poj2553强连通+缩点 如果让我说：我只能说，实力决定一切

思路：tarjan()算法，记录每个点的出度值(缩点之后）,把所以出度为0的点找出来，然后进行从小到大排序

 

#include<iostream>#include<algorithm>#include<cstdio>#include<string.h>#include<vector>#include<stack>#define  maxn 5111using namespace std;int V,E,sum,tem;int dfn[maxn],low[maxn],vis[maxn],outd[maxn],belong[maxn];int ans[maxn];vector<int>gra[maxn];stack<int>S;int MIN(int a,int b){    return a>b?b:a;}void tarjan(int pox){    vis[pox]=2;    dfn[pox]=low[pox]=++sum;    S.push(pox);    for(int i=0; i<gra[pox].size(); i++)    {        int t=gra[pox][i];        if(!dfn[t])        {            tarjan(t);            low[pox]=MIN(low[pox],low[t]);        }        else if(vis[t]==2)        {            low[pox]=MIN(low[pox],dfn[t]);        }    }    if(dfn[pox]==low[pox])    {        ++tem;        while(!S.empty())        {            int gh=S.top();            S.pop();            outd[gh]=0;            belong[gh]=tem;            vis[gh]=1;            if(gh==pox)                break;        }    }}int main(){    int a,b;    while(scanf("%d",&V),V)    {        tem=sum=0;        memset(ans,0,sizeof(ans));        memset(dfn,0,sizeof(dfn));        memset(low,0,sizeof(low));        memset(vis,0,sizeof(vis));        memset(belong,0,sizeof(belong));        for(int i=0; i<maxn; i++)            gra[i].clear();        while(!S.empty())            S.pop();        scanf("%d",&E);        for(int i=0; i<E; i++)        {            scanf("%d%d",&a,&b);            gra[a].push_back(b);        }        for(int i=1; i<=V; i++)            if(!dfn[i])                tarjan(i);        for(int i=1; i<=V; i++)            for(int j=0; j<gra[i].size(); j++)            {                if(belong[i]!=belong[gra[i][j]])                {                    outd[belong[i]]++;                }            }        int as=0;        for(int i=1; i<=V; i++)            if(!outd[belong[i]])                ans[as++]=i;        if(as==0)        {            printf("\n");            continue;        }        sort(ans,ans+as);        for(int i=0; i<as; i++)            if(i==0)                printf("%d",ans[i]);            else                printf(" %d",ans[i]);        printf("\n");    }    return 0;}