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
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]。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 using namespace std; 5 #define N 30006 6 int fa[N]; 7 int under[N];//下边的个数 8 int cnt[N];//所在堆的堆个数 9 void init(){ 10 for(int i=0;i<N;i++){ 11 fa[i]=i; 12 under[i]=0; 13 cnt[i]=1; 14 } 15 } 16 int find(int son){ 17 if(fa[son]!=son){ 18 int t=find(fa[son]); 19 under[son]+=under[fa[son]]; 20 fa[son]=t; 21 } 22 return fa[son]; 23 //return fa[x]==x?x:fa[x]=find(fa[x]); 24 } 25 void merge(int x,int y){ 26 int root1=find(x); 27 int root2=find(y); 28 if(root1==root2)return; 29 under[root1]=cnt[root2]; 30 cnt[root2]+=cnt[root1]; 31 fa[root1]=root2; 32 } 33 int main() 34 { 35 int n; 36 while(scanf("%d",&n)==1){ 37 init(); 38 char s[3]; 39 int x,y; 40 for(int i=0;i<n;i++){ 41 scanf("%s",s); 42 if(s[0]=='M'){ 43 scanf("%d%d",&x,&y); 44 merge(x,y); 45 } 46 else{ 47 scanf("%d",&x); 48 find(x); 49 printf("%d ",under[x]); 50 } 51 } 52 } 53 return 0; 54 }
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