HDU 3635 Dragon Balls

这题的并查集是看题解的，想不出怎么算转移次数，题解好像用的是并入的次序来计算的，这真没想到。。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
#include <map>
#include <queue>
#include <ctime>
#include <set>
#define ll long long
#define MP make_pair
#define PB push_back
#define SZ(x) ((int)(x).size())
#define REP(i,n) for (int i=0; (i)<(int)n; ++(i))
#define FOR(it,c) for ( __typeof((c).begin()) it=(c).begin(); it!=(c).end(); ++it )
using namespace std;

const int N=10010;
int p[N],cnt[N],a[N];

int find(int x){
    if(p[x]==x) return x;
    int t=p[x];
    p[x]=find(p[x]);
    a[x]+=a[t];
    return p[x];
}

int main(){
#ifndef ONLINE_JUDGE
    freopen("a.in","r",stdin);
    freopen("a.out","w",stdout);
#endif
    int t,ca=0;
    cin>>t;
    while(t--){
        memset(a,0,sizeof(a));
        int n,m;
        printf("Case %d:\n",++ca);
        scanf("%d%d",&n,&m);
        fill(cnt,cnt+n+1,1);
        for(int i=1;i<=n;i++) p[i]=i;
        while(m--){
            char op[10];
            scanf("%s",op);
            if(op[0]=='T'){
                int x,y;
                scanf("%d%d",&x,&y);
                int tx=find(x),ty=find(y);
                if(tx!=ty){
                    p[tx]=ty;
                    a[tx]=1;
                    cnt[ty]+=cnt[tx];
                }
            }
            else if(op[0]=='Q'){
                int x;
                scanf("%d",&x);
                int tx=find(x);
                printf("%d %d %d\n",tx,cnt[tx],a[x]);
            }
        }
    }
    return 0;
}