poj 1703

http://poj.org/problem?id=1703

rank 用于记录祖先的等级

opp 用于记录敌人。。

#include <iostream>
#include<cstdio>
using namespace std;
int f[100005];
int rank[100005],opp[100005];
char str[3];
void init(int n){
    for(int i=1;i<=n;i++){
        f[i] = i;
        rank[i]=opp[i]=0;
    }
}

int find(int x){
    if(x!=f[x])
        f[x] = find(f[x]);
    return f[x];
}

void unin(int a,int b){
    int fa = find(a);
    int fb = find(b);
    if(fa==fb)
        return ;
    if(rank[fa]>rank[fb])
        f[fb] = fa;
    else{
        f[fa] = f[fb];
        if(rank[fa]==rank[fb])
            rank[fb]++;
    }
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--){
        int n,m;
        scanf("%d%d",&n,&m);
        init(n);
        int a,b;
        for(int i=1;i<=m;i++){
            scanf("%s%d%d",str,&a,&b);
            int fa = find(a);
            int fb = find(b);
            if(str[0]=='A'){
                if(fa==fb){
                    printf("In the same gang.
");
                }
                else if(fa==find(opp[fb]))//如果fa！=fb  但是fa和b的敌人的祖先相同，那么a，b。不同族群
                    printf("In different gangs.
");
                else
                    printf("Not sure yet.
");
            }
            else{
                if(opp[a]==0)
                    opp[a] = fb;
                if(opp[b]==0)
                    opp[b]=fa;
                unin(a,opp[fb]);//合并
                unin(b,opp[fa]);
            }
        }
    }
    return 0;
}