点双连通分量（V-DCC）

原理

见《进阶指南》第402页。

模板题

洛谷-T103492-【模板】点双连通分量

#include <iostream>
#include <cstring>
#include <stack>
#include <vector>
using namespace std;

const int N = 50010, M = 300010;
struct edge {
    int to, next;
};
edge e[2 * M];
int idx, head[N];
int n, m, root;
int cnt, dfn[N], low[N];
bool cut[N];
stack<int> s;
int tot;
vector<int> dcc[N];

void add_edge (int u, int v) {
    e[idx].to = v;
    e[idx].next = head[u];
    head[u] = idx ++;
}
void tarjan (int cur) {
    low[cur] = dfn[cur] = ++ cnt;
    s.push(cur);
    if (cur == root && head[cur] == -1) { // 孤立点
        dcc[++ tot].push_back(cur);
        return;
    }
    int sum = 0;
    for (int i = head[cur]; i != -1; i = e[i].next) {
        int to = e[i].to;
        if (dfn[to] == 0) {
            tarjan(to);
            low[cur] = min(low[cur], low[to]);

            if (low[to] >= dfn[cur]) {
                ++ sum;
                if (cur != root || sum >= 2)
                    cut[cur] = true;
                ++ tot;
                int t;
                do {
                    t = s.top();
                    s.pop();
                    dcc[tot].push_back(t);
                } while (t != to);
                dcc[tot].push_back(cur);
            }
        } else {
            low[cur] = min(low[cur], dfn[to]);
        }
    }
}

int main () {
    memset(head, -1, sizeof(head));
    scanf("%d%d", &n, &m);
    for (int i = 1, u, v; i <= m; ++ i) {
        scanf("%d%d", &u, &v);
        add_edge(u, v);
        add_edge(v, u);
    }

    for (int i = 1; i <= n; ++ i) {
        if (dfn[i] == 0) {
            root = i;
            tarjan(i);
        }
    }
    for (int i = 1; i <= tot; ++ i) {
        for (int j = 0; j < dcc[i].size(); ++ j)
            printf("%d ", dcc[i][j]);
        printf("\n");
    }
    return 0;
}