E - Cube Stacking

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   

#include<iostream>
#include<cstdio>
using namespace std;
#define ma 31005
int fa[ma],sum[ma],under[ma];//sum用来记录相应堆的方块数量，而under[i]记录方块i下有方块多少块；但在进行合并时under记录的是该堆根结点的下方
int findfa(int a)           //块的数量所以在进行合并时都是1；而不是真正意义上准确的值，还要在调用findfa函数时进行叠加得到最后的值；
{
    if(a==fa[a])return a;
    int t=findfa(fa[a]);
    under[a]+=under[fa[a]];
    fa[a]=t;
    return fa[a];
}
void Merge(int a,int b)
{
    int pa=findfa(a);
    int pb=findfa(b);
    if(pa==pb)return;
    fa[pb]=pa;
    under[pb]=sum[pa];
    sum[pa]+=sum[pb];
}
int main()
{
    int n;///freopen("1.txt","r",stdin);
    while(cin>>n)
    {
        for(int i=1;i<ma;i++)
        {
            fa[i]=i;under[i]=0;sum[i]=1;
        }
        for(int i=0;i<n;i++)
        {
            char ch;cin>>ch;
            if(ch=='M')
            {
                int a,b;cin>>a>>b;
                Merge(b,a);
            }
            else
            {
                int a;cin>>a;
                findfa(a);
                cout<<under[a]<<endl;
            }
        }
    }
}

虽然是个模板题，但并查集真正的难点还是在于如何开设数组，如何根据题目的要求去写自己的集合合并函数与路径压缩函数；