer].lt;
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}
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/// Push Down (updated)
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if(tree[_indexer].attr!=-1)
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{
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/// Sync the Attribute to sub-tree
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tree[_indexer<<1].attr=tree[_indexer<<1|1].attr=tree[_indexer].attr;
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tree[_indexer].attr=-1;
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tree[_indexer<<1].rsum=tree[_indexer<<1].lsum=tree[_indexer<<1].sum= tree[_indexer<<1].attr==1 ? 0 : tree[_indexer<<1].rt-tree[_indexer<<1].lt+1;
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tree[_indexer<<1|1].rsum=tree[_indexer<<1|1].lsum=tree[_indexer<<1|1].sum= tree[_indexer<<1|1].attr==1 ? 0 : tree[_indexer<<1|1].rt-tree[_indexer<<1|1].lt+1;
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}
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int Mid=(tree[_indexer].rt+tree[_indexer].lt)>>1;
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/// Left
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if(tree[_indexer<<1].sum>=Val)
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{
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return query(Val,_indexer<<1);
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}
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/// Both Left and Right
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else if(tree[_indexer<<1].rsum+tree[_indexer<<1|1].lsum >= Val)
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{
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return Mid-tree[_indexer<<1].rsum+1;
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}
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else /// Right
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{
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return query(Val,_indexer<<1|1);
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}
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}
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}/// End of namespace AttributeMergeSegmentTree
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最长连续上升子段(LCIS)与线段树结合(HDU 3308) 模板
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/// 最长连续上升字串 与线段树结合
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namespace LCISSegmentTree
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{
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const int MAXN = 1000100;
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const int MAXTREENODE = MAXN<<2;
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int seq[MAXN];
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struct node
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{
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/// Be Sure That "BounderLen" always equal to "RightBounder - LeftBounder + 1"
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/// And Bounder Never change in one single test.
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int leftbounder,rightbounder,bounderlen;
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int leftseqlen,rightseqlen,mergedseqlen; /// From HDU 3308
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int leftvalue,rightvalue;
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};
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node tree[MAXTREENODE];
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/// _internal_v is a indexer of SegmentTree. It guides the procedure to the right node.
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void pushup(int _internal_v)
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{
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/// Left == Left.Left
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tree[_internal_v].leftseqlen=tree[_internal_v<<1].leftseqlen;
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tree[_internal_v].leftvalue=tree[_internal_v<<1].leftvalue;
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/// Right == Right.Right
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tree[_internal_v].rightseqlen=tree[_internal_v<<1|1].rightseqlen;
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tree[_internal_v].rightvalue=tree[_internal_v<<1|1].rightvalue;
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/// Merged SeqLen is the max one of two sub-tree.MergedSeqLen
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tree[_internal_v].mergedseqlen=max(tree[_internal_v<<1].mergedseqlen,tree[_internal_v<<1|1].mergedseqlen);
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/// If LeftSon.RightValue < RightSon.LeftValue, a longer Seq may exist.
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if(tree[_internal_v<<1].rightvalue<tree[_internal_v<<1|1].leftvalue)
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{
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/// If LeftSon.LeftSeqLen == LeftSon.BounderLen ...
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if(tree[_internal_v<<1].leftseqlen == tree[_internal_v<<1].bounderlen )
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{
|
||
/// ... ThisNode.LeftSeqLen += RightSon.LeftSeqLen
|
||
tree[_internal_v].leftseqlen+=tree[_internal_v<<1|1].leftseqlen;
|
||
}
|
||
|
||
/// If RightSon.RightSeqLen == RightSon.BounderLen ...
|
||
if(tree[_internal_v<<1|1].rightseqlen == tree[_internal_v<<1|1].bounderlen )
|
||
{
|
||
/// ... ThisNode.RightSeqLen += Left.RightSeqLen
|
||
tree[_internal_v].rightseqlen+=tree[_internal_v<<1].rightseqlen;
|
||
}
|
||
|
||
/// ThisNode.MergedSeqLen is the max one between itself and ...
|
||
/// ... LeftSon.RightSeqLen + RightSon.LeftSeqLen
|
||
tree[_internal_v].mergedseqlen=
|
||
max(tree[_internal_v].mergedseqlen,
|
||
tree[_internal_v<<1].rightseqlen+tree[_internal_v<<1|1].leftseqlen);
|
||
}
|
||
}
|
||
|
||
void build(int L,int R,int _internal_v=1) /// Build a tree, _internal_v is 1 by default.
