Building Block


Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3196    Accepted Submission(s): 957

Problem Description


John are playing with blocks. There are N blocks (1 <= N <= 30000) numbered 1...N。Initially, there are N piles, and each pile contains one block. Then John do some operations P times (1 <= P <= 1000000). There are two kinds of operation:

M X Y : Put the whole pile containing block X up to the pile containing Y. If X and Y are in the same pile, just ignore this command.
C X : Count the number of blocks under block X

You are request to find out the output for each C operation.


 


Input


The first line contains integer P. Then P lines follow, each of which contain an operation describe above.



Output


Output the count for each C operations in one line.


 


Sample Input


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



Sample Output


1 0 2 


/*
2818 带权并查集 
题意：
有N个积木编号为1、、、N，分成N堆，每堆只包含一个积木，然后做P次操作，操作分为2种，
M X Y：把包含X的一堆放到包含Y的一堆上，如果XY同在一堆上，不做处理
C X：计算出X积木下边有多少个积木
每次遇到C操作，输出数量

up[x]表示x上有多少个元素
sum[x]表示这一堆总的数目
*/
#include<iostream>
#include<stdio.h>
using namespace std;
#define N 30010

int pre[N],up[N],sum[N];

void init()
{
  int i;
  for(i=0;i<N;i++)
  {
    pre[i]=i;
    sum[i]=1;
    up[i]=0;
  } 
}

int find(int x)
{ 
  int t;
  if(x!=pre[x])
  {
    t=find(pre[x]);
    up[x]+=up[pre[x]];
    pre[x]=t;
  }
  return pre[x];
}

void merge(int x,int y)
{
  int fx=find(x);
  int fy=find(y);
  if(fx!=fy)
  {
    pre[fy]=fx;
    up[fy]=sum[fx];
    sum[fx]+=sum[fy]; 
  }
}
int main()
{
  int n,a,b;
  char ch[2];
  
//  freopen("test.txt","r",stdin);
  while(scanf("%d",&n)!=EOF)
  {
    init();
    while(n--)
    {
      scanf("%s",ch);
      if(ch[0]=='M')
      {
        scanf("%d%d",&a,&b);
        merge(a,b);
      }
      else
      {
        scanf("%d",&a);
        b=find(a);
        printf("%d\n",sum[b]-up[a]-1);
      } 
    }
  }
}