65 lines
2.3 KiB
C++
65 lines
2.3 KiB
C++
#include <bits/stdc++.h>
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using namespace std;
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const int N = 5e5 + 10, M = 4e6 + 10;
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int n, m;
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int dfn[N], low[N], ts, root, stk[N], top;
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vector<int> dcc[N];
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int dcc_cnt;
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//链式前向星
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int e[M], h[N], idx, w[M], ne[M];
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void add(int a, int b, int c = 0) {
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e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
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}
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//点双连通分量 模板
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void tarjan(int u) {
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dfn[u] = low[u] = ++ts; //分配时间戳
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stk[++top] = u; // u入栈
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if (u == root && h[u] == -1) { // 根，而且没有出边，孤立点
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dcc[++dcc_cnt].push_back(u); // 分量数++
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return;
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}
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for (int i = h[u]; ~i; i = ne[i]) { //枚举u的每条边
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int j = e[i];
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if (!dfn[j]) { //如果j没有被走过
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tarjan(j); // dfs
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low[u] = min(low[u], low[j]); //用儿子的low[j]尝试更新low[u]
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if (low[j] >= dfn[u]) { //如果u是割点
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dcc_cnt++; //连通块增加
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int x; //利用栈，维护此连通块中所有的点
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do {
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x = stk[top--];
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dcc[dcc_cnt].push_back(x);
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} while (x != j);
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dcc[dcc_cnt].push_back(u); // u是割点，也必须属于其它连通块，不能用了拉倒，需要再次入栈
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}
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} else
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low[u] = min(low[u], dfn[j]);
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}
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}
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int main() {
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memset(h, -1, sizeof h);
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scanf("%d %d", &n, &m);
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for (int i = 0; i < m; i++) {
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int a, b;
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scanf("%d %d", &a, &b);
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if (a != b) add(a, b), add(b, a); //此题目需要判断一下自环，否则1和11会挂
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}
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//给定一个n个点m条边的无向图，求该图中的所有点双连通分量(v-dcc)
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for (root = 1; root <= n; root++) //每个点做为根结点进行枚举
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if (!dfn[root]) tarjan(root);
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// printf("%d\n", dcc_cnt);
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for (int i = 1; i <= dcc_cnt; i++) { //枚举每个双连通分量
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// printf("%d ", dcc[i].size());
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for (int j = 0; j < dcc[i].size(); j++) //输出双连通分量中的点有哪些
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printf("%d ", dcc[i][j]);
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puts("");
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}
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return 0;
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} |