poj2186 最受欢迎的牛（求最受欢迎的牛的数量）

#include <string.h>
#include <stdio.h>
#define V    10500
#define E    50500

struct edge
{
    int to, next;
}Edge[E];
int head[V], e, n;

int indeg[V], outdeg[V]; //点的入度和出度数
int belong[V], low[V], dfn[V], scc, cnt;//dfn[]:遍历到u点的时间; low[]:u点可到达的各点中最小的dfn[v]
int S[V], top;
bool vis[V];//v是否在栈中

int addedge(int u, int v)
{
    Edge[e].to = v;
    Edge[e].next = head[u];
    head[u] = e++;
    return 0;
}
void tarjan(int u)
{
    int v;
    dfn[u] = low[u] = ++cnt;//开始时dfn[u] == low[u]
    S[top++] = u;//不管三七二十一进栈
    vis[u] = true;
    for (int i=head[u]; i!=-1; i=Edge[i].next)
    {
        v = Edge[i].to;
        if (dfn[v] == 0)//如果v点还未遍历
        {
            tarjan(v);//向下遍历
            low[u] = low[u] < low[v] ? low[u] : low[v];//确保low[u]最小
        }
        else if (vis[v] && low[u] > dfn[v])//v在栈中，修改low[u]
            low[u] = dfn[v];
    }
    if (dfn[u] == low[u])//u为该强连通分量中遍历所成树的根
    {
        ++scc;
        do
        {
            v = S[--top];//栈中所有到u的点都属于该强连通分量，退栈
            vis[v] = false;
            belong[v] = scc;
        } while (u != v);
    }

}

int solve(){
    scc = top = cnt = 0;
    memset(dfn, 0, sizeof(dfn));
    memset(vis, false, sizeof(vis));
    for (int u=1; u<=n; ++u)
        if (dfn[u] == 0)
            tarjan(u);
    return scc;
}

void count_deg()
{
    memset(indeg, 0, sizeof(indeg));
    memset(outdeg, 0, sizeof(outdeg));
    for (int u=1; u<=n; ++u)
        for (int i=head[u]; i!=-1; i=Edge[i].next)
        {
            int v = Edge[i].to;
            if (belong[u] != belong[v])
            {
                indeg[belong[v]]++;
                outdeg[belong[u]]++;
            }
        }
}

int main()
{
    int u, v, i;
    int m;
    while (~scanf("%d%d", &n,&m))
    {
        e = 0;
        memset(head, -1, sizeof(head));
        for (int i=1;i<=m;i++){
            scanf("%d%d",&u,&v);    
                addedge(u, v);
        }
        solve();
            count_deg();

          int num=0,tmp;
        for(int i = 1; i <= scc; i++)  
        {  
            if(!outdeg[i])  
            {  
                num++;  
                tmp = i;  
            }  
                  
        }  
          int ans=0;
        if(num == 1)  
        {  
            for(int i = 1; i <= n; i++)  
            {  
                if(belong[i] == tmp)  
                    ans++;  
            }  
            printf("%d\n", ans);  
        }  
        else  
        {  
            printf("0\n");  
        }  

        
    }
    return 0;
}