割点算法

时间复杂度：O(N+M) 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 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

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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]);
                }
            }
        }
    }
}

割边算法

时间复杂度：O(N+M) 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 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

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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]);
                }
            }
        }
    }
}