pku 2186 Popular Cows (tarjan缩点)

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

将所有最大连通分量缩点，然后统计缩点后每个点的出度，出度为0的肯定就是了可是这个点可能是缩出来的，所以要记录这个点真正包含的点数。如果出度为0的点大于1个说明不存在

稍微尝试了一下stl的vector和stack

#include <cstdio>
#include <cstring>
#include <vector>
#include <stack>
#include <iostream>
#define maxn 10017
using namespace std;
int low[maxn],dfn[maxn],val[maxn],out[maxn];
bool instack[maxn];
int belong[maxn];
int bcnt,index;
int n,m;
vector<int>g[maxn];
stack<int>s;
void init()
{
    for (int i = 0; i < maxn; ++i)
    {
        g[i].clear();
        low[i] = dfn[i] = belong[i] = val[i] = out[i] = 0;
        instack[i] = false;
    }
    index = bcnt =0;
}
void tarjan(int i)
{
    int j,k;
    low[i] = dfn[i] = ++index;
    s.push(i);
    instack[i] = true;
    for (k = 0; k < g[i].size(); ++k)
    {
        j = g[i][k];
        if (!dfn[j])
        {
            tarjan(j);
            low[i] = min(low[i],low[j]);
        }
        else if (instack[j])
        {
            low[i] = min(low[i],dfn[j]);
        }
    }
    if(low[i] == dfn[i])
    {
        bcnt++;
        do
        {
            j = s.top();
            s.pop();
            instack[j] = false;
            belong[j] = bcnt;
            val[bcnt]++;//记录缩点后的点真正拥有的点
        } while (j != i);
    }
}
void solve()
{
    int count = 0,i,j;
    for (i = 1; i <= n; ++i)
    {
        if (!dfn[i])  tarjan(i);
    }
    for (i = 1; i <= n; ++i)
    {
        for (j = 0; j < g[i].size(); ++j)
        {
            int k = g[i][j];
            if (belong[i] != belong[k])
            out[belong[i]]++;
        }
    }
    int t = 0;
    for (i = 1; i <= bcnt; ++i)
    {
        if (!out[i])
        {
            count++;
            t = i;
        }
    }
    if (count > 1) puts("0");
    else
    printf("%d\n",val[t]);
}
int main()
{
    //freopen("d.txt","r",stdin);
    int x,y,i;
    while (cin>>n>>m)
    {
        init();
        for (i = 1; i <= m; ++i)
        {
            cin>>x>>y;
            g[x].push_back(y);
        }
        solve();
    }
    return 0;
}