Dragon Balls

Dragon Balls

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1665    Accepted Submission(s): 675

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

 

#include <iostream>
using namespace std;
const int MAX=10000;
int total[MAX+10];
int times[MAX+10];
int parent[MAX+10];
int n,q,t;
int GetParent(int a)
{
	int k=parent[a];
	if(parent[a]!=a)
	{
		parent[a]=GetParent(parent[a]);
		times[a]+=times[k];
	}
	return parent[a];
}
void Merge(int a,int b)
{
	int p1=GetParent(a);
	int p2=GetParent(b);
	if(p1==p2)
		return ;
	total[p2]+=total[p1];
	total[p1]=0;
	parent[p1]=p2;
	times[p1]++;
}
void Query(int a)
{
	int city=GetParent(a);
	printf("%d %d %d
",city,total[city],times[a]);
}
int main()
{
	scanf("%d",&t);
	for(int j=1;j<=t;j++)
	{
		printf("Case %d:
",j);
		scanf("%d%d",&n,&q);
		for(int i=1;i<=n;i++)
		{
			total[i]=1;
			times[i]=0;
			parent[i]=i;
		}
		char cmd;
		int a,b;
		for(int i=0;i<q;i++)
		{
			getchar();
			scanf("%c",&cmd);
			if(cmd=='T')
			{
				scanf("%d%d",&a,&b);
				Merge(a,b);
			}
			else 
			{
				scanf("%d",&a);
				Query(a);
			}
		}
	}
	return 0;
}
/**
*
搞了一下午，发现它的模型其实就是个基础的并查集，我想太多了。。注意的是times的求法，一个递归，，每次递归都把我搞得晕头转向。。。
*
**/