[USACO06JAN]牛的舞会 Tarjan模版

[USACO06JAN]牛的舞会

只要找强联通分量大于一的个数就行

#include <bits/stdc++.h>
using namespace std;
const int N = 100000;
struct E{int v,nxt;};
E edge[N];
int dfn[N],low[N],cnt,visnum,num[N],belong[N],head[N];
int n,m,vis[N];
stack<int> q;
void tarjan(int u){
	dfn[u] = low[u] = ++visnum;
	vis[u] = 1;
	q.push(u);
	for(int v,i=head[u];i;i=edge[i].nxt){
		v = edge[i].v;
		if(!vis[v]){
			tarjin(v);
			low[u] = min(low[u],low[v]);
		}else low[u] = min(low[u],dfn[v]);
	}
	if(dfn[u]==low[u]){
		++cnt;
		int v;
		do{
			v = q.top();q.pop();
			belong[v] = cnt;
			num[cnt]++;
			vis[v] = 0;
		}while(u!=v);//回溯 
	}
}
int main()
{
	scanf("%d %d",&n,&m);
	for(int i=1;i<=m;i++){
		int u,v;
		scanf("%d %d",&u,&v);
		edge[i].v = v; edge[i].nxt = head[u]; head[u] = i;
	}
	for(int i=1;i<=n;i++){
		if(!dfn[i]){
			tarjin(i);
		}
	}
	int ans=0;
	for(int i=1;i<=cnt;i++){
		if(num[i]>1)ans++;
	}printf("%d\n",ans);
	return 0;
 }