Cube Stacking P0J 1988（加权并查集）

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 Outpu1

0
2

题目意思：有n块方块，p次操作，两种类型，'M a b' 意思是将含有方块a的堆挪到b上：‘C a’意思是查询方块a下面有几个方块。
解题思路：这道题要使用并查集来进行计算。首先为什么会考虑使用并查集呢？因为每一次操作都是把含有a的堆挪到b上，这个时候将会把b包含到这个堆里面，重新组合成一个新的堆，这是什么？状态划分。这道题的难点主要在于虽然在了同一个集合中，但要询问这个集合内的关系，查询a下面几个方块，就可以看成其在集合中的深度。sum[n]数组来统计当前堆n中方块个数。under[n]数组来统计当前n下面的方块个数。在进行计算的时候，路径压缩和合并均更新under的值。未进行路径压缩的时候，方块n所记录的under并非final value， 仅仅只是包含了到root[n]的方块个数。 通过路径压缩的过程，不断的增加当前n的根节点（递归）的under值，求出最终的under值。进行路径合并的时候，更新sum值和under值。当一个堆放在另一个堆上时，被移动的堆的under值一定为0, 此时更新under值为另一个堆的根的sum值，即计算出了此处的部分under值。然后更新合并堆的sum值。

#include<cstdio>
#include<cstring>
#define maxs 30010
using namespace std;
int pre[maxs];
int sum[maxs];
int under[maxs];
void init()
{
    int i;
    for(i=0; i<maxs; i++)
    {
        pre[i]=i;
        under[i]=0;
        sum[i]=1;
    }
}
int Find(int x)
{
    int t;
    if(x==pre[x])
    {
        return x;
    }
    t=Find(pre[x]);
    under[x]+=under[pre[x]];
    pre[x]=t;
    return pre[x];
}
void Union(int x,int y)
{
    int a=Find(x);
    int b=Find(y);
    if(a==b)
    {
        return ;
    }
    pre[a]=b;
    under[a]=sum[b];///将a堆放在b堆之上,更新此时a堆下面的数量
    sum[b]+=sum[a];///将a堆和b堆合为一个整体，更新整体新堆的数量
}

int main()
{
    int n;
    char q;
    int a,b;
    scanf("%d",&n);
    getchar();
    init();
    while(n--)
    {
        scanf("%c",&q);
        if(q=='M')
        {
            scanf("%d%d",&a,&b);
            getchar();
            Union(a,b);
        }
        else if(q=='C')
        {
            scanf("%d",&a);
            getchar();
            b=Find(a);
            printf("%d
",under[a]);
        }
    }
    return 0;
}