割点：对于一个连通图来说，如果删除某个点之后图不再连通，这个点就称为割点
割点算法
时间复杂度：O(N+M)
但是下面给出的算法时间复杂度为O(N^2)，这是因为下面的存储结构都是邻接矩阵，这样的话用该算法就完全失去了意义，毕竟如果使用邻接矩阵完全可以通过删除一个点，然后进行DFS/BFS遍历即可了。所以在实际应用中还是要使用邻接表来保存数据，这样时间复杂度可以达到O(N+M)
Input:
6 7
1 4
1 3
4 2
3 2
2 5
2 6
5 6

Output:
2

import java.util.Scanner;
public class cutPoint {static int n, m;static int[][] e = new int[9][9];static int[] num = new int[9];static int[] low = new int[9];static int[] flag = new int[9];static int index,root;static Scanner input = new Scanner(System.in);public static void main(String[] args) {n = input.nextInt();m = input.nextInt();for (int i = 1; i <= n; i++) {for (int j = 1; j <= n; j++) {e[i][j] = 0;}}for (int i = 1; i <= m; i++) {int x = input.nextInt();int y = input.nextInt();e[x][y] = 1;e[y][x] = 1;}root = 1;dfs(1, root);for (int i = 1; i <= n; i++) {if (flag[i] == 1) {System.out.print(i + " ");}}}private static void dfs(int cur, int father) {int child = 0;index++;num[cur] = index;low[cur] = index;for (int i = 1; i <= n; i++) {if (e[cur][i] == 1) {if (num[i] == 0) {child++;dfs(i, cur);low[cur] = Math.min(low[cur], low[i]);if (cur != root && low[i] >= num[cur]) {flag[cur] = 1;}if (cur == root && child == 2) {flag[cur] = 1;}} else if (i != father){low[cur] = Math.min(low[cur], num[i]);}}}}
}

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割边：对于一个连通图来说，如果删除某个边之后图不再连通，这个点就称为割边
割边算法
时间复杂度：O(N+M)
同割点算法一样，在实际应用中还是要使用邻接表来保存数据，这样时间复杂度可以达到O(N+M)
Input:
6 7
1 4
1 3
4 2
3 2
2 5
2 6
5 6

Output:
5-6
2-5

import java.util.Scanner;
public class cutEdge {static int n, m;static int[][] e = new int[9][9];static int[] num = new int[9];static int[] low = new int[9];static int index,root;static Scanner input = new Scanner(System.in);public static void main(String[] args) {n = input.nextInt();m = input.nextInt();for (int i = 1; i <= n; i++) {for (int j = 1; j <= n; j++) {e[i][j] = 0;}}for (int i = 1; i <= m; i++) {int x = input.nextInt();int y = input.nextInt();e[x][y] = 1;e[y][x] = 1;}root = 1;dfs(1, root);}private static void dfs(int cur, int father) {index++;num[cur] = index;low[cur] = index;for (int i = 1; i <= n; i++) {if (e[cur][i] == 1) {if (num[i] == 0) {dfs(i, cur);low[cur] = Math.min(low[i], low[cur]);if (low[i] > num[cur]) {System.out.println(cur + "-" + i);}} else if (i != father) {low[cur] = Math.min(low[cur], num[i]);}}}}
}