poj 2553 强连通分支与缩点

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,-1,sizeof(index));
    memset(degree,0,sizeof(degree));
    memset(dfn,0,sizeof(dfn));
    memset(low,0,sizeof(low));
    memset(vi,0,sizeof(vi));
    memset(mark,0,sizeof(mark));
    memset(be,0,sizeof(be));
    e=lab=top=num=list=0;
}
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]=0;
        mark[i]=1;
    }
    while(i!=u);
}
int dfs(int u)
{
    dfn[u]=low[u]=++lab;
    stack[top++]=u;
    vi[u]=1;
    int i,j,temp;
    for(i=index[u];i!=-1;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 0;
}
int solve()
{
    int i,j,temp;
    for(i=1;i<=n;i++)
        if(!dfn[i])
            dfs(i);
    memset(vi,1,sizeof(vi));
    for(i=1;i<=n;i++)
    {
        for(j=index[i];j!=-1;j=edge[j].next)
        {
            temp=edge[j].to;
            if(be[i]!=be[temp])
            {
                vi[be[i]]=0;
            }
        }
    }
    for(i=1;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("
");
    return 0;
}
int main()
{
    int m,i,j,a,b;
    while(scanf("%d",&n)!=EOF,n)
    {
        scanf("%d",&m);
        init();
        for(i=1;i<=m;i++)
        {
            scanf("%d%d",&a,&b);
            addedge(a,b);
        }
        solve();
    }
    return 0;
}