Poj 2186 Popular Cows

Popular Cows
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 29240 Accepted: 11831
Description
Every cow’s dream is to become the most popular cow in the herd. In a herd of N (1 <= N <= 10,000) cows, you are given up to M (1 <= M <= 50,000) ordered pairs of the form (A, B) that tell you that cow A thinks that cow B is popular. Since popularity is transitive, if A thinks B is popular and B thinks C is popular, then A will also think that C is
popular, even if this is not explicitly specified by an ordered pair in the input. Your task is to compute the number of cows that are considered popular by every other cow.
Input
* Line 1: Two space-separated integers, N and M
* Lines 2..1+M: Two space-separated numbers A and B, meaning that A thinks B is popular.
Output
* Line 1: A single integer that is the number of cows who are considered popular by every other cow.
Sample Input
3 3
1 2
2 1
2 3
Sample Output
1
Hint
Cow 3 is the only cow of high popularity.
Source
USACO 2003 Fall
传送门

/*
Tarjan模板+缩点 
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#define MAXN 50001
using namespace std;
struct data
{
    int v,next;
}
e[MAXN];
int n,m,cut,belong[MAXN],head[MAXN],dfn[MAXN],stack[MAXN],low[MAXN],in[MAXN],top,tot;
void init()
{
    cut=tot=top=0;
    memset(head,-1,sizeof(head));
    memset(stack,0,sizeof(stack));
    memset(dfn,0,sizeof(dfn));
    memset(low,0,sizeof(low));
    memset(in,0,sizeof(in));
}
void add_edge(int u,int v)
{
    e[tot].v=v;
    e[tot].next=head[u];
    head[u]=tot++;


}
void tarjan(int u)
{
    int v;
    low[u]=dfn[u]=++cut;
    in[u]=1;//当前节点进栈
    stack[top++]=u; 
    for(int i=head[u];i!=-1;i=e[i].next)//遍历边
    {
        v=e[i].v;
        if(!dfn[v])//未被访问 
        {
            tarjan(v);
            low[u]=min(low[u],low[v]);
        }
        else if(in[v])
        {
            low[u]=min(low[u],dfn[v]);
        }
    }
    if(dfn[u]==low[u])//找到一个强联通分量的根 
    {
        tot++;
        do
        {
            v=stack[--top];//退栈
            in[v]=0;
            belong[v]=tot;//tot为连通分量编号 
        }
        while(u!=v);
    }
}
int main()
{
    while(scanf("%d %d",&n,&m)!=EOF)
    {
        int x,y;
        init();
        for(int i=1;i<=m;i++)
        {
            cin>>x>>y;
            add_edge(x,y);
        }
        tot=0;
        for(int i=1;i<=n;i++)
        {
            if(!dfn[i])
              tarjan(i);
        }
        int out[MAXN]={0},u,v;
        for(int i=1;i<=n;i++)
        {
            for(int j=head[i];j!=-1;j=e[j].next)
            {
                u=belong[i],v=belong[e[j].v];
                if(u!=v) out[u]++;
            }
         }
         int flag=0;
        for(int i=1;i<=tot;i++)
        {
            if(!out[i])//找出度为0的连通块 
            {
                flag++;
                x=i;
            }
        }
        if(flag>1) printf("0
");
        else
        {
            int ans=0;
            for(int i=1;i<=n;i++)
            {
                if(belong[i]==x)//属于该连通块的点即为答案的贡献
                  ans++;
            }
            printf("%d
",ans);
        }
    }
    return 0;
}