HDU 3635 Dragon Balls （並查集）

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

這個題的第三個結果搞了我一晚上，第二天早上去醫院的路上想明白了。

題意：一共有n個龍珠和n個城市，第i顆龍珠在第i座城市。下面有兩種操作：
T A B 將A龍珠所在的城市的所有龍珠移動到第B個龍珠所在的城市
Q A 詢問A龍珠所在的城市，這座城市有多少顆龍珠，A龍珠被移動了多少次

思路：詢問的前兩個問題都很好解決，有難度的是第三個問題，龍珠被移動的次數。這個次數的問題，本來如果不適用壓縮路徑的話，次數就是節點到根節點的路徑長度（這個我早上才想明白，嚓），那麼現在使用了壓縮路徑了，就要在壓縮的時候順便把路徑長度也存下來.

#include
#include
#include
#include
#include
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的節點數增加a的節點數
                    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;
}