Cube Stacking

Description
Farmer 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.

Output
Print the output from each of the count operations in the same order as the input file.

Sample Input

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

Sample Output

1
0
2

题解：带权并查集。想要知道一个节点下面有多少个，可以用该节点所在集合的总个数 - 该节点上面的个数 - 自身（就是1）。用up保存上面的数量，sum保存一个集合的总数量就行了。如果不画图理解的话很困难，动手才是硬道理啊。

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

int pre[300005]; //保存前驱
int up[300005]; //该点的上面总共的个数
int sum[300005]; //该集合的总个数

int find(int x)
{
if(x == pre[x]) //根上面没有节点了，up也不需要更新
{
return x;
}
int fa = pre[x];
pre[x] = find(pre[x]);
up[x] += up[fa]; //该节点上面的个数等于以前上面的个数加上父节点上面的个数 (父节点保存在以前中)

return pre[x];
}

int main()
{
int n;
while(scanf("%d",&n) != EOF)
{
for(int i = 0;i <= n;i++)
{
pre[i] = i;
up[i] = 0;
sum[i] = 1;
}
char s[10];
int x,y;
for(int i = 0;i < n;i++)
{
scanf("%s",s);
if(s[0] == 'M')
{
scanf("%d%d",&x,&y);
int l = find(x); //找到跟，随你便更新了x集合的up
int r = find(y);
if(l != r) //不在同一个集合
{
pre[r] = l; //合并
up[r] += sum[l]; //y的集合的up增加x集合的数量
sum[l] += sum[r];//x集合数量增加,y集合变成了x集合
}
}
else
{
scanf("%d",&x);
y = find(x);//此时x可能没有更新up，再说我也需要知道x集合的数量.
printf("%d\n",sum[y] - up[x] - 1);//去掉自己
}
}
}

return 0;
}