hdu 2818 Building Block (带权并查集，很优美的题目)

Problem Description

John are playing with blocks. There are N blocks ( <= N <= ) numbered ...N。Initially, there are N piles, and each pile contains one block. Then John do some operations P times ( <= P <= ). 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

M
C
M
M
C
C 

Sample Output

 

Source

2009 Multi-University Training Contest 1 - Host by TJU

 

题目大意：有N个piles（按序编号），每次有两种操作：

 

M x y表示把x所在的那一堆全部移到y所在的那一堆

 

C x 询问在x之下有多少个方块

 

解决方法：使用并查集（路径压缩）实现，然后用count[X]表示X所在的那一堆总共多少个piles，under[x]表示x之下有多少个piles。

 

首先，每次操作我们合并两个集合（如果原来在同一集合中除外），count[X]是每次操作可以直接实现的,就是把两堆的数目相加，很容易（初始值为1）。那么当某次移动操作发生时，首先确定x所在的那一堆最底部的X以及y所在那一堆最底部的Y，那么under[X]的数目就是另外一堆piles的总数count[Y]，有了这个条件，在接下去的操作中，就可以根据FIND（x）递归去一边寻找根一边更新其他未知的under[x]。

 

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 using namespace std;
 #define N 30006
 int fa[N];
 int under[N];//下边的个数
 int cnt[N];//所在堆的堆个数
 void init(){
     for(int i=;i<N;i++){
         fa[i]=i;
         under[i]=;
         cnt[i]=;
     }
 }
 int find(int son){
     if(fa[son]!=son){
         int t=find(fa[son]);
         under[son]+=under[fa[son]];
         fa[son]=t;
     }
     return fa[son];
     //return fa[x]==x?x:fa[x]=find(fa[x]);
 }
 void merge(int x,int y){
     int root1=find(x);
     int root2=find(y);
     if(root1==root2)return;
     under[root1]=cnt[root2];
     cnt[root2]+=cnt[root1];
     fa[root1]=root2;
 }
 int main()
 {
     int n;
     while(scanf("%d",&n)==){
         init();
         char s[];
         int x,y;
         for(int i=;i<n;i++){
             scanf("%s",s);
             if(s[]=='M'){
                 scanf("%d%d",&x,&y);
                 merge(x,y);
             }
             else{
                 scanf("%d",&x);
                 find(x);
                 printf("%d\n",under[x]);
             }
         }
     }
     return ;
 }

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