24分求调必关

#include<bits/stdc++.h>
using namespace std;
const int N=1e4+10,M=1e5+10;
int n,m,dfn[N],low[N],tot=0,stk[N],instk[N],top=0,scc[N],siz[N],cnt;
vector<int>e[N];
void tarjan(int x){
	dfn[x]=low[x]=++tot;
	stk[++top]=x;
	instk[x]=1;
	for(auto&y:e[x]){
		if(!dfn[y]){
			tarjan(y);
			low[x]=min(low[x],dfn[y]);
		}else if(instk[y]){
			low[x]=min(low[x],dfn[y]);
		}
	}
	if(dfn[x]==low[x]){
		int y;
		++cnt;
		do{
			y=stk[top--];
			instk[y]=0;
			scc[y]=cnt;
			++siz[cnt];
		}while(y!=x);
	}
} 
int main(){
	cin>>n>>m;
	for(int i=1;i<=m;i++){
		int a,b;
		cin>>a>>b;
		e[a].push_back(b);
	}
	for(int i=1;i<=n;i++){
		if(!dfn[i]){
			tarjan(i);
		}
	}
	int ans=0;
	for(int i=1;i<=cnt;i++){
		if(siz[i]>1){
			ans++;
		}
	}
	cout<<ans<<endl;
	return 0;
}
//24分求调必关