HDU2818:Building Block

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
   
   
   
   

 

 

#include <stdio.h>

int pre[30005];
int high[30005];//以该点为根节点的最大高度
int low[30005];//在该集合中在该点下面的个数

int find(int x)
{
    if(pre[x] == x)
        return x;
    int z = find(pre[x]);
    low[x] = low[pre[x]]+low[x];
    return pre[x] = z;
}

void set()
{
    int i;
    for(i = 0; i<30005; i++)
    {
        pre[i] = i;
        high[i] = 1;
        low[i] = 0;
    }
}

int main()
{
    int p;
    char c;
    scanf("%d%*c",&p);
    set();
    while(p--)
    {
        int x,y;
        scanf("%c",&c);
        if(c == 'M')
        {
            scanf("%d%d%*c",&x,&y);
            int rx,ry;
            rx = find(x);
            ry = find(y);
            if(rx==ry)
                continue;
            pre[rx] = ry;
            low[rx] = high[ry];
            high[ry] = high[rx] + high[ry];
        }
        else if(c == 'C')
        {
            int k,i;
            scanf("%d%*c",&k);
            find(k);
            printf("%d\n",low[k]);
        }

    }

    return 0;
}