poj 2492 A Bug's Life

// 题意:有种虫只有两种性别，它们只和不同性别的虫配对，给出n对异性虫,判断是否冲突

#include <iostream>        //并查集
using namespace std;
int p[2002],diff[2002],n,m;        //diff[i]=j表示编号为j跟编号为i的人不在同一组
void init()
{
    for(int i=1;i<=n;++i)
    {
        p[i]=i;
        diff[i]=-1;
    }
}
int find(int x)
{
    return p[x]==x ? x : p[x]=find(p[x]);
}
void Union(int a,int b)      
{
    int x=find(a),y=find(b);
    if(x!=y)
        p[x]=y;
}
int main()
{
    int cases;
    cin>>cases;
    for(int k=1;k<=cases;++k)
    {
        cin>>n>>m;
        init();
        int suc=1,a,b;
        while(m--)
        {
            scanf("%d%d",&a,&b);
            if(!suc)
                continue;
            int fa=find(a),fb=find(b);
            if(fa==fb)
            {
                suc=0;
            }
            else
            {
                if(diff[a]!=-1)
                    Union(b,diff[a]);
                if(diff[b]!=-1)
                    Union(a,diff[b]);
                diff[a]=b;diff[b]=a;
            }
        }
        printf("Scenario #%d:\n",k);
        if(suc)
            printf("No suspicious bugs found!\n\n");
        else
            printf("Suspicious bugs found!\n\n");
    }
    return 0;
}