题目：
| Cube Stacking| Time Limit: 2000MS | | Memory Limit: 30000K |
| :--------------------------: | - | ------------------------ |
| Total Submissions: 24296 | | Accepted: 8517 |
| Case Time Limit: 1000MS | | |DescriptionFarmer John and Betsy are playing a game with N (1 <= N <= 30,000)identical cubes labeled 1 through N. They start with N stacks, each containing a single cube. Farmer John asks Betsy to perform P (1<= P <= 100,000) operation. There are two types of operations: moves and counts. * In a move operation, Farmer John asks Bessie to move the stack containing cube X on top of the stack containing cube Y. * In a count operation, Farmer John asks Bessie to count the number of cubes on the stack with cube X that are under the cube X and report that value. Write a program that can verify the results of the game. Input* Line 1: A single integer, P * Lines 2..P+1: Each of these lines describes a legal operation. Line 2 describes the first operation, etc. Each line begins with a 'M' for a move operation or a 'C' for a count operation. For move operations, the line also contains two integers: X and Y.For count operations, the line also contains a single integer: X. Note that the value for N does not appear in the input file. No move operation will request a move a stack onto itself. OutputPrint the output from each of the count operations in the same order as the input file. Sample Input6 M 1 6 C 1 M 2 4 M 2 6 C 3 C 4 Sample Output```
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代码：
```cpp
#include <stdio.h>
#include <string.h>
#include <stack>
#include <queue>
#include <algorithm>
#define mem(a,b) memset(a,b,sizeof(a))
#define shu 20
using namespace std;
int n,fa[shu],r[shu],mx[shu];
//r[x]表示x到根节点的距离，mx[x]表示集合中有多少个盒子
void init()
{
int i;
for(i=1; i<shu; i++)
{
fa[i]=i;
r[i]=0;
mx[i]=1;
}
}
int find(int x)//查找并查集祖先
{
int fx=fa[x];
if(fa[x]!=x)
{
fx=find(fa[x]);
r[x]+=r[fa[x]];
}
return fa[x]=fx;
}
void u(int x,int y)//将两个集合合并
{
int fx=find(x);
int fy=find(y);
fa[fy]=fx;
r[fy]+=mx[fx];//合并时更改相应的值
mx[fx]+=mx[fy];
}
int main()
{
char op[3];
int i,j;
while(~scanf("%d",&n))
{
init();
while(n--)
{
scanf("%s",op);
if(op[0]=='C')
{
scanf("%d",&i);
int f=find(i);
printf("%d\n",mx[f]-r[i]-1);//该集合盒子个数减去在i盒子之上的个数
}
else
{
scanf("%d%d",&i,&j);
u(i,j);
}
}
}
return 0;
}
x所在集合的元素个数 - 在它上面的个数 - 它本身 就是在它之下的个数
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