poj 3180 The Cow Prom

http://poj.org/problem?id=3180

题意：给一个无向图，求出节点数据大于1的边通分量有多少个。

思路：直接tarjan缩点统计。

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<stack>
#include<iostream>
#include<utility>
using namespace std;
const int maxn = 10005;
struct nd
{
        int v,next;
}edge[maxn*5];
int head[maxn],vis[maxn],dfn[maxn],low[maxn],c[maxn];
int ecnt,idx,cnt,n,m;
stack<int>st;

void add(int u,int v)
{
        edge[ecnt].v = v;
        edge[ecnt].next = head[u];
        head[u] = ecnt++;
}

void tarjan(int u)
{
        int i,v;
        dfn[u] = low[u] = ++idx;
        st.push(u);
        vis[u] = 1;
        for(i = head[u]; i != -1; i = edge[i].next)
        {
                v = edge[i].v;
                if(!dfn[v]){
                        tarjan(v);
                        low[u] = min(low[v],low[u]);
                }else if(vis[v]) low[u] = min(dfn[v],low[u]);
        }
        if(dfn[u]==low[u]){
                cnt++;
                do{
                        v = st.top();
                        st.pop();
                        vis[v] = 0;
                        c[cnt]++;
                }while(v!=u);
        }
}

int main()
{
        int i,j,k,u,v,w;
        while(scanf("%d %d",&n,&m)==2)
        {
                memset(head,-1,sizeof(head));
                memset(dfn,0,sizeof(dfn));
                memset(vis,0,sizeof(vis));
                memset(c,0,sizeof(c));
                ecnt = idx = cnt = 0;
                for(i = 0; i < m; ++ i)
                {
                        scanf("%d %d",&u,&v);
                        add(u,v);
                }
                for(i = 1; i <= n; ++ i) if(!dfn[i]) tarjan(i);
                k = 0;
                for(i = 1; i <= cnt; ++ i) if(c[i]>1)k++;
                printf("%d\n",k);
        }
        return 0;
}