T103489 【模板】边双连通分量

题目传送门

知识点

• 无向边双连通分量模板，注意边是无向边，需要开双倍\(M\)
• 没了...

#include <bits/stdc++.h>

using namespace std;
const int N = 5 * 1e4 + 10, M = 6 * 1e5 + 10;

int n, m;
int h[N], e[M], ne[M], idx;
int dfn[N], low[N], timestamp;
int stk[N], top;
int id[N], dcc_cnt;
bool is_bridge[M];
int d[N];

void add(int a, int b) {
    e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}
//无向图边双连通性模板,缩点
void tarjan(int u, int from) {
    dfn[u] = low[u] = ++timestamp;
    stk[++top] = u;
    for (int i = h[u]; ~i; i = ne[i]) {
        int j = e[i];
        if (!dfn[j]) {
            tarjan(j, i);
            low[u] = min(low[u], low[j]);
            if (dfn[u] < low[j]) is_bridge[i] = is_bridge[i ^ 1] = true;
        } else if (i != (from ^ 1))
            low[u] = min(low[u], dfn[j]);
    }
    if (dfn[u] == low[u]) {
        ++dcc_cnt;
        int y;
        do {
            y = stk[top--];
            id[y] = dcc_cnt;
        } while (y != u);
    }
}

int main() {
    cin >> n >> m;
    memset(h, -1, sizeof h);
    for (int i = 0; i < m; i++) {
        int a, b;
        cin >> a >> b;
        add(a, b), add(b, a);
    }
    for (int i = 1; i <= n; i++)
        if (!dfn[i]) tarjan(i, -1);

    printf("%d\n", dcc_cnt);
    return 0;
}