Dragon Balls

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4384    Accepted Submission(s):
1673

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

题意：遇到T时输入两个数x，y代表把x城市的龙珠转移到y城市，遇到Q输入a  求第a个龙珠所在的城市b，求b城市的龙珠总数，求

第a个龙珠的转移次数

题解：最难在于求龙珠的转移次数，当龙珠的父节点转移时，龙珠也跟着转移

#include<stdio.h>
#include<string.h>
#define MAX 20000
int set[MAX];
int path[MAX];
int time[MAX];
int find(int fa)
{
	int t;
    if(fa==set[fa])
    return fa;
     t=set[fa];
    set[fa]=find(set[fa]);
    time[fa]+=time[t];
    return set[fa];
}
void mix(int x,int y)
{
	int fx;
	int fy;
	fx=find(x);
	fy=find(y);
	if(fx!=fy)
	{
	    set[fx]=fy;
	    path[fy]+=path[fx];
	    time[fx]++;
	}
}
int main()
{
	int t,n,m,x,y,b,i;
	char a;
	scanf("%d",&t);
	int k=0;
	while(t--)
	{
		scanf("%d%d",&n,&m);
		printf("Case %d:\n",++k);
		for(i=1;i<=n;i++)
		{
			set[i]=i;
			path[i]=1;
			time[i]=0;
		}
		while(m--)
		{
			getchar();
			scanf("%c %d",&a,&x);
			if(a==‘T‘)
			{
				scanf("%d",&y);
				mix(x,y);
			}
			else
			{
				y=find(x);//此处必须用一个变量值来表示find（x）
				printf("%d %d %d\n",y,path[y],time[x]);
			}
		}
	}
	return 0;
}