（连通图 缩点 强联通分支）Popular Cows -- poj --2186

http://poj.org/problem?id=2186

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

先缩点（把强联通分量看为一个点）， 判断出度为 0 的点有几个，如果大于 1 则输出 0， 否则输出 出度为零的点的个数

代码：

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<queue>
#include<stack>
#include<algorithm>
using namespace std;

#define N 50005

struct node
{
    int v, next;
}a[N];

int Head[N], cnt;
int dfn[N], low[N], Time, bnt, belong[N];
int Stack[N], InStack[N], top;

void Init()
{
    cnt = Time = bnt = top = 0;
    memset(Head, -1, sizeof(Head));
    memset(dfn, 0, sizeof(dfn));
    memset(low, 0, sizeof(low));
    memset(Stack, 0, sizeof(Stack));
    memset(InStack, 0, sizeof(InStack));
}

void Add(int u, int v)
{
    a[cnt].v = v;
    a[cnt].next = Head[u];
    Head[u] = cnt++;
}

void Tarjar(int u)
{
    int v;
    low[u] = dfn[u] = ++Time;
    InStack[u] = 1;
    Stack[top++] = u;

    for(int j=Head[u]; j!=-1; j=a[j].next)
    {
        v = a[j].v;
        if(!dfn[v])
        {
            Tarjar(v);
            low[u] = min(low[u], low[v]);
        }
        else if(InStack[v])
            low[u] = min(low[u], dfn[v]);
    }

    if(dfn[u]==low[u])
    {
        bnt++;
        do
        {
            v = Stack[--top];
            InStack[v] = 0;
            belong[v] = bnt;
        }while(u!=v);
    }
}

int main()
{
    int n, m;
    while(scanf("%d%d", &n, &m)!=EOF)
    {
        int i, u, v;

        Init();
        for(i=1; i<=m; i++)
        {
            scanf("%d%d", &u, &v);
            Add(u, v);
        }

        for(i=1; i<=n; i++)
        {
            if(!dfn[i])
            Tarjar(i);
        }

        int Out[N]={0};
        for(int i=1; i<=n; i++)
        {
            for(int j=Head[i]; j!=-1; j=a[j].next)
            {
                u = belong[i], v = belong[a[j].v];
                if(u!=v)
                    Out[u]++;
            }
        }

        int flag=0, Index;
        for(i=1; i<=bnt; i++)
        {
            if(!Out[i])
            {
                flag++;
                Index = i;
            }
        }

        if(flag>1)
            printf("0
");
        else
        {
            int ans = 0;
            for(i=1; i<=n; i++)
            {
                if(belong[i]==Index)
                    ans++;
            }
            printf("%d
", ans);
        }
    }
    return 0;
}