hdu3635 Dragon Balls

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

这题考查了并查集的路径压缩，这道题让我对路径压缩有了进一步了解，题意是初始时，有n个龙珠，编号从1到n，分别对应的放在编号从1到n的城市中。2种操作：

T A B，表示把A球所在城市全部的龙珠全部转移到B城市。
Q A，表示查询A。要求得到的信息分别是：A现在所在的城市，A所在城市的龙珠数目，A转移到该城市移动的次数（如果没有移动就输出0）

这里难点是求出询问的这个球转移的次数，这里设一个函数zhuanyi[n],初始化为0，每次移动的时候，这个球的祖先转移次数为1（其实以每个城市为祖先的移动最多只有一次，其他都是跟着自己的祖先移动），然后用并查集递归压缩路径的方法，使得当前这个点加上自己所有祖先的转移次数，然后使自己的祖先变为当前转移的城市，这样就能保证下次不会重复加。

#include<stdio.h>
#include<string.h>
int pre[10006],zhuanyi[10006],num[10006];
char s[10];
int find(int x)
{
int temp;
if(x==pre[x])return x;
temp=pre[x];
pre[x]=find(pre[x]);
zhuanyi[x]+=zhuanyi[temp];
return pre[x];
}

int main()
{
int T,n,m,i,j,a,b,c,t1,t2,num1=0;
scanf("%d",&T);
while(T--)
{
num1++;
printf("Case %d:\n",num1);
scanf("%d%d",&n,&m);
for(i=1;i<=n;i++){
pre[i]=i;zhuanyi[i]=0;num[i]=1;  //pre[]表示球的父亲，zhuangyi[]表示球转移的次数，num[i]表示球i所在城市的球的个数 
}
for(i=1;i<=m;i++){
scanf("%s",s);
if(s[0]=='T'){
scanf("%d%d",&a,&b);
t1=find(a);t2=find(b);
if(t1==t2)continue;
pre[t1]=t2;
num[t2]+=num[t1];
zhuanyi[t1]=1;
}
else if(s[0]=='Q'){
scanf("%d",&a);
t1=find(a);
printf("%d %d %d\n",t1,num[t1],zhuanyi[a]);
}
}
}
return 0;
}