|
||
{
|
||
tree[_internal_v].leftbounder=L;
|
||
tree[_internal_v].rightbounder=R;
|
||
tree[_internal_v].bounderlen=R-L+1;
|
||
if(L==R)
|
||
{
|
||
tree[_internal_v].leftvalue=tree[_internal_v].rightvalue=seq[L];
|
||
/** SeqLen of Single Position is 1 , of course*/
|
||
tree[_internal_v].leftseqlen=
|
||
tree[_internal_v].rightseqlen=
|
||
tree[_internal_v].mergedseqlen=1;
|
||
return;
|
||
}
|
||
int mid=(L+R)>>1;
|
||
build(L,mid,_internal_v<<1);
|
||
build(mid+1,R,_internal_v<<1|1);/// x<<1 == x*2; x<<1|1 == x*2+1; (faster == slower)
|
||
|
||
/// Push Up
|
||
pushup(_internal_v);
|
||
}
|
||
|
||
void update(int Pos,int Val,int _internal_v=1)/// Update a position, _internal_v is 1 by default.
|
||
{
|
||
/// Reach a clearly node with same LeftBounder and RightBounder
|
||
if(tree[_internal_v].leftbounder==tree[_internal_v].rightbounder)
|
||
{
|
||
tree[_internal_v].leftvalue=tree[_internal_v].rightvalue=Val;
|
||
return;
|
||
}
|
||
/// Calculate Mid
|
||
int mid=(tree[_internal_v].leftbounder+tree[_internal_v].rightbounder)>>1;
|
||
/// If in left then try update in left
|
||
if(Pos <= mid)
|
||
update(Pos,Val,_internal_v<<1);
|
||
else /// Else try update in right
|
||
update(Pos,Val,_internal_v<<1|1);
|
||
/// And then push it up !
|
||
pushup(_internal_v);
|
||
}
|
||
|
||
int query(int L,int R,int _internal_v=1)
|
||
{
|
||
/// This Node ( and the segment which is under its control )
|
||
/// is included in query area.
|
||
if(L<=tree[_internal_v].leftbounder && tree[_internal_v].rightbounder <= R)
|
||
{
|
||
return tree[_internal_v].mergedseqlen;
|
||
}
|
||
/// Calculate Mid
|
||
int mid=(tree[_internal_v].leftbounder+tree[_internal_v].rightbounder)>>1;
|
||
/// Answer saved in 'ans'
|
||
int ans=0;
|
||
|
||
/// Query If Segment L~R has common area with ThisNode.LeftBounder~Mid
|
||
if(L<=mid)
|
||
{
|
||
ans=max(ans,query(L,R,_internal_v<<1));
|
||
}
|
||
|
||
/// Query If Segment L~R has common area with Mid+1 ~ ThisNode.RightBounder
|
||
if(mid<R)
|
||
{
|
||
ans=max(ans,query(L,R,_internal_v<<1|1));
|
||
}
|
||
|
||
/// Besides these conditions, the following condition is more complex...
|
||
/// If LeftNode.RightValue < RightNode.LeftValue
|
||
/// (looks like Push Up, but why not push up here ?
|
||
/// Is the amount of query action so huge ? )
|
||
if(tree[_internal_v<<1].rightvalue<tree[_internal_v<<1|1].leftvalue)
|
||
{
|
||
/// Here comes the most complex logic.
|
||
/// Answer is the max one of ...
