HDOJ-----3635带权并查集

Five hundred years later, the number of dragon balls will increase unexpectedly, so it's too difficult for Monkey King(WuKong) to gather all of the dragon balls together.

His country has N cities and there are exactly N dragon balls in the world. At first, for the ith dragon ball, the sacred dragon will puts it in the ith city. Through long years, some cities' dragon ball(s) would be transported to other cities. To save physical strength WuKong plans to take Flying Nimbus Cloud, a magical flying cloud to gather dragon balls.
Every time WuKong will collect the information of one dragon ball, he will ask you the information of that ball. You must tell him which city the ball is located and how many dragon balls are there in that city, you also need to tell him how many times the ball has been transported so far.

 

Input

The first line of the input is a single positive integer T(0 < T <= 100).
For each case, the first line contains two integers: N and Q (2 < N <= 10000 , 2 < Q <= 10000).
Each of the following Q lines contains either a fact or a question as the follow format:
  T A B : All the dragon balls which are in the same city with A have been transported to the city the Bth ball in. You can assume that the two cities are different.
  Q A : WuKong want to know X (the id of the city Ath ball is in), Y (the count of balls in Xth city) and Z (the tranporting times of the Ath ball). (1 <= A, B <= N)

 

Output

For each test case, output the test case number formated as sample output. Then for each query, output a line with three integers X Y Z saparated by a blank space.

 

Sample Input

2
3 3
T 1 2
T 3 2
Q 2
3 4
T 1 2
Q 1
T 1 3
Q 1

 

Sample Output

Case 1:
2 3 0
Case 2:
2 2 1
3 3 2


七龙珠~~~
有好多龙珠，最开始A龙珠在A城，B龙珠在B城。。。类推，，，
第一行一个数T代表测试实例， 第二行两个数N, Q，N代表城市数目,Q代表操作次数，接下来Q行以T开头的
后边两个数A,B,代表把龙珠A所在城市所有龙珠移到龙珠B所在城市
以Q开头，后边输入一个数A，然后输出A在哪个城市(X)，X城市有多少个龙珠（Y），A目前为止移动了几次（Z）



#include<cstdio>
#include<cstring>
int pre[10002], cnt[10002], sum[10002];
struct node{
    int x, y, z;
}s[10002];
int find(int a){
    if(a == pre[a]){
        return a;
    }
    int x = pre[a];
    pre[a] = find(pre[a]);
    sum[a] += sum[x];//a移动次数等于本身移动次数加上父节点移动次数，递归
    return pre[a];
}
void merge(int a, int b){
    int fx = find(a), fy = find(b);
    if(fx != fy){
        pre[fx] = fy;
        cnt[fy] += cnt[fx];//a城市全体龙珠转移到b城，a城清零
        cnt[fx] = 0;
        sum[fx] = 1;//a移动次数加一
    }
}
int main(){
    int t, m, n, a, b, ans, bbs;
    char N;
    scanf("%d", &t);
    ans = 0;
    for(ans = 1; ans <= t; ans++){
        printf("Case %d:\n", ans);
        scanf("%d%d", &m, &n);
        for(int i = 0; i <= m; i++){//初始每个城市一个龙珠，每个龙珠移动次数为零
            pre[i] = i;
            cnt[i] = 1;
            sum[i] = 0;
        }
        bbs = 0;
        while(n--){
            scanf(" %c", &N);
            if(N == 'T'){
                scanf("%d%d", &a, &b);
                merge(a, b);
            }
            else{
                scanf("%d", &a);
                s[bbs].x = find(a);
                s[bbs].y = cnt[find(a)];
                s[bbs++].z = sum[a];
            }
        }
        for(int i = 0; i < bbs; i++){
            printf("%d %d %d\n", s[i].x, s[i].y, s[i].z);
        }
    }
    return 0;
}