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题目大意：题目大意是：在一个牧群中，有N个奶牛，给定M对关系（A,B）表示A仰慕B，而且仰慕关系有传递性，问被所有奶牛（除了自己）仰慕的奶牛个数
解题思路：找出所有的连通分量，如果只有一个连通分量的出度为0，那么输出那个连通分量的点的个数即可,如果不唯一就输出0
因为连通分量的出度有两个的话，那么肯定至少存在一头牛不仰慕另外一头牛，所以我们至少保证要出度为0的连通分量唯一

/*
tarjan
Memory 872K
Time   32MS
*/
#include 
using namespace std;
#define MAXM 50010
#define MAXV 10010
#define min(a,b) (a>b?b:a)
 
typedef struct{
	int s,t,next;
}Edge;
 
Edge edge[MAXM];
int n,m,headlist[MAXV];
 
int dfn[MAXV];									//第一次访问的步数
int low[MAXV];									//子树中最早的步数
int stap[MAXV],stop;							//模拟栈
bool instack[MAXV];								//是否在栈中
int count;										//记录连通分量的个数
int cnt;										//记录搜索步数
int belong[MAXV];								//属于哪个连通分量
 
void init(){
	count=stop=cnt=0;
	memset(instack,false,sizeof(instack));
	memset(dfn,0,sizeof(dfn));
}
 
void tarjan(int x){
	int i;
	dfn[x]=low[x]=++cnt;
	stap[stop++]=x;
	instack[x]=true;
	for(i=headlist[x];i!=-1;i=edge[i].next){
		int a=edge[i].t;
		if(!dfn[a]){
			tarjan(a);
			low[x]=min(low[a],low[x]);
		}else if(instack[a])
			low[x]=min(dfn[a],low[x]);
	}
	
	if(low[x]==dfn[x]){
		count++;
		while(1){
			int tmp=stap[--stop];
			belong[tmp]=count;
			instack[tmp]=false;
			if(tmp==x) break;
		}
	}
}
 
void work(){
	init();
	for(int i=1;i<=n;i++)
		if(!dfn[i]) tarjan(i);
}
 
void output(){
	int i,outdegree[MAXV]={0},outzero=0,ans=0;
	for(i=0;i
		if(belong[edge[i].s]!=belong[edge[i].t]){
			outdegree[belong[edge[i].s]]++;
		}
	}
	for(i=1;i<=count;i++){
		if(!outdegree[i]){
			outzero++;
			cnt=i;
		}
	}
	for(i=1;i<=n;i++)
		if(belong[i]==cnt) ans++;
	if(outzero==1)	printf("%d\n",ans);
	else printf("0\n");
}
 
int main(){
	int i,a,b;
	while(~scanf("%d%d",&n,&m)){
		memset(headlist,-1,sizeof(headlist));
		for(i=0;i
			scanf("%d%d",&a,&b);
			edge[i].s=a;
			edge[i].t=b;
			edge[i].next=headlist[a];
			headlist[a]=i;
		}
		work();
		output();
	}
	return 0;
}

=============================================================================

/*
kosaraju
Memory 1036K
Time    63MS
*/
#include 
using namespace std;
#define MAXM 50010
#define MAXV 10010
#define min(a,b) (a>b?b:a)
 
typedef struct{
	int s,t,next,next2;
}Edge;
 
Edge edge[MAXM];
int n,m,headlist[MAXV],headlist2[MAXV];
 
int order[MAXV],belong[MAXV];
int num,count;
bool vis[MAXV];
 
void dfs(int x){
	int i,a;
	vis[x]=1;
	for(i=headlist[x];i!=-1;i=edge[i].next){
		a=edge[i].t;
		if(!vis[a]) dfs(a);
	}
	order[++num]=x;
}
 
void dfst(int x){
	int i,a;
	belong[x]=count;		//记录结点属于哪个连通分量
	vis[x]=1;
	for(i=headlist2[x];i!=-1;i=edge[i].next2){		//要将边反过来遍历一遍
		a=edge[i].s;
		if(!vis[a]) dfst(a);
	}
}
 
void kosaraju(){
	int i;
	memset(vis,0,sizeof(vis));
	num=count=0;
	for(i=1;i<=n;i++)
		if(!vis[i]) dfs(i);
	memset(vis,0,sizeof(vis));
		
	for(i=n;i>=1;i--)
		if(!vis[order[i]]){
			count++;
			dfst(order[i]);
		}
}
 
void output(){
	int i,outdegree[MAXV]={0},outzero=0,ans=0,cnt;
	for(i=0;i
		if(belong[edge[i].s]!=belong[edge[i].t]){
			outdegree[belong[edge[i].s]]++;
		}
	}
	for(i=1;i<=count;i++){
		if(!outdegree[i]){
			outzero++;
			cnt=i;
		}
	}
	for(i=1;i<=n;i++)
		if(belong[i]==cnt) ans++;
		if(outzero==1)	printf("%d\n",ans);
		else printf("0\n");
}
 
int main(){
	int i,a,b;
	while(~scanf("%d%d",&n,&m)){
		memset(headlist,-1,sizeof(headlist));
		memset(headlist2,-1,sizeof(headlist2));
		for(i=0;i
			scanf("%d%d",&a,&b);
			edge[i].s=a;
			edge[i].t=b;
			edge[i].next=headlist[a];
			headlist[a]=i;
			edge[i].next2=headlist2[b];		//记录反边
			headlist2[b]=i;
		}
		kosaraju();
		output();
	}
	return 0;
}