POJ 2553 强连通分量+缩点+求出度

求强连通分量，缩点求出出度为0的点，从小到大输出。

英语不好这题读起来真拗口。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstring>
#include <queue>
#include <set>
#include <vector>
#include <stack>
#include <map>
#include <iomanip>
#define PI acos(-1.0)
#define Max 100005
#define inf 1<<28
#define LL(x) (x<<1)
#define RR(x) (x<<1|1)
#define FOR(i,s,t) for(int i=(s);i<=(t);++i)
#define ll long long
#define mem(a,b) memset(a,b,sizeof(a))
#define mp(a,b) make_pair(a,b)
using namespace std;

struct kdq
{
    int e ,next ;
}ed[Max] ;
int head[Max] ;
int num1 = 0 ;
int dfn[Max] ,low[Max] , st[Max] ,vis[Max] , belong[Max] ;
int in[Max] ,out[Max] ;
int tp = 0 ,dp = 0 ;
int num = 0 ;
set<int>ans[5005] ;
void add(int a,int b)
{
    ed[num1].e = b ;
    ed[num1].next =head[a] ;
    head[a] = num1 ++ ;
}
void init(int n )
{
    mem(dfn,-1) ;
    mem(vis,0) ;
    mem(low,0) ;
    mem(st,0) ;
    mem(belong,0) ;
    mem(in,0) ;
    mem(out,0) ;
    mem(head,-1) ;
    num = 0 ;
    num1 = 0 ;
    tp = 0 ;
    dp = 0 ;
    for (int i = 0 ;i <= n ;i ++ )ans[i].clear() ;
}
void tarjan(int now)
{
    vis[now] = 1 ;
    st[tp ++ ] = now ;
    dfn[now] = low[now] = dp ++ ;
    for (int i = head[now] ; i != -1 ;i = ed[i].next )
    {
        int v = ed[i].e ;
        if(dfn[v] == -1)
        tarjan(v) ,
        low[now] = min(low[now], low[v]) ;
        else if(vis[v])
        low[now] = min(low[now], dfn[v]) ;
    }
    if(low[now] == dfn[now])
    {
        int xx ;
        num ++ ;
        do
        {
            xx = st[-- tp] ;
            vis[xx] = 0 ;
            belong[xx] = num ;
            ans[num].insert(xx) ;
        }
        while(xx != now) ;
    }
}
int n , m ;
void solve()
{
    while(cin >> n ,n )
    {
        init(n) ;
        cin >> m ;
        for (int i = 0 ; i < m ;i ++)
        {
            int a, b ;
            scanf("%d%d",&a,&b) ;
            add(a,b) ;
        }
        for (int i = 1 ;i <= n ;i ++ )
        if(dfn[i] == -1)tarjan(i) ;
        for (int i = 1 ;i <= n ;i ++ )
        {
            for (int j = head[i] ;j != -1 ;j = ed[j].next )
            {
                int x = belong[i] ;
                int y = belong[ed[j].e] ;
                if(x != y)
                {
                    out[x] = 1 ;
                    in[y] = 1 ;
                }
            }
        }
        set<int>::iterator it ;
        for (int i = 1 ;i <= num ;i ++)
        {
            if(out[i] == 0)//出度为0
            {
                for (it = ans[i].begin() ; it != ans[i].end() ;it ++ )
                {
                    belong[*it] = 0 ;//置为0，从小到大输出
                }
            }
        }
        for (int i = 1 ;i <= n; i ++)if(!belong[i])cout <<i <<" ";
        cout <<endl;

    }
}
int main()
{
    solve() ;
    return 0;
}