55 lines
1.6 KiB
C++
55 lines
1.6 KiB
C++
#include <bits/stdc++.h>
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using namespace std;
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const int N = 2e4 + 10, M = 2e5 + 10;
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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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int n, m; // n个顶点,m条边
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int dfn[N], low[N], timestamp, root; // tarjan算法
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vector<int> v; //割点集合
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//割点
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void tarjan(int u) {
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low[u] = dfn[u] = ++timestamp;
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int son = 0;
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for (int i = h[u]; ~i; i = ne[i]) {
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int j = e[i];
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if (!dfn[j]) {
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son++;
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tarjan(j);
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low[u] = min(low[u], low[j]);
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//根据割点分析，low[j]>=dfn[u]的时候，u是割点
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//由于u可能有多个子节点，都会产生回边，u可能进入多次，后面需要去重
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if (u != root && low[j] >= dfn[u]) v.push_back(u);
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}
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low[u] = min(low[u], dfn[j]);
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}
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//特例：如果u是根，并且有两个以上的儿子节点，则u为割点
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if (u == root && son > 1) v.push_back(u);
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}
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int main() {
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scanf("%d %d", &n, &m);
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memset(h, -1, sizeof h);
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for (int i = 1; i <= m; i++) {
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int a, b;
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scanf("%d %d", &a, &b);
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add(a, b), add(b, a);
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}
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for (root = 1; root <= n; root++)
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if (!dfn[root]) tarjan(root);
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//排序+去重
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sort(v.begin(), v.end());
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v.erase(unique(v.begin(), v.end()), v.end());
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//输出割点个数
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printf("%d\n", v.size());
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//由小到大输出割点
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for (int i : v) printf("%d ", i);
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return 0;
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} |