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.

 題意：
給出仰慕關係， 求有幾個奶牛被除了自己的所有奶牛仰慕。

看著題解才勉強看懂。 。

先對給出的關係運用targan演算法+縮點將強連通分量進行縮點。 。

如果縮點為1個， 則說明這n個奶牛相互崇拜。 。。

然後求他們的出度。

如果有大於1個的點出度為0， 說明所有的奶牛都不符合條件。 。為神魔呢， 因為出度大於1的話， 說明至少有一隻奶牛誰也不崇拜， 所以不滿足條件。

如果只有一個點出度為1， 則說明這個縮點的所有點都符合條件， 只要求出此縮點有多少個點即可。 。

程式碼如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <stack>
#include <vector>
using namespace std;
const int maxn=10005;
vector <int>ve[maxn];
int n,m;
int dfn[maxn],low[maxn],ccnc[maxn],on[maxn];
int tot,ccnum;
stack<int>s;
void init ()
{
    ccnum=tot=0;
    memset (dfn,0,sizeof(dfn));
    memset (low,0,sizeof(low));
    memset (ccnc,0,sizeof(ccnc));
    memset (on,0,sizeof(on));
    for (int i=0;i<=n;i++)
        ve[i].clear();
    while (!s.empty())
          s.pop();
}
void targan (int x)
{
    dfn[x]=low[x]=++tot;
    s.push(x);
    for (int i=0;i<ve[x].size();i++)
    {
        int v=ve[x][i];
        if(!dfn[v])
        {
            targan(v);
            low[x]=min(low[x],low[v]);
        }
        else if(!ccnc[v])
            low[x]=min(low[x],dfn[v]);
    }
    if(low[x]==dfn[x])
    {
        ccnum++;
        while (1)
        {
            int now=s.top();
            s.pop();
            ccnc[now]=ccnum;
            if(now==x)
                break;
        }
    }
}
void finds ()
{
    if(ccnum==1)
    {
        printf("%d\n",n);
        return;
    }
    for (int i=1;i<=n;i++)
    {
        for (int j=0;j<ve[i].size();j++)
        {
            int v=ve[i][j];
            if(ccnc[i]!=ccnc[v])
            {
                on[ccnc[i]]++;
            }
        }
    }
    int num=0,loc,ans=0;
    for (int i=1;i<=ccnum;i++)
          if(!on[i])
          {
              num++;
              loc=i;
          }
    if(num>1)
    {
        printf("0\n");
        return;
    }
    for (int i=1;i<=n;i++)
        if(ccnc[i]==loc)
            ans++;
    printf("%d\n",ans);
}
int main()
{
    while (scanf("%d%d",&n,&m)!=EOF)
    {
        init();
        while (m--)
      {
        int x,y;
        scanf("%d%d",&x,&y);
        ve[x].push_back(y);
      }
       for (int i=1;i<=n;i++)
           if(!dfn[i])
              targan(i);
       finds();
    }
    return 0;
}