割点——tarjan
1 #include <bits/stdc++.h>
2 using namespace std;
3
4 const int MAXN = 20001;
5 const int MAXM = 100001;
6 int n, m;
7 int ans;//个数
8
9
10 int head[MAXN], cnt, v[2 * MAXM], nxt[2 * MAXM];
11 void add(int x, int y) {
12 nxt[++cnt] = head[x];
13 head[x] = cnt;
14 v[cnt] = y;
15 }
16
17
18 int dfn[MAXN], low[MAXN], ind;
19 int cut[MAXN];
20
21
22 void tarjan(int now, int fa) {
23 dfn[now] = low[now] = ++ind;
24 int child = 0;
25 for (int i = head[now]; i ; i = nxt[i]) {
26 if(!dfn[v[i]]) {
27 tarjan(v[i], now);
28 low[now] = min(low[now], low[v[i]]);
29
30 if(low[v[i]] >= dfn[now] && now != fa) cut[now] = 1;//不为根节点
31 if(now == fa) child++; //为根节点，子树统计
32 }
33 else {
34 low[now] = min(low[now], dfn[v[i]]);
35 }
36 }
37 if(now == fa && child >= 2) cut[now] = 1;//为根节点且有两个及以上的子节点
38 }
39 int main() {
40
41 scanf("%d%d", &n, &m);
42 for (int i = 1; i <= m; i++) {
43 int x, y;
44 scanf("%d%d", &x, &y);
45 add(x, y);
46 add(y, x);
47 }
48
49 for (int i = 1; i <= n; i++)
50 if(!dfn[i])
51 tarjan(i, i);
52
53 for (int i = 1; i <= n; i++)
54 if(cut[i] == 1)
55 ans++;
56 printf("%d\n", ans);
57 for (int i = 1; i <= n; i++)
58 if(cut[i] == 1)
59 printf("%d ", i);
60
61
62 return 0;
63 }
