HDU 3635 Dragon Balls （并查集）

Dragon Balls

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3360    Accepted Submission(s): 1303

Problem Description

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

 

Author

possessor WC

 

这个题的第三个结果搞了我一晚上，第二天早上去医院的路上想明白了。

题意：一共有n个龙珠和n个城市，第i颗龙珠在第i座城市。下面有两种操作：
T A B 将A龙珠所在的城市的所有龙珠移动到第B个龙珠所在的城市
Q A 询问A龙珠所在的城市，这座城市有多少颗龙珠，A龙珠被移动了多少次

思路：询问的前两个问题都很好解决，有难度的是第三个问题，龙珠被移动的次数。这个次数的问题，本来如果不适用压缩路径的话，次数就是节点到根节点的路径长度（这个我早上才想明白，嚓），那么现在使用了压缩路径了，就要在压缩的时候顺便把路径长度也存下来.

#include<iostream>
#include<string.h>
#include<stdlib.h>
#include<string>
#include<stdio.h>
using namespace std;
int pre[10010];
int trans[10010];
int ranks[10010];
int union_find(int node)
{
    int temp;
    if(node==pre[node])return node;
    else
    {
        temp=pre[node];
        pre[node]=union_find(pre[node]);
        trans[node]+=trans[temp];
    }
    return pre[node];
}
int main(int argc, char *argv[])
{
//    freopen("3635.in","r",stdin);
    int T,N,Q,a,b;
    scanf("%d",&T);
    char s[10];
    int total=T;
    while(T--)
    {
        scanf("%d %d",&N,&Q);
        printf("Case %d:\n",total-T);
        memset(pre, 0, sizeof(pre));
        memset(trans, 0, sizeof(trans));
        memset(ranks, 0, sizeof(ranks));
        for(int i=1;i<=N;++i){
            pre[i]=i;
            ranks[i]=1;
        }
        while(Q--)
        {
            scanf("%s",s);
            if(s[0]=='T')
            {
                scanf("%d %d",&a,&b);
                int p=union_find(a);
                int q=union_find(b);
                if(p!=q){
                    ranks[q]+=ranks[p];//b的节点数增加a的节点数
                    pre[p]=q;//a的根嫁接到b的根上
                    trans[p]++;
                    ranks[p]=0;
                }
            }
            else
            {
                scanf("%d",&a);
                union_find(a);
                printf("%d %d %d\n",pre[a],ranks[pre[a]],trans[a]);
            }
        }
    }
    return 0;
}