POJ 3180-The Cow Prom (图论-有向图强联通tarjan算法）

题目大意：有n个牛在一块， m条单项绳子， 有m个链接关系， 问有多少个团体内部任意两头牛可以相互可达

解题思路：有向图强连通分量模版图

代码如下:

#include<stdio.h>
#include<vector>
#include<map>
#include<set>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 10003;

vector<int>G[N], DQ;
int low[N], dfn[N], tot;
bool mk[N];
int n, m, ans;

void init()
{
    ans = tot = 0;
    DQ.clear();
    for(int i=1; i<=n; ++ i)
    {
        G[i].clear();
        low[i] = dfn[i] = -1;
        mk[i] = false;
    }
}

void tarjan(int u, int f)
{
    dfn[u] = low[u] = ++ tot;
    DQ.push_back(u);
    mk[u] = true;
    for(int i = 0; i<G[u].size(); ++ i)
    {
        int v = G[u][i];
        if(dfn[v] == -1)
        {
            tarjan(v, u);
            low[u] = min(low[u], low[v]);
        }
        else if(mk[v])
            low[u] = min(low[u], dfn[v]);
    }

    if(dfn[u] == low[u])
    {
        int s;
        int k = 0;
        do
        {
            s = DQ.back();
            k ++;
            DQ.pop_back();
            mk[s] = false;
        }
        while(u != s);
        if(k > 1)
            ans ++;
    }
}

void solve()
{
    for(int i=1; i<=n; ++ i)
    {
        if(dfn[i] == -1)
            tarjan(i, -1);
    }
    printf("%d
", ans);
}

int main()
{
    while(~scanf("%d%d", &n, &m))
    {
        init();
        for(int i=1; i<=m; ++ i)
        {
            int u, v;
            scanf("%d%d", &u, &v);
            G[u].push_back(v);
        }
        solve();
    }
    return 0;
}