POJ 1988 Cube Stacking（并查集）

最新推荐文章于 2022-03-03 13:43:07 发布

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#人工智能

本文介绍了一个名为CubeStacking的游戏算法实现，该游戏涉及立方体的堆叠与计数操作。通过描述游戏规则及提供具体代码实现，展示了如何处理立方体的移动与计数操作，以验证游戏结果。

Cube Stacking

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 15900		Accepted: 5419
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

Sample Output

1
0
2

Source

USACO 2004 U S Open

/*
POJ 1988
*/
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <string.h>
using namespace std;
const int MAXN=30010;
int F[MAXN];
int num[MAXN];//堆的数目
int val[MAXN];//下面的个数
int find(int x)
{
    if(F[x]==-1)return x;
    int tmp=find(F[x]);
    val[x]+=val[F[x]];
    return F[x]=tmp;
}
void bing(int u,int v)//将u所在的堆放在v上面，注意顺序
{
    int t1=find(u),t2=find(v);
    if(t1!=t2)
    {
        F[t1]=t2;
        val[t1]=num[t2];
        num[t2]+=num[t1];
    }
}
int main()
{
    int P;
    int u,v;
    char op[10];
    while(scanf("%d",&P)==1)
    {
        for(int i=0;i<MAXN;i++)
        {
            F[i]=-1;
            val[i]=0;
            num[i]=1;
        }
        while(P--)
        {
            scanf("%s",&op);
            if(op[0]=='C')
            {
                scanf("%d",&u);
                find(u);
                printf("%d\n",val[u]);
            }
            else
            {
                scanf("%d%d",&u,&v);
                bing(u,v);
            }
        }
    }
    return 0;
}