？HDU 2818 Building Block 求大神讲解 (并查集)

#include <stdio.h>  #include <iostream>using namespace std;int ff[30005];//ff[x]表示x的父节点 int high[30005];int low[30005];void ii(int n)    //初始化   {  int i;    for(i=0;i<=n;i++)            //从1开始也是错误的{         ff[i]=i;    low[i] = 0;high[i] = 1;    } }    int dd(int x)    //带路径压缩的查找{      if(x==ff[x])return x;int z=dd(ff[x]);low[x] = low[ff[x]]+low[x];return ff[x]=z;     //return z;  则会超时 ？？？} int main()  {      int i,t,k,j,N,m,n,a,b,A,B,x,y;int count,max=0,ans;char ope;    scanf("%d%*c",&N);ii(30001);    while(N--) {   scanf("%c",&ope);if (ope == 'M'){scanf("%d%d%*c",&x,&y);            int dx,dy;dx=dd(x);dy=dd(y);if(dx==dy)continue;ff[dx] = dy;low[dx] = high[dy];high[dy] += high[dx]; }else if(ope == 'C'){            scanf("%d%*c",&x);            dd(x);printf("%d\n",low[x]);}    }      return 0;  } 

分析：  开始时就得考虑将ff[x] 指向y  而不是反过来指，因为后面要将下层的block加起来，还要考虑下层的block是否还有下层，所以要用到low数组来存放，high数组的作用就是将所要求的下层的block数加起来的总和，

问题：对于代码中的dd的return 不是很理解 为什么不能改成 return z 或者 return ff[x]  尽管都能得到正解，可还是会栈溢出 或者 超时

求大神讲解  O(∩_∩)O谢谢

Building Block

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3719    Accepted Submission(s): 1128

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

6M 1 6C 1M 2 4M 2 6C 3C 4

 

Sample Output

102

 

Source

2009 Multi-University Training Contest 1 - Host by TJU