264 lines
7.5 KiB
Markdown
264 lines
7.5 KiB
Markdown
<!-- GFM-TOC -->
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* [二分图](#二分图)
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* [1. 判断是否为二分图](#1-判断是否为二分图)
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* [拓扑排序](#拓扑排序)
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* [1. 课程安排的合法性](#1-课程安排的合法性)
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* [2. 课程安排的顺序](#2-课程安排的顺序)
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* [并查集](#并查集)
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* [1. 冗余连接](#1-冗余连接)
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<!-- GFM-TOC -->
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# 二分图
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如果可以用两种颜色对图中的节点进行着色，并且保证相邻的节点颜色不同，那么这个图就是二分图。
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## 1. 判断是否为二分图
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[785. Is Graph Bipartite? (Medium)](https://leetcode.com/problems/is-graph-bipartite/description/)
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```html
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Input: [[1,3], [0,2], [1,3], [0,2]]
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Output: true
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Explanation:
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The graph looks like this:
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0----1
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3----2
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We can divide the vertices into two groups: {0, 2} and {1, 3}.
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```
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```html
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Example 2:
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Input: [[1,2,3], [0,2], [0,1,3], [0,2]]
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Output: false
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Explanation:
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The graph looks like this:
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0----1
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| \ |
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| \ |
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3----2
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We cannot find a way to divide the set of nodes into two independent subsets.
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```
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```java
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public boolean isBipartite(int[][] graph) {
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int[] colors = new int[graph.length];
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Arrays.fill(colors, -1);
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for (int i = 0; i < graph.length; i++) { // 处理图不是连通的情况
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if (colors[i] == -1 && !isBipartite(i, 0, colors, graph)) {
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return false;
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}
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}
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return true;
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}
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private boolean isBipartite(int curNode, int curColor, int[] colors, int[][] graph) {
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if (colors[curNode] != -1) {
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return colors[curNode] == curColor;
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}
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colors[curNode] = curColor;
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for (int nextNode : graph[curNode]) {
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if (!isBipartite(nextNode, 1 - curColor, colors, graph)) {
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return false;
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}
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}
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return true;
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}
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```
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# 拓扑排序
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常用于在具有先序关系的任务规划中。
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## 1. 课程安排的合法性
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[207. Course Schedule (Medium)](https://leetcode.com/problems/course-schedule/description/)
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```html
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2, [[1,0]]
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return true
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```
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```html
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2, [[1,0],[0,1]]
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return false
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```
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题目描述：一个课程可能会先修课程，判断给定的先修课程规定是否合法。
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本题不需要使用拓扑排序，只需要检测有向图是否存在环即可。
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```java
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public boolean canFinish(int numCourses, int[][] prerequisites) {
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List<Integer>[] graphic = new List[numCourses];
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for (int i = 0; i < numCourses; i++) {
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graphic[i] = new ArrayList<>();
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}
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for (int[] pre : prerequisites) {
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graphic[pre[0]].add(pre[1]);
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}
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boolean[] globalMarked = new boolean[numCourses];
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boolean[] localMarked = new boolean[numCourses];
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for (int i = 0; i < numCourses; i++) {
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if (hasCycle(globalMarked, localMarked, graphic, i)) {
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return false;
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}
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}
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return true;
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}
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private boolean hasCycle(boolean[] globalMarked, boolean[] localMarked,
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List<Integer>[] graphic, int curNode) {
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if (localMarked[curNode]) {
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return true;
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}
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if (globalMarked[curNode]) {
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return false;
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}
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globalMarked[curNode] = true;
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localMarked[curNode] = true;
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for (int nextNode : graphic[curNode]) {
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if (hasCycle(globalMarked, localMarked, graphic, nextNode)) {
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return true;
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}
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}
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localMarked[curNode] = false;
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return false;
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}
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```
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## 2. 课程安排的顺序
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[210. Course Schedule II (Medium)](https://leetcode.com/problems/course-schedule-ii/description/)
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```html
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4, [[1,0],[2,0],[3,1],[3,2]]
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There are a total of 4 courses to take. To take course 3 you should have finished both courses 1 and 2. Both courses 1 and 2 should be taken after you finished course 0. So one correct course order is [0,1,2,3]. Another correct ordering is[0,2,1,3].
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```
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使用 DFS 来实现拓扑排序，使用一个栈存储后序遍历结果，这个栈的逆序结果就是拓扑排序结果。
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证明：对于任何先序关系：v->w，后序遍历结果可以保证 w 先进入栈中，因此栈的逆序结果中 v 会在 w 之前。
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```java
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public int[] findOrder(int numCourses, int[][] prerequisites) {
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List<Integer>[] graphic = new List[numCourses];
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for (int i = 0; i < numCourses; i++) {
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graphic[i] = new ArrayList<>();
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}
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for (int[] pre : prerequisites) {
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graphic[pre[0]].add(pre[1]);
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}
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Stack<Integer> postOrder = new Stack<>();
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boolean[] globalMarked = new boolean[numCourses];
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boolean[] localMarked = new boolean[numCourses];
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for (int i = 0; i < numCourses; i++) {
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if (hasCycle(globalMarked, localMarked, graphic, i, postOrder)) {
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return new int[0];
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}
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}
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int[] orders = new int[numCourses];
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for (int i = numCourses - 1; i >= 0; i--) {
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orders[i] = postOrder.pop();
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}
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return orders;
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}
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private boolean hasCycle(boolean[] globalMarked, boolean[] localMarked, List<Integer>[] graphic,
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int curNode, Stack<Integer> postOrder) {
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if (localMarked[curNode]) {
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return true;
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}
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if (globalMarked[curNode]) {
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return false;
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}
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globalMarked[curNode] = true;
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localMarked[curNode] = true;
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for (int nextNode : graphic[curNode]) {
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if (hasCycle(globalMarked, localMarked, graphic, nextNode, postOrder)) {
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return true;
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}
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}
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localMarked[curNode] = false;
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postOrder.push(curNode);
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return false;
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}
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```
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# 并查集
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并查集可以动态地连通两个点，并且可以非常快速地判断两个点是否连通。
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## 1. 冗余连接
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[684. Redundant Connection (Medium)](https://leetcode.com/problems/redundant-connection/description/)
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```html
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Input: [[1,2], [1,3], [2,3]]
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Output: [2,3]
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Explanation: The given undirected graph will be like this:
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1
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/ \
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2 - 3
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```
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题目描述：有一系列的边连成的图，找出一条边，移除它之后该图能够成为一棵树。
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```java
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public int[] findRedundantConnection(int[][] edges) {
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int N = edges.length;
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UF uf = new UF(N);
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for (int[] e : edges) {
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int u = e[0], v = e[1];
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if (uf.connect(u, v)) {
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return e;
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}
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uf.union(u, v);
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}
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return new int[]{-1, -1};
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}
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private class UF {
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private int[] id;
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UF(int N) {
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id = new int[N + 1];
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for (int i = 0; i < id.length; i++) {
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id[i] = i;
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}
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}
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void union(int u, int v) {
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int uID = find(u);
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int vID = find(v);
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if (uID == vID) {
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return;
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}
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for (int i = 0; i < id.length; i++) {
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if (id[i] == uID) {
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id[i] = vID;
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}
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}
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}
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int find(int p) {
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return id[p];
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}
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boolean connect(int u, int v) {
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return find(u) == find(v);
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}
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}
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```
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