[HAOI2006] Popular cattle, Luoguzhi's improved experience, strong connected component

The subject

      [HAOI2006] Popular cow

      In fact, this question is to find the size of the ring with an in-degree of 0 after the contraction point.

      After we run Tarjan for a short time, we can record the number and size of the ring where each point is located.

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<stack>
#include<vector>
using namespace std;

int n,m;
int first[10010];
struct edge{
	int x,y,next;
}s[100010];
struct node{
	int dfn,low;
}on[10010];
int len ​​= 0;
bool we[10010];
bool tf[10010];
int where[10010];
int diss[10010];
int node=0;
int tot=0,now=0;
vector<int> T[10010];
stack<int> f;

void ins(int x,int y){
	len ++;
	s[len].x=x;s[len].y=y;s[len].next=first[x];first[x]=len;
}

void Tarjan(int x){
	tf[x]=true;f.push(x);
	op[x].dfn=op[x].low=++now;
	for(int i=first[x];i!=0;i=s[i].next){
		int y=s[i].y;
		if(op[y].dfn==0){
			Tarjan (y);
			if (on [y] .low <on [x] .low) on [x] .low = on [y] .low;
		}
		else if(tf[y])
			if(op[y].dfn<op[x].low) op[x].low=op[y].dfn;
	}
	if(op[x].dfn==op[x].low){
		to++;
		while(1){
			int k=f.top();
			f.pop();
			T[tot].push_back(k);
			where[k]=tot;
			tf[k]=false;
			if(k==x) break;
		}
	}
}

int main(){
	scanf("%d %d",&n,&m);
	for(int i=1;i<=m;i++){
		int x,y;
		scanf("%d %d",&x,&y);
		ins(x,y);
	}
	for(int i=1;i<=n;i++)
		if(op[i].dfn==0) Tarjan(i);
	for(int i=1;i<=m;i++){
		int x=s[i].x,y=s[i].y;
		if(where[x]!=where[y])
			diss[where[x]]++;
	}
	int ans=0;
	for(int i=1;i<=tot;i++){
		if(diss[i]==0){
			if(ans==0) ans=T[i].size();
			else {
				printf("0");
				return 0;	
			}	
		}
	}
	printf("%d\n",ans);
}