poj 1988

Cube Stacking

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 11443		Accepted: 3804
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

Source

USACO 2004 U S Open

 

解题思路：

题意看了一会，有些拗口。大致意思： 有几个stack，初始里面有一个cube。支持两种操作：1.move x y： 将x所在的stack移动到y所在stack的顶部。2.count x：数在x所在stack中，在x之下的cube的个数。

加入了count和deep数组。count，set所有的cube的个数。 deep：cube在stack所处的位置。

关系需要推倒以下，关系推导错误，wa了几次。

#include <iostream>
#include <cstdio>

using namespace std;

const int N=30010;

int parent[N+1];
int deep[N+1];
int count[N+1];

void createSet(int x)
{
    parent[x]=x;
    deep[x]=0;
    count[x]=1;
}

int find(int x)
{
    int temp;

    if(x!=parent[x])
    {
        temp=parent[x];
        parent[x]=find(parent[x]);
        deep[x]+=deep[temp];
    }
    return parent[x];
}

void unionSet(int a, int b)
{
    int pa, pb;

    pa=find(a);
    pb=find(b);

    parent[pb]=pa;
    deep[pb]=count[pa];
    count[pa]+=count[pb];
}

int main()
{
    int p;
    int a,b;
    char op;
    int pa,pb;
    int cnt;

    while(scanf("%d",&p)!=EOF)
    {
        for(int i=1; i<=N; i++)
        {
            createSet(i);
        }
        while(p--)
        {
            scanf("\n%c ",&op);
            if(op=='M')
            {
                scanf(" %d %d",&a,&b);
                pa=find(a);
                pb=find(b);
                if(pa!=pb)
                {
                    unionSet(a,b);
                }
            }
            else if(op=='C') //op = C
            {
                scanf(" %d",&a);
                cnt=0;
                pa=find(a);
                cnt=count[pa]-deep[a]-1;
                printf("%d\n",cnt);
            }
        }
    }

    return 0;
}