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

题目大意：有若干个方块，经p次操作后，在x方块下面的方块有多少个。
M操作—>将包含x方块的堆移到含y方块的堆上
C操作—>输出x方块下方方块的数目

解题思路：以堆最底部的方块作为父节点，dis[a]表示a到父节点的距离，rank[y]表示以y为底部的方块堆的大小，即含有多少个方块，借用find()更新dis。

Code：

#include<iostream>
using namespace std;
int n,x,y,pre[100005],dis[100005],rank[100005];
char ch;//以最底部的为父节点，dis[a]表示a到父节点的距离 ，rank[]表示堆的大小 

int find(int x){
	if(pre[x]==x) return x;
	else {
		int r = pre[x];
		pre[x]=find(r);
		dis[x]+=dis[r];//dis[r] 表示x的父节点到堆底部的大小，dis[x]表示x到父节点的大小 
	} 
	return pre[x];
}
void join(int x,int y){
	int fx=find(x),fy=find(y);
	if(fx!=fy){
		pre[fx]=fy;
		dis[fx]=rank[fy];
		rank[fy]+=rank[fx];
	}
}

int main(){
	while(cin>>n){
		for(int i=1;i<=n;i++) pre[i]=i;
		for(int i=1;i<=n;i++) rank[i]=1,dis[i]=0;
		for(int i=1;i<=n;i++){
			cin>>ch;
			if(ch=='M'){
				cin>>x>>y;
				if(find(x)!=find(y)){
					join(x,y);
				}
			}
			else if(ch=='C'){
				cin>>x;
				find(x);
				cout<<dis[x]<<endl;
			}
		}
	}
	return 0;
}