|
||
ans=max(ans,
|
||
/// the minimum one of "Mid - L + 1" (Actually Left Bounder)
|
||
/// and LeftSon.RightSeqLen
|
||
min(mid-L+1,tree[_internal_v<<1].rightseqlen)
|
||
/// and
|
||
+
|
||
/// the minimum one of "R - Mid" (Actually Right Bounder)
|
||
/// and RightSon.LeftSeqLen
|
||
min(R-mid,tree[_internal_v<<1|1].leftseqlen)
|
||
);
|
||
}
|
||
|
||
/// Return ans. Ans is at least 1
|
||
return ans;
|
||
}
|
||
|
||
}/// End of namespace LCISSegmentTree
|
||
|
||
|
||
ACM模板——KMP算法
|
||
#include <string>
|
||
#include <iostream>
|
||
#include <cstring>
|
||
using namespace std;
|
||
|
||
void getfill(string s,int* f)
|
||
{
|
||
//memset(f,0,sizeof(f)); //根据其前一个字母得到
|
||
for(size_t i=1;i<s.size();i++)
|
||
{
|
||
int j=f[i];
|
||
while(j && s[i]!=s[j])
|
||
j=f[j];
|
||
f[i+1]=(s[i]==s[j])?j+1:0;
|
||
}
|
||
}
|
||
|
||
int KMP(string a,string s)
|
||
{
|
||
int* f=new int[s.size()+32];
|
||
memset(f,0,sizeof(int)*s.size());
|
||
getfill(s,f);size_t j=0;
|
||
for(size_t i=0;i<a.size();i++)
|
||
{
|
||
while(j && a[i]!=s[j])
|
||
j=f[j];
|
||
if(a[i]==s[j])
|
||
j++;
|
||
if(j==s.size()){
|
||
delete[] f;return i-s.size()+1;
|
||
}
|
||
}
|
||
delete[] f;
|
||
return -1;
|
||
}
|
||
|
||
KMP (int)
|
||
|
||
注: N为数组T的长度, M为数组P的长度. Next数组长度应稍大于P的长度
|
||
void MakeNext(int* P,int M,int* Next){
|
||
Next[0] = -1;
|
||
int i = 0, j = -1;
|
||
while(i<M){
|
||
if(j==-1||P[i]==P[j]){
|
||
i++,j++;
|
||
if(P[i]!=P[j])Next[i] = j;
|
||
else Next[i] = Next[j];
|
||
}
|
||
else j = Next[j];
|
||
}
|
||
}
|
||
int KMP(int* T,int N,int* P,int M)
|
||
{
|
||
MakeNext(P,M,Next);
|
||
int i=0,j=0;
|
||
while(i<N&&j<M){
|
||
if(T[i]==P[j]||j==-1)i++,j++;
|
||
else j = Next[j];
|
||
}
|
||
if(j==M)return i-M;
|
||
else return -2;
|
||
}
|
||
|
||
ACM模板——区间问题(线段树 RMQ-ST)模板
|
||
找到了一个非常好用的模板，应该主要用于线段树的维护。其中算法部分只需要修改algo_delegate和ValueType即可，极其方便！
|
||
#include <cstdio>
|
||
#include <cstring>
|
||
#include <cmath>
|
||
#include <algorithm>
|
||
#include <functional>
|
||
#define INF 0x3f3f3f3f
|
||
using namespace std;
|
||
const int maxn = 10010;
|
||
struct node
|
||
{
|
||
int lt,rt,v;
|
||
};
|
||
node tree[maxn<<2];
|
||
|
||
/** ========== Conditional =========== */
|
||
using ValueType = int;
|
||
|
||
inline ValueType algo_delegate(ValueType a,ValueType b)
|
||
{
|
||
return min(a,b);
|
||
}
|
||
/****************************************/
|
||
|
||
void build(int lt,int rt,int v)
|
||
{
|
||
tree[v].lt = lt;
|
||
tree[v].rt = rt;
|
||
if(lt == rt)
|
||
{
|
||
scanf("%d",&tree[v].v);
|
||
return;
|
||
}
|
||
int mid = (lt + rt)>>1;
|
||
build(lt,mid,v<<1);
|
||
build(mid+1,rt,v<<1|1);
|
||
tree[v].v = algo_delegate(tree[v<<1].v,tree[v<<1|1].v);
|
||
}
|
||
void update(int p,int val,int v)
|
||
{
|
||
if(tree[v].lt == tree[v].rt)
|
||
{
|
||
tree[v].v = val;
|
||
return;
|
||
}
|
||
if(p <= tree[v<<1].rt) update(p,val,v<<1);
|
||
if(p >= tree[v<<1|1].lt) update(p,val,v<<1|1);
