hdu 2818 Building Block

Building Block

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1982    Accepted Submission(s): 598

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

题目大意:有n个blocks,定义两种操作，一种是M x y 将block x加入到block y 下面，包括x的下面所有blocks。另一个是C x，计算block  x 下面block 的数量并输出来。

分析：这道题可以通过并查集来解决，首先每个block的跟自己归在一类，，当遇到M操作时，将x 集合加入到y集合下面，并将y的数量加上x的数量，当碰到C 操作时，只要输出相应的数量即可。

#include<iostream>
#include<cstdio>
#include<cstring>
#define maxlen 30010
using namespace std;
int father[maxlen],all[maxlen],sum[maxlen];
void Init()
{
    for(int i=0; i<maxlen; ++i)
    {
        father[i]=i;
        all[i]=1;
        sum[i]=0;
    }
}

int Find(int x)
{
    if(x == father[x])
        return father[x];
    int temp=Find(father[x]);
    sum[x]+=sum[father[x]];
    father[x]=temp;
    return father[x];
}

void Union(int x,int y)
{
    int a=Find(x);
    int b=Find(y);
    if(a==b)
        return;
    father[a]=b;
    sum[a]+=all[b];
    all[b]+=all[a];
}

int main()
{
    int n,a,b;
    char s;
    Init();
    scanf("%d",&n);
    while(n--)
    {
        getchar();
        scanf("%c",&s);
        if(s == 'M')
        {
            scanf("%d%d",&a,&b);
            Union(a,b);
        }
        else
        {
            scanf("%d",&a);
            Find(a);
            printf("%d
",sum[a]);
        }
    }
    return 0;
}