HDU - 2818

Building Block

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3563    Accepted Submission(s): 1072

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

 

Recommend

gaojie

/***
题意：给出一些数，并给对这些数进行操作；
          'M a b'代表把a堆加到b堆上面，
          ‘C a ' 代表查询当前点a下面有多少个
做法：并查集。
***/
#include<iostream>
#include<string.h>
#include<algorithm>
#include<stdio.h>
#include<cmath>
using namespace std;
#define maxn  300000 + 3000
int fa[maxn];                 ///记录父节点
int sum[maxn];            ///记录当前堆的总和
int under[maxn];      ///带表当前节点下有多少pile
int n;
void Init()
{
    for(int i=; i<=n; i++)
    {
        sum[i] = ;
        fa[i] = i;
    }
    memset(under,,sizeof(under));
}
int Find(int u)
{
    int tmp;
    if(u != fa[u])
    {
        tmp = Find(fa[u]);
        under[u] +=  under[fa[u]];
        fa[u] = tmp;
    }
    return fa[u];
}
void Union(int x,int y)
{
    int X,Y;
    X = Find(x);
    Y = Find(y);
    if (X!=Y)
    {
        under[X] = sum[Y];             //X是当前堆(集合)中最底部的，直接更新under[]
        sum[Y] += sum[X];            //直接更新Y这堆(集合)的高度(总共多少个Piles)
        fa[X] = Y;                   //合并
    }
}
int main()
{
    while(~scanf("%d",&n))
    {
        Init();
        char ch[];
        int u,v,w;
        while(n--)
        {
            scanf("%s",ch);
            if(ch[] == 'M')
            {
                scanf("%d %d",&u,&v);
                Union(u,v);
            }
            else
            {
                scanf("%d",&w);
                Find(w);
                printf("%d\n",under[w]);
            }
        }
    }
    return ;
}