poj 1988 Cube Stacking（带权并查集）

Cube Stacking

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 19073		Accepted: 6644
Case Time Limit: 1000MS

Description

Farmer John and Betsy are playing a game with N (1 <= N <= 30,000)identical cubes labeled 1 through N. They start with N stacks, each containing a single cube. Farmer John asks Betsy to perform P (1<= P <= 100,000) operation. There are two types of operations: 
moves and counts. 
* In a move operation, Farmer John asks Bessie to move the stack containing cube X on top of the stack containing cube Y. 
* In a count operation, Farmer John asks Bessie to count the number of cubes on the stack with cube X that are under the cube X and report that value. 

Write a program that can verify the results of the game. 

Input

* Line 1: A single integer, P 

* Lines 2..P+1: Each of these lines describes a legal operation. Line 2 describes the first operation, etc. Each line begins with a 'M' for a move operation or a 'C' for a count operation. For move operations, the line also contains two integers: X and Y.For count operations, the line also contains a single integer: X. 

Note that the value for N does not appear in the input file. No move operation will request a move a stack onto itself. 

Output

Print the output from each of the count operations in the same order as the input file. 

Sample Input

6
M 1 6
C 1
M 2 4
M 2 6
C 3
C 4

Sample Output

1
0
2

题意： John和Betsy用N个完全相同的立方体玩游戏，John让Betsy完成P步操作，操作分为两种：

           1.移动操作（ M ），移动含立方体 X 所在的堆加到 Y 所在的堆上头；

           2.计数操作（ C ），计算立方体 X 下的立方体的个数，并输出；

分析：带权并查集。用三个数组，一个数组储存该树的总长度；另外一个数组储存该立方体X离根节点的长度；            还有一个数组储存立方体X的根节点；

#include <stdio.h>
#include <string.h>

int father[30010];//储存根节点
int dis[30010];   //记录以该节点为根的树的总节点数
int cnt[30010];   //记录该节点与根节点的距离
                  

int find(int x)//查根并加权
{
	
	if(x!=father[x])
	{
	  int ans=father[x];
	  father[x]=find(father[x]);
	  cnt[x]=cnt[x]+cnt[ans];
	  //x与x的父亲的距离加上
	  //现在父亲与其“顶头上司”的距离
	  
	}
	return father[x];
}

int main()
{
	int n;
	char a[2];
	int b,c,x,y;
	   scanf("%d",&n);
  		for(int i=0;i<=30000;i++)//初始化
		{
			father[i]=i;
			dis[i]=1;
			cnt[i]=0;
		}
		while(n--)
		{
			scanf("%s",&a);
			if(a[0]=='M')
			{
              scanf("%d%d",&b,&c);
              x=find(b);
              y=find(c);
              if(x!=y)//合并树
              {
              	father[y]=x;//将x作为y的根节点
                cnt[y]+=dis[x];//因为x所在的堆要放在y上
                dis[x]+=dis[y];//将y上的节点数赋给x（进行更新）
              }
			}
			else 
			{
               scanf("%d",&b);
               x=find(b);
               printf("%d\n",dis[x]-cnt[b]-1);
			}
		}
	return 0;
}