poj-2553（强连通分量）

//强连通分量
//题意：求从u出发能到v，并且从v出发也能到u的这些点，并且将其输出；
//求出一个图的若干个强连通分量并且如果这个强连通分量中的缩点出度为0的话那么这个强连通分量里面的点就都满足题意。
//kosaraju
//poj 2553
#include <iostream>
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 solve()
{
	int i,j,k;
	int sum=0;
	int outgree[5000];
	memset(outgree,0,sizeof(outgree));
	for(i=1;i<=n;i++)
	  for(j=headlist[i];j!=-1;j=edge[j].next)
		  if(belong[i]!=belong[edge[j].t])
			  outgree[belong[i]]++;
	for(i=1;i<=n;i++)
		if(outgree[belong[i]]==0)
			printf("%d ",i);

	printf("\n");
}
int main()
{
	int i;
	int x,y;
	while(scanf("%d",&n)!=EOF && n)
	{
		scanf("%d",&m);
		memset(headlist,-1,sizeof(headlist));
		memset(headlist2,-1,sizeof(headlist2));
		for(i=0;i<m;i++)
		{
			scanf("%d%d",&x,&y);
			edge[i].s=x;
			edge[i].t=y;
			edge[i].next=headlist[x];
			headlist[x]=i;
			edge[i].next2=headlist2[y];
			headlist2[y]=i;
		}
		kosaraju();
		solve();
	}
}