poj 1988 Cube Stacking

Cube Stacking

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 13922		Accepted: 4695
Case Time Limit: 1000MS

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

并查集的题目，但是需要加一个小技巧，用deep来保存该节点在整个一叠冰块中从上到下的位置，用count保存以该节为根时下面的子节点数目
每次更新时需要更新deep的值，而count的值只要更新根节点的那一个就可以了，中间节点不用管。
做这种题尽量用按秩合并的方法，不然很麻烦 还会很慢
就这么稍微变了一下又费了我n久。。最后还是学了网上的代码才过了。。。
cin>>非常慢，差点超时，用scanf好些

#include <iostream>  
#include <cstdio>  
using namespace std;  
const int N=30010;  
int father[N+1];  
int deep[N+1];  
int count[N+1];  
void make_set(int x)  
{  
    father[x]=x;  
    deep[x]=0;  
    count[x]=1;  
}
int find_father(int x)  
{  
    int temp;  
    if(x!=father[x])  
    {  
        temp=father[x];  
        father[x]=find_father(father[x]);  
        deep[x]+=deep[temp];  
    }  
    return father[x];  
}
void union_set(int a, int b)  
{  
    int pa, pb; 
    pa=find_father(a);  
    pb=find_father(b);  
    father[pb]=pa;  
    deep[pb]=count[pa];  
    count[pa]+=count[pb];  
}  
int main()  
{  
    int p;  
    int a,b;  
    char op;  
    int pa,pb;  
    int cnt; 
    freopen("e:\\data.txt", "r", stdin);        
    freopen("e:\\out.txt", "w", stdout); 

    cin>>p;
    for(int i=1; i<=N; i++)  
    {  
        make_set(i);  
    }  
    while(p--)  
    {  
        cin>>op; 
        if(op=='M')  
        {  
            cin>>a>>b;
            pa=find_father(a);  
            pb=find_father(b);  
            if(pa!=pb)  
            {  
                union_set(a,b);  
            }  
            cout<<"count"<<endl;
            for(int i=1;i<10;i++)
                cout<<count[i]<<" ";
            cout<<endl;
            cout<<"deep"<<endl;
            for(int i=1;i<10;i++)
                cout<<deep[i]<<" ";
            cout<<endl;
        }  
        else if(op=='C')
        {  
            cin>>a;
            cnt=0;  
            pa=find_father(a);  
            cnt=count[pa]-deep[a]-1;  
            cout<<cnt<<endl; 
        }  
    } 
    return 0;  
}