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

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

 

Sample Output

102

 

 

#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;}