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题目大意：如果v点能够到的点，反过来能够到达v点，则称这个点为sink点，输出所有的sink点
解题思路：求连通分量，然后出度为0的连通分量里面的点就是sink点

/*
kosaraju
Memory 524K
Time   63MS
*/
#include 
using namespace std;
#define MAXM 50010
#define MAXV 5010
#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,j,outdegree[MAXV]={0};
	for(i=1;i<=n;i++)
		for(j=headlist[i];j!=-1;j=edge[j].next)
			if(belong[i]!=belong[edge[j].t]){
				outdegree[belong[i]]++;
			}
			
	memset(vis,0,sizeof(vis));
	for(i=1;i<=n;i++)
		if(!outdegree[belong[i]]) vis[i]=1;
				
	for(i=1;i<=n;i++)
		if(vis[i]) printf("%d ",i);
	printf("\n");
}
 
 
int main(){
	int i,a,b;
	while(scanf("%d",&n) && n){
		memset(headlist,-1,sizeof(headlist));
		memset(headlist2,-1,sizeof(headlist2));
		scanf("%d",&m);
		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;
}

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

/*
tarjan
Memory 544K
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;
}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,j,outdegree[MAXV]={0};
	for(i=1;i<=n;i++)
		for(j=headlist[i];j!=-1;j=edge[j].next)
			if(belong[i]!=belong[edge[j].t]){
				outdegree[belong[i]]++;
			}
			
	memset(instack,0,sizeof(instack));
	for(i=1;i<=n;i++)
		if(!outdegree[belong[i]]) instack[i]=1;
				
	for(i=1;i<=n;i++)
		if(instack[i]) printf("%d ",i);
	printf("\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;
}