POJ2186 Popular Cows (强连通缩点)

大意：给定n头牛，牛之间为有向的羡慕关系并且关系可以传递，问最后有多少个牛被所有的牛羡慕。

思路：羡慕的关系链或者环都可以进行缩点，使得图抽象DAG（有向无环图）。每个节点就是一个联通分量，我们统计每个联通分量的出度，当出度为0的连通分量数目大于1时就表明不是所以牛都羡慕其中的一头牛，结果为0。否则，我们直接找到最后缩点后的图中有多少个点就是最终的结果。

#include<map>
#include<queue>
#include<cmath>
#include<cstdio>
#include<stack>
#include<iostream>
#include<cstring>
#include<algorithm>
#define LL int
#define inf 0x3f3f3f3f
#define eps 1e-8
#include<vector>
#define ls l,mid,rt<<1
#define rs mid+1,r,rt<<1|1

using namespace std;
const int Ma = 1000005;

int head[Ma],dfn[Ma],num[Ma],du[Ma],stk[Ma],vis[Ma],low[Ma];
int cnt,top,time,css;

struct node{
    int to,next;
}q[Ma];

void bu(int a,int b){
    q[cnt].to = b;
    q[cnt].next = head[a];
    head[a] = cnt++;
}

void init(){
    time = 1;
    css = top = cnt = 0;
    memset(dfn,0,sizeof(dfn));
    memset(low,0,sizeof(low));
    memset(vis,0,sizeof(vis));
    memset(head,-1,sizeof(head));
    memset(num,0,sizeof(num));
    memset(du,0,sizeof(du));
}

void Tarjan(int u){
    low[u] = dfn[u] = time++;
    vis[u] = 1;
    stk[top++] = u;
    for(int i = head[u]; ~i ; i = q[i].next){
        int v = q[i].to;
        if(!dfn[v]){
            Tarjan(v);
            low[u] = min(low[u],low[v]);
        }
        else if(vis[v])
            low[u] = min(low[u],dfn[v]);
    }
    if(low[u] == dfn[u]){//找到极大联通分量
        css++;
        while(top > 0&&stk[top] != u){
            top --;
            vis[stk[top] ] = 2;
            num[stk[top] ] = css;
        }
    }
}
int main(){
    int n,m,i,j,k,a,b;
    while(~scanf("%d%d",&n,&m)){
            init();
        for(i = 0;i < m;++ i){
            scanf("%d%d",&a,&b);
            bu(a,b);
        }
        for(i = 1;i <= n;++ i){
            if(!dfn[i])
                Tarjan(i);
        }

        for(i = 1;i <= n;++ i){
            for(j = head[i]; ~j ; j = q[j].next){
                if(num[i]!=num[q[j].to ])
                    du[num[i] ]++;
            }
        }

        int sum = 0,tmp;
        for(i = 1;i <= css;++ i){
            if(!du[i]){
                sum++;
                tmp = i;
            }
        }
        if(sum==1){
            sum = 0;
            for(i = 1;i <= n;++ i)
                if(num[i] == tmp)
                    sum++;
            printf("%d\n",sum);
        }
        else
            printf("0\n");
    }
    return 0;
}