poj 3180 The Cow Prom 强接通分量

poj 3180 The Cow Prom 强连通分量

水题，直接贴代码。

//poj 3180
//sep9
#include <iostream>
#include <stack>
using namespace std;
const int maxN=10024;
const int maxM=50048;
int sum[maxN];
int head[maxN],dfn[maxN],low[maxN],ins[maxN],belong[maxN]; 
stack<int> s;
struct Edge{
	int v,next;
}edge[maxM];
int n,m,e,t,cnt;

void tarjan(int u){
	dfn[u]=low[u]=++t;
	s.push(u);
	ins[u]=1;
	for(int i=head[u];i!=-1;i=edge[i].next){
		int v=edge[i].v;
		if(dfn[v]==0){
			tarjan(v);
			low[u]=min(low[u],low[v]);
		}else if(ins[v]==1)
			low[u]=min(low[u],dfn[v]);		
	}
	int k;
	if(low[u]==dfn[u]){
		++cnt;
		do{
			k=s.top();
			s.pop();
			ins[k]=0;	
			belong[k]=cnt;
		}while(low[k]!=dfn[k]);	
	}	
	return ;		
}

int main()
{
	int n,m;
	scanf("%d%d",&n,&m);
 	e=0;  
  	memset(head,-1,sizeof(head));  
   	memset(dfn,0,sizeof(dfn));  
    memset(ins,0,sizeof(ins));  
	while(m--){
		int a,b;
		scanf("%d%d",&a,&b);
		edge[e].v=b;edge[e].next=head[a];head[a]=e++;
	}
	t=cnt=0;
	for(int i=0;i<n;++i)
		if(!dfn[i])
			tarjan(i);   
	memset(sum,0,sizeof(sum));
	for(int i=0;i<n;++i)
		++sum[belong[i]];
	int ans=0;
	for(int i=1;i<=cnt;++i)
		if(sum[i]>1)
			++ans;
	printf("%d",ans);
	return 0;	
}