tarjan算法求無向圖的橋、邊雙連通份量並縮點

 

// tarjan算法求無向圖的橋、邊雙連通份量並縮點
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
const int SIZE = 100010;
int head[SIZE], ver[SIZE * 2], Next[SIZE * 2];
int dfn[SIZE], low[SIZE], c[SIZE];
int n, m, tot, num, dcc, tc;
bool bridge[SIZE * 2];
int hc[SIZE], vc[SIZE * 2], nc[SIZE * 2];

void add(int x, int y) {
    ver[++tot] = y, Next[tot] = head[x], head[x] = tot;
}

void add_c(int x, int y) {
    vc[++tc] = y, nc[tc] = hc[x], hc[x] = tc;
}

void tarjan(int x, int in_edge) {
    dfn[x] = low[x] = ++num;
    for (int i = head[x]; i; i = Next[i]) {
        int y = ver[i];
        //當前點未走過
        if (!dfn[y]) {
            tarjan(y, i);
            low[x] = min(low[x], low[y]);
            if (low[y] > dfn[x])
                //i 與 i^1是橋
                bridge[i] = bridge[i ^ 1] = true;
        }
        //反向邊更新
        else if (i != (in_edge ^ 1))
            low[x] = min(low[x], dfn[y]);
    }
}

void dfs(int x) {
    c[x] = dcc;
    for (int i = head[x]; i; i = Next[i]) {
        int y = ver[i];
        if (c[y] || bridge[i]) continue;
        dfs(y);
    }
}

int main() {
    cin >> n >> m;
    tot = 1;
    for (int i = 1; i <= m; i++) {
        int x, y;
        scanf("%d%d", &x, &y);
        add(x, y), add(y, x);
    }
    for (int i = 1; i <= n; i++)
        if (!dfn[i]) tarjan(i, 0);
    for (int i = 2; i < tot; i += 2)
        if (bridge[i])//當前橋，輸出連通橋的兩點
            printf("%d %d\n", ver[i ^ 1], ver[i]);
        //求邊雙連通份量（不存在橋）
    for (int i = 1; i <= n; i++)
        if (!c[i]) {
            ++dcc;
            dfs(i);
        }
    printf("There are %d e-DCCs.\n", dcc);
    for (int i = 1; i <= n; i++)
        printf("%d belongs to DCC %d.\n", i, c[i]);

    //縮點
    tc = 1;
    for (int i = 2; i <= tot; i++) {
        int x = ver[i ^ 1], y = ver[i];
        if (c[x] == c[y]) continue;
        add_c(c[x], c[y]);
    }
    printf("縮點以後的森林，點數 %d，邊數 %d\n", dcc, tc / 2);
    for (int i = 2; i < tc; i += 2)
        printf("%d %d\n", vc[i ^ 1], vc[i]);
}