poj1703 Find them, Catch them

     poj1182是此题的加强版。若两人属于不同集合则他们所属帮派未知。若属于同一集合，设ra=find(a),若rank[a]==0,则a与ra为同一帮派，反之rank[a]==1,a与ra则分属于不同帮派。另外测下这组数据 1 2 1 A 1 2 答案应该是In different gangs.（在discuss里看到的）。

#include<cstdio>
#include<string.h>
#include<iostream>
using namespace std;
const int maxn=100001;
int p[maxn],rank[maxn],n;

void init()
{
    int i;
    for(i=0;i<=n;i++)
    {
        p[i]=i;
        rank[i]=0;
    }
}

int find(int x)
{
    int temp=p[x];
    if(x!=p[x])
    {
        p[x]=find(p[x]);
        rank[x]=(rank[x]+rank[temp])%2;
    }
    return p[x];
}

void unions(int x,int y,int rx,int ry)
{
    p[ry]=rx;
    rank[ry]=(rank[x]-rank[y]+3)%2;
}
int main()
{
    //freopen("test.txt","r",stdin);
    int T,k;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&k);
        init();
        int i,a,b,ra,rb;
        char c[2];
        for(i=0;i<k;i++)
        {
            scanf("%s%d%d",c,&a,&b);
            ra=find(a);
            rb=find(b);
            if(n==2&&c[0]=='A') { printf("In different gangs.\n");continue;}
            if(c[0]=='D') unions(a,b,ra,rb);
            if(c[0]=='A')
            {
                if(ra==rb)
                {
                    if(rank[a]==rank[b]) printf("In the same gang.\n");
                    else printf("In different gangs.\n");
                }
                else printf("Not sure yet.\n");
            }
        }
    }
    return 0;
}