思路：将所有强连通分支找出来，并进行缩点，然后找其中所有出度为0的连通分支，就是题目要求的。
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#define Maxn 5100
#define Maxm Maxn*100
#define inf 0x7fffffff
using namespace std;
int index[Maxn],vi[Maxn],stack[Maxn],dfn[Maxn],low[Maxn],e,n,lab,top,num,list,mark[Maxn],degree[Maxn],be[Maxn];
struct Edge{
int from,to,next;
}edge[Maxm];
void addedge(int from,int to)
{
edge[e].from=from;
edge[e].to=to;
edge[e].next=index[from];
index[from]=e++;
}
void init()
{
memset(index,-,sizeof(index));
memset(degree,,sizeof(degree));
memset(dfn,,sizeof(dfn));
memset(low,,sizeof(low));
memset(vi,,sizeof(vi));
memset(mark,,sizeof(mark));
memset(be,,sizeof(be));
e=lab=top=num=list=;
}
void Out(int u)//将该连通分量的点进行统计
{
int i,j;
list++;
do{
i=stack[--top];
be[i]=list;
vi[i]=;
mark[i]=;
}
while(i!=u);
}
int dfs(int u)
{
dfn[u]=low[u]=++lab;
stack[top++]=u;
vi[u]=;
int i,j,temp;
for(i=index[u];i!=-;i=edge[i].next)
{
temp=edge[i].to;
if(!dfn[temp])
{
dfs(temp);
low[u]=min(low[u],low[temp]);
}
if(vi[temp])
low[u]=min(low[u],dfn[temp]);
}
if(low[u]==dfn[u])//找到连通分量
Out(u);
return ;
}
int solve()
{
int i,j,temp;
for(i=;i<=n;i++)
if(!dfn[i])
dfs(i);
memset(vi,,sizeof(vi));
for(i=;i<=n;i++)
{
for(j=index[i];j!=-;j=edge[j].next)
{
temp=edge[j].to;
if(be[i]!=be[temp])
{
vi[be[i]]=;
}
}
}
for(i=;i<=n;i++)
{
if(mark[i])
{
if(vi[be[i]])
{
printf("%d",i);
break;
}
}
}
for(i++;i<=n;i++)
{
if(mark[i])
{
if(vi[be[i]])
printf(" %d",i);
}
}
printf("\n");
return ;
}
int main()
{
int m,i,j,a,b;
while(scanf("%d",&n)!=EOF,n)
{
scanf("%d",&m);
init();
for(i=;i<=m;i++)
{
scanf("%d%d",&a,&b);
addedge(a,b);
}
solve();
}
return ;
}