|
||
tree[v].v = algo_delegate(tree[v<<1].v,tree[v<<1|1].v);
|
||
}
|
||
int query(int lt,int rt,int v)
|
||
{
|
||
if(tree[v].lt >= lt && tree[v].rt <= rt)
|
||
return tree[v].v;
|
||
int a = INF,b = INF;
|
||
if(lt <= tree[v<<1].rt) a = query(lt,rt,v<<1);
|
||
if(rt >= tree[v<<1|1].lt) b = query(lt,rt,v<<1|1);
|
||
return algo_delegate(a,b);
|
||
}
|
||
|
||
最长公共上升子序列 LCIS
|
||
/// LCIS 最长公共上升子序列
|
||
namespace LCIS
|
||
{
|
||
|
||
const int MAXLEN_A = 500;
|
||
const int MAXLEN_B = 500;
|
||
int dp[MAXLEN_A+5][MAXLEN_B+5];
|
||
int deal(const char* a,const char* b)
|
||
{
|
||
int lena=strlen(a);
|
||
int lenb=strlen(b);
|
||
for(int i=1;i<=lenb;i++)
|
||
{
|
||
int k=0;
|
||
for(int j=1;j<=lena;j++)
|
||
{
|
||
dp[i][j]=dp[i-1][j];/// when b[i-1] != a[j-1]
|
||
if(b[i-1]>a[j-1]) k=max(k,dp[i-1][j]);
|
||
else if(b[i-1]==a[j-1]) dp[i][j]=k+1;
|
||
}
|
||
}
|
||
int ans=0;
|
||
for(int i=1;i<=lena;i++) ans=max(ans,dp[lenb][i]);
|
||
return ans;
|
||
}
|
||
|
||
}
|
||
//End of namespace LCIS
|
||
|
||
最长公共子序列 LCS (Updated On 20160819)
|
||
/// LCS 最长子序列
|
||
namespace LCS
|
||
{
|
||
const int MAXLEN_A = 512;
|
||
const int MAXLEN_B = 512;
|
||
int dp[MAXLEN_A][MAXLEN_B];
|
||
int deal(const char* a,const char* b)
|
||
{
|
||
int lena=strlen(a);
|
||
int lenb=strlen(b);
|
||
for(int i=0;i<=lenb;i++)
|
||
{
|
||
for(int j=0;j<=lena;j++)
|
||
{
|
||
if(i==0) dp[i][j]=0;
|
||
else if(j==0) dp[i][j]=0;
|
||
else if(b[i-1]==a[j-1])
|
||
{
|
||
dp[i][j]=dp[i-1][j-1]+1;
|
||
}
|
||
else
|
||
{
|
||
dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
|
||
}
|
||
}
|
||
}
|
||
return dp[lenb][lena];
|
||
}
|
||
}//End of namespace LCS
|
||
|
||
最长上升子序列(LIS) 更好的方案（经过比赛检测...） 其中lower_bound来自<vector>
|
||
//最长上升子序列 Longest Increasing Subsequence O(nlogn)
|
||
int b[N];
|
||
int LIS(int a[], int n) {
|
||
int len = 1; b[0] = a[0];
|
||
for (int i = 1; i < n; i++) {
|
||
b[a[i] > b[len - 1] ? len++ : lower_bound(b, b + len, a[i]) - b] = a[i]; //非降换为>=和upper_bound
|
||
}
|
||
return len;
|
||
}
|
||
|
||
Floyd算法是用于求解所有点对之间的最短距离，如果只需要求一个起点到所有其他点的最短距离应该使用Dijstra算法。
|
||
|
||
Floyd核心Logic
|
||
|
||
注：m[f][t] 意为 从f出发到t点的距离. 输入可能是边的形式或者是图的形式，需要灵活处理。
|
||
|
||
for(int k=1;k<=n;k++)
|
||
{
|
||
for(int f=1;f<=n;f++)
|
||
{
|
||
for(int t=1;t<=n;t++)
|
||
{
|
||
if(f==t||f==k||t==k) continue;
|
||
if(m[f][k]!=INF&&m[k][t]!=INF)
|
||
{
|
||
int total=m[f][k]+m[k][t];
|
||
if(total<m[f][t]||m[f][t]==INF)
|
||
{
|
||
m[f][t]=total;
|
||
}
|
||
}
|
||
}
|
||
}
|
||
}
|
||
|
||
ACM模板——快速判断素数
|
||
by Coffee
|
||
//Written by Coffee. 判断素数
|
||
bool isPrime(int num)
|
||
{
|
||
if (num == 2 || num == 3)
|
||
{
|
||
return true;
|
||
}
|
||
if (num % 6 != 1 && num % 6 != 5)
|
||
{
|
||
return false;
|
||
}
|
||
for (int i = 5; i*i <= num; i += 6)
|
||
{
|
||
if (num % i == 0 || num % (i+2) == 0)
|
||
{
|
||
return false;
|
||
}
|
||
}
|
||
return true;
