边双连通分量(e-DCC)的求法+e-DCC缩点

#include <bits/stdc++.h>

using namespace std;
const int SIZE = 1e5 + 10;
int head[SIZE], ver[SIZE * 2], Next[SIZE * 2];
int dfn[SIZE], low[SIZE], n, m, tot, num;
bool bridge[SIZE * 2];
int c[SIZE], dcc;
int hc[SIZE], vc[SIZE * 2], nc[SIZE * 2], tc;

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];
        if (!dfn[y]) {
            tarjan(y, i);
            low[x] = min(low[x], low[y]);
            if (low[y] > dfn[x])
                bridge[i] = bridge[i ^ 1] = true;
        } 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 = 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]);
    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("缩点之后的森林，点数%d，边数%d（可能有重边）\n", dcc, tc / 2);
    for (int i = 2; i < tc; i += 2)
        printf("%d %d\n", vc[i ^ 1], vc[i]);
    return 0;
}

input:
9 11
1 2
2 3
3 4
4 5
5 1
5 2
1 6
6 9
6 8
9 8
6 7
output:
There are 3 e-DCCs.
1 belongs to DCC 1
2 belongs to DCC 1
3 belongs to DCC 1
4 belongs to DCC 1
5 belongs to DCC 1
6 belongs to DCC 2
7 belongs to DCC 3
8 belongs to DCC 2
9 belongs to DCC 2
缩点之后的森林，点数3，边数2（可能有重边）
1 2
2 3