hdu 2818 Building Block（并查集，有点点复杂）

Building Block

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5811    Accepted Submission(s): 1790

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   |   We have carefully selected several similar problems for you:  2819 2817 2824 2821 2820 

 

题解：

因为要查询x的下面有多少blocks，所以我们在普通并查集的基础上在维护两个域num[x]和under[x]，分别表示x所在堆的大小以及x下面的元素。

在合并的时候，我们分别取x,y的堆的最下面一块，也就是他们的根a,b.a和b相等就不用处理了。如果不相等，那么就让fa[a] = b.而在这之前，我们要维护size和under，所有x原来所在的堆的每个元素的under都要增加num[b],如果全都修改会超时，所以我们之修改under[a],把其它修改放在压缩里面，要查哪一个再更新。同时，为了方便我们只把size存在根上，也就是num[b]+=num[a],num[a] = 0。

在find的时候，我们进行压缩，这时候更新under[x],under[x]+=under[fx]就可以了。

注意：这题一直wa，原因是在刚开始赋值时，应该从0下标开始，然而题目中标号是从1号开始的，不懂。。。。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
using namespace std;
int fa[],under[],num[];
int p;
char ch[];
int findfa(int k)
{
    if (fa[k]==k) return k;
    int fa1=fa[k]; //找到之前的那个父亲，这个父亲的under值在后面find的过程中也会变，所以只要加之前的父亲节点的under就行了。
    fa[k]=findfa(fa[k]);
    under[k]+=under[fa1];
    return fa[k];
}
void uni(int x,int y)
{
    int fx=findfa(x);
    int fy=findfa(y);
    if (fx!=fy)
    {
       under[fx]+=num[fy];
       num[fy]+=num[fx];
       num[fx]=;
       fa[fx]=fy;
    }
    return;
}

int main()
{
        for(int i=;i<;i++)
        { fa[i]=i; under[i]=; num[i]=;}

        scanf("%d",&p);
        for(;p>;p--)
        {
            scanf("%s",&ch);
            if (ch[]=='M')
            {
                int x,y;
                scanf("%d%d",&x,&y);
                uni(x,y);
            } else
            {
                int x;
                scanf("%d",&x);
                findfa(x);
                printf("%d\n",under[x]);
            }
        }

    return ;
}