POJ 1988 Cube Stacking

题意：有编号为1~N的N个小木块，有两种操作

1. M x y 将木块x所在的堆放到木块y所在的堆的上面
2. C x 询问木块x下面有多少块木块

代码巧妙就巧妙在GetParent函数中在进行路径压缩的同时，也计算好了该木块对应的under值

这个需要好好体会

 //#define LOCAL
 #include <iostream>
 #include <cstdio>
 #include <cstring>
 using namespace std;

 const int maxn =  + ;
 int parent[maxn], under[maxn], sum[maxn];

 int GetParent(int a)
 {
     if(parent[a] == a)
         return a;
     int t = GetParent(parent[a]);
     under[a] += under[parent[a]];
     parent[a] = t;
     return t;
 }

 void Merge(int a, int b)
 {
     int pa = GetParent(a);
     int pb = GetParent(b);
     if(pa == pb)    return;
     parent[pb] = pa;
     under[pb] = sum[pa];
     sum[pa] += sum[pb];
 }

 int main(void)
 {
     #ifdef LOCAL
         freopen("1988in.txt", "r", stdin);
     #endif

     for(int i = ; i < maxn; ++i)
     {
         parent[i] = i;
         under[i] = ;
         sum[i] = ;
     }
     int n;
     scanf("%d", &n);
     getchar();
     for(int i = ; i < n; ++i)
     {
         char p;
         scanf("%c", &p);
         if(p == 'M')
         {
             int a, b;
             scanf("%d%d", &a, &b);
             getchar();
             Merge(b, a);
         }
         else
         {
             int a;
             scanf("%d", &a);
             getchar();
             GetParent(a);
             printf("%d\n", under[a]);
         }
     }
     return ;
 }

代码君