|
||
}
|
||
|
||
|
||
//From Baidu. 快速幂
|
||
int PowerMod(int a, int b, int c)
|
||
{
|
||
int ans = 1;
|
||
a = a % c;
|
||
while(b>0)
|
||
{
|
||
|
||
if(b % 2 == 1)
|
||
ans = (ans * a) % c;
|
||
b = b/2;
|
||
a = (a * a) % c;
|
||
}
|
||
return ans;
|
||
}
|
||
|
||
///约瑟夫问题,n个人,查m个数
|
||
int JosephusProblem_Solution4(int n, int m)
|
||
{
|
||
if(n < 1 || m < 1)
|
||
return -1;
|
||
|
||
vector<int> f(n+1,0);
|
||
for(unsigned i = 2; i <= n; i++)
|
||
f[i] = (f[i-1] + m) % i;
|
||
|
||
return f[n];
|
||
}
|
||
|
||
DP方法求解最长子段,复杂度O(N)
|
||
|
||
不需要求最长子段起终点
|
||
int MaxSum(int n,int *a)
|
||
{
|
||
int sum=NINF,b=0;
|
||
for(int i=0; i<n; i++)
|
||
{
|
||
if(b>0)
|
||
{
|
||
b+=a[i];
|
||
}
|
||
else
|
||
{
|
||
b=a[i];
|
||
}
|
||
if(b>sum)
|
||
{
|
||
sum = b;
|
||
}
|
||
}
|
||
return sum;
|
||
}
|
||
需要求解最长子段起终点
|
||
typedef struct
|
||
{
|
||
int result,start,ends;
|
||
}PACK;
|
||
PACK MaxSum(int N,int* a)
|
||
{
|
||
int sum=-1;
|
||
int tmp=0;
|
||
int start=0;
|
||
int ends=0;
|
||
int tmpstart=0;
|
||
int tmpends=0;
|
||
for(int i=0;i<N;i++)
|
||
{
|
||
if(tmp>0)
|
||
{
|
||
tmp+=a[i];
|
||
tmpends=i;
|
||
}
|
||
else
|
||
{
|
||
tmp=a[i];
|
||
tmpstart=i;
|
||
}
|
||
if(tmp>sum)
|
||
{
|
||
sum=tmp;
|
||
start=tmpstart;
|
||
ends=tmpends;
|
||
}
|
||
}
|
||
if(ends<start)
|
||
{
|
||
ends=start;
|
||
}
|
||
PACK c;
|
||
c.result=sum;
|
||
c.start=start;
|
||
c.ends=ends;
|
||
return c;
|
||
}
|
||
DP方法求最大子矩阵
|
||
#define MAXN 128
|
||
typedef int ARRAY[MAXN][MAXN];
|
||
int MaxSumRect(int m,int n,ARRAY& a)
|
||
{
|
||
int sum = NINF;
|
||
int* b = new int[n+1];
|
||
for(int i=0; i<m; i++)//枚举行
|
||
{
|
||
memset(b,0,sizeof(int)*(n+1));
|
||
|
||
for(int j=i; j<m; j++) //枚举初始行i,结束行j
|
||
{
|
||
for(int k=0; k<n; k++)
|
||
{
|
||
b[k] += a[j][k];//b[k]为纵向列之和
|
||
}
|
||
int max = MaxSum(n,b);
|
||
if(max>sum)
|
||
{
|
||
sum = max;
|
||
}
|
||
|
||
}
|
||
}
|
||
delete[] b;
|
||
return sum;
|
||
}
|
||
|
||
正方形矩阵求解
|
||
int MaxSumSquare(int N,ARRAY& a)
|
||
{
|
||
return MaxSumRect(N,N,a);
|
||
}
|
||
|
||
无穷常数
|
||
const int INF = 0x3f3f3f3f;
|
||
const int NINF = 0xc0c0c0c0;
|
||
|
||
#include <iostream>
|
||
#include <algorithm>
|
||
using namespace std;
|
||
|
||
其他代码(GitHub: kiritow/OJ-Problems-Source)
|
||
|
||
|
||
namespace RMQ_ST
|
||
{
|
||
const int MAXN=10000;
|
||
int f[MAXN][MAXN];
|
||
int a[MAXN];
|
||
int n;
|
||
void init()
|
||
{
|
||
for(int i = 1;i<=n;i++)
|
||
{
|
||
f[i][0]=a[i];
|
||
}
|
||
for(int j=1;(1<<j)<=n;j++)
|
||
{
|
||
for(int i=1;i+(i<<j)-1<=n;i++)
|
||
{
|
||
f[i][j]=max(f[i][j-1],f[i+(1<<j-1)][j-1]);
|
||
}
|
||
}
|
||
}
|
||
|
||
int rmq(int L,int R)
|
||
{
|
||
int k=0;
|
||
while((1<<(k+1)<=R-L+1)) k++;
|
||
return max(f[L][k],f[R-(1<<k)+1][k]);
|
||
}
|
||
|
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}/// End of namespace RMQ_ST
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