输入输出样例

输入 #1复制

6 7
1 2
1 3
1 4
2 5
3 5
4 5
5 6

输出 #1复制

1 
5

说明/提示

对于全部数据，n \le 20000n≤20000，m \le 100000m≤100000

点的编号均大于00小于等于nn。

tarjan图不一定联通。

一道tarjan求割点的裸题

#include<bits/stdc++.h>
using namespace std;
const int maxn = 3e6 + 10;
int head[maxn];
int dfn[maxn];
int low[maxn];
bool vis[maxn];
int tot;
int dex;
int cnt = 0;
int n, m;

struct node{
	int to;
	int next;
	node() {}
	node(int a, int b) : to(a), next(b) {}
}edge[maxn];
 
void edgeadd(int a, int b){
	edge[tot] = node(b, head[a]);
	head[a] = tot++;
}
 
void init(){
	memset(head, -1, sizeof(head));
	memset(vis, 0, sizeof(vis));
	memset(dfn, 0, sizeof(dfn));
	tot = 0;
	dex = 0;
}
set<int> t;

void tarjan(int u, int rt){
	dfn[u] = low[u] = ++dex;
	int ans = 0;
	for(int i = head[u]; i != -1; i = edge[i].next){
		int v = edge[i].to;
		if(dfn[v] == 0){
			ans++;
			tarjan(v, rt);
			low[u] = min(low[u], low[v]);
			if(u != rt && low[v] >= dfn[u])
				t.insert(u);
		} 
		low[u] = min(low[u], dfn[v]);
	}
	if(u == rt && ans >= 2)
		t.insert(u);
}
 
 
int main(){
	ios::sync_with_stdio(false);
	cin >> n >> m;
	init();
	int x, y;
	for(int i = 1; i <= m; i++){
		cin >> x >> y;
		edgeadd(x, y);
		edgeadd(y, x);
	}
	for(int i = 1; i <= n; i++){
		if(!dfn[i])
			tarjan(i, i);
	}
	int r;
	r =  t.size();
	cout << r << endl;
	set<int>::iterator it;            
	for(it = t.begin(); it != t.end(); it++)  //使用迭代器进行遍历 
	{
		cout << *it << " ";
	}
	cout << endl;
	return 0;
}