HDOJ 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): 4741    Accepted Submission(s): 1799

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： 读懂这道题真不容易啊。先说每个城市和龙珠都有编号，对应的第i个龙珠放在第i个城市。T A B表示把A号龙珠所在的城市的所有龙珠全部搬运到B号龙珠所在的城市。 Q A表示要求出X（第A号龙珠所在的城市编号），Y

（第X号城市存放的龙珠个数），Z（第A号龙珠被搬运的次数）。

题解： 不管怎样搬运龙珠，拥有龙珠的城市必然有一个和城市编号相对应的龙珠，因此可以将其作为父节点（龙珠所在城市编号）。根据题意无论哪个节点移动时，都是整棵树移动，所以直接可以用一个数组citynum[]表示城市存放的龙珠数，tranum[]表示龙珠移动的次数。

具体代码如下：

#include<cstdio>
#include<cstring>
#define maxn 10010
int tree[maxn];
int citynum[maxn];//记录每个城市的龙珠个数 
int tranum[maxn];//记录每个龙珠移动的次数 

int find(int x)//while循环的路径压缩不知道怎么处理移动次数，无奈递归了 
{
	int t;
	if(x!=tree[x])
	{
		int t=tree[x];
		tree[x]=find(tree[x]);
		tranum[x]+=tranum[t];//加上父节点的移动次数的次数 
	}
	return tree[x];
} 

void megre(int x,int y)
{
	int fx=find(x);
	int fy=find(y);
	if(fx!=fy)
	{
		tree[fx]=fy;
		citynum[fy]+=citynum[fx];//把子节点的龙珠全部给父节点 
		tranum[fx]++;//转移次数加一 
	}
}

int main()
{
	int T,k=1,n,q,i,a,b;
	char s[3];
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d%d",&n,&q);
		for(i=1;i<=n;++i)
		{
			tree[i]=i;
			citynum[i]=1;
			tranum[i]=0;
		}
		printf("Case %d:\n",k++);
		while(q--)
		{
			scanf("%s",s);
			if(s[0]=='T')
			{
				scanf("%d%d",&a,&b);
				megre(a,b);
			}
			else
			{
				scanf("%d",&a);
				int t=find(a);
				printf("%d %d %d\n",t,citynum[t],tranum[a]);
			}
		}
	}
	return 0;
}