Popular Cows//强连通分支Kosaraju加缩点

#include <iostream>
#include <vector>
using namespace std;
const int size = 1e4 + 1;
vector<int> graph[size];
vector<int> reverse[size];
int order[size];//栈
int pos = 0;//用于order[]的索引
bool marked[size];
bool marked2[size];
int ans = 0;//连通分支数
int DAG[size];//分离强连通分量
int out[size];//DAG的出度,一共只有强连通分支数那么多个点

void init(int N)//初始化
{
    pos = ans = 0;
    for(int i = 1; i <= N; i++)
    {
        marked[i] = false;
        marked2[i] = false;
        graph[i].clear();
        reverse[i].clear();
        DAG[i] = 0;
        out[i] = 0;
    }
}

void dfsReverseOrder(int x)
{
    marked[x] = true;
    for(int i = 0; i < reverse[x].size(); i++)
    {
        if(!marked[reverse[x][i]])
            dfsReverseOrder(reverse[x][i]);
    }
    order[pos++] = x;
}

void dfs(int x)
{
    DAG[x] = ans;
    marked2[x]  = true;
    for(int i = 0; i < graph[x].size(); i++)
    {
        if(!marked2[graph[x][i]])
        {
            dfs(graph[x][i]);
        }
        else if(DAG[graph[x][i]] != DAG[x])//和计算出度有关
            out[ans] = 1;
    }
}

void Kosaraju(int N)
{
    for(int i = 1; i <= N; i++)
    {
        if(!marked[i])
            dfsReverseOrder(i);
    }
    pos--;

    for(; pos >= 0;pos--)
    {
        if(!marked2[order[pos]])
        {
            dfs(order[pos]);
            ans++;
        }
    }
}

int main()
{
    int N, M;
    int a, b;
    while(cin >> N >> M)
    {
        init(N);
        //输入
        for(int i = 0; i < M; i++)
        {
            cin >> a >> b;
            graph[a].push_back(b);
            reverse[b].push_back(a);    
        }
        Kosaraju(N);
        int count = 0;
        int flag;
        for(int i = 0; i < ans; i++)//找出度为0的强连通分支的个数
        {
            if(out[i] == 0)
            {
                count++;
                flag = i;
            }
        }
        int num = 0;
        if(count == 1)//若只有一个出度为0的强连通分支
        {
            for(int i = 1; i <= N; i++)//计算该强连通分支内点的个数
            {
                if(DAG[i] == flag)
                    num++;
            }
            printf("%d\n", num);
        }
        else 
            printf("%d\n", 0);
    }
    return 0;
}