Building Block

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3352    Accepted Submission(s): 1003

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

Source

2009 Multi-University Training Contest 1 - Host by TJU

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<cstdlib>
#include<algorithm>
using namespace std;
#define maxn 300100
int n,fa[maxn],num[maxn],under[maxn];
char s[5];
int find(int x)
{
      if(x==fa[x])
            return x;
      int k=fa[x];
      fa[x]=find(fa[x]);
      under[x]+=under[k];
      return fa[x];
}
void Union(int x,int y)
{
      int fx,fy;
      fx=find(x),fy=find(y);
      if(fx!=fy)
      {
            under[fx]=num[fy];
            num[fy]+=num[fx];
            fa[fx]=fy;
      }
}
int main()
{
      int x,y;
      while(scanf("%d",&n)!=EOF)
      {
            for(int i=0;i<=n;i++)
                  fa[i]=i,num[i]=1,under[i]=0;
            for(int i=0;i<n;i++)
            {
                scanf("%s",s);
                if(s[0]==‘M‘)
                {
                      scanf("%d%d",&x,&y);
                      Union(x,y);
                }
                else if(s[0]==‘C‘)
                {
                      scanf("%d",&x);
                      find(x);
                      printf("%d\n",under[x]);
                }
            }
      }
      return 0;
}