Cube Stacking

最新推荐文章于 2022-03-03 13:43:07 发布

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最新推荐文章于 2022-03-03 13:43:07 发布

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分类专栏：并查集

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本文深入解析CubeStacking问题，利用并查集解决游戏策略，结合具体输入输出实例，详细阐述解题过程与算法应用，同时探讨问题背后的数学逻辑与算法优化。

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Cube Stacking

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 17959		Accepted: 6208
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

代码：

#include<iostream>
#include<cstdio>
#define Maxn 30010
using namespace std;

int fa[Maxn],dis[Maxn],height[Maxn];
int findset(int x){
    if(fa[x]!=x){
        int t=fa[x];
        fa[x]=findset(fa[x]);
        dis[x]+=dis[t];
    }
    return fa[x];
}
void unionset(int x,int y){
    x=findset(x),y=findset(y);
    if(x==y) return;
    fa[y]=x;
    dis[y]=height[x];
    height[x]+=height[y];
}
int main()
{
    int p,a,b;
    char s;
    cin>>p;
    for(int i=0;i<Maxn;i++) fa[i]=i,dis[i]=0,height[i]=1;
    for(int i=0;i<p;i++){
        cin>>s;
        if(s=='M') {cin>>a>>b;unionset(a,b);}
        else{
            cin>>a;
            int t=findset(a);
            cout<<height[t]-dis[a]-1<<endl;
        }
    }
    return 0;
}

注：这题又是一道种类并查集，思想和食物链那道类似，关键这题的突破口是想到每个元素与并查集根的距离，通过每堆总结点数，即可换算出需要的结果，维护距离是在路径压缩时，这样可以做的原因，在我看来是，需要用到这点，这点必被查找必然更新。像这种并查集有带向量运算感，因此叫偏移向量也不为过。这题很明显距离是直接可加的，而食物链那题并不明显，那几个公式都要枚举所用情况，总结得到的，而且还带有模运算，但最终的公式回归到了向量运算。

我一直在想这是巧合吗？还是必然……其实我到现在还没完全想明白，但我更偏向巧合，因为食物链那题只有三种关系，并且必须设同类关系为0，否则就没有矢量性，而且巧合的是三种不同类生物构成一个矢量三角形……如果想要推广到n种关系，那么就不止要满足矢量n边形了，最重要的是必须满足加法性质，而且加完一圈可以绕回来，甚至从n点中任取m（m>=2）点，都需满足上述性质，这样才会有偏移矢量。

以上纯属个人看法，不喜勿喷！欢迎大牛前来指正！