POJ 2951 Cake Cutting（动态规划）

最新推荐文章于 2021-02-01 21:45:47 发布

piekey1994

最新推荐文章于 2021-02-01 21:45:47 发布

阅读量717

点赞数

分类专栏： ACM集训-动态规划文章标签：动态规划 POJ 2951 Cake Cutting

ACM集训-动态规划专栏收录该内容

2 篇文章

订阅专栏

本文探讨了如何将一个矩形蛋糕通过指定数量的直线切割分成多个矩形部分，使得最大的那块面积尽可能小。采用记忆化搜索和动态规划的方法，通过枚举切割的方向和位置来寻找最优解。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Description

You are given a rectangular cake of integral dimensions w × h. Your goal is to divide this cake into m rectangular pieces of integral dimensions such that the area of the largest piece is minimal. Each cut must be a straight line parallel to one of the sides of the original cake and must divide a piece of cake into two new pieces of positive area. Note that since a cut divides only a single piece, exactly m−1 cuts are needed.

Examples

If w = 4, h = 4, and m = 4, then the following cuts minimize the area of the largest piece:

However, if w = 4, h = 4, and m = 3, then the following cuts are optimal:

Input

The input test file will contain multiple test cases, each of which consists of three integers w, h, m separated by a single space, with 1 <= w, h,m <= 20 and m <= wh. The end-of-file is marked by a test case with w = h = m = 0 and should not be processed.

Output

For each test case, write a single line with a positive integer indicating the area of the largest piece.

Sample Input

4 4 4
4 4 3
0 0 0

Sample Output

4
6

总之要想办法构造出状态转移方程，这道题是：

记忆化搜索，dp[k][i][j]为将一块i*j的蛋糕切k刀的最大面积的最小值，枚举下一刀的横竖以及切的位置和其中一份的刀数

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int inf=0x3f3f3f3f;
int dp[405][25][25];
int dfs(int m,int w,int h)
{
    if(dp[m][w][h]!=inf)
        return dp[m][w][h];
    else if(m==0)
        return dp[m][w][h]=w*h;
    else
    {
        int ans=inf;
        for(int i=1,lim=w/2+1;i<=lim;i++)
            for(int k=0;k<=m-1;k++)
                ans=min(ans,max(dfs(k,i,h),dfs(m-1-k,w-i,h)));
        for(int i=1,lim=h/2+1;i<=lim;i++)
            for(int k=0;k<=m-1;k++)
                ans=min(ans,max(dfs(k,w,i),dfs(m-1-k,w,h-i)));
        return dp[m][w][h]=ans;
    }
}
int main()
{
    int w,h,m;
    memset(dp,0x3f,sizeof(dp));
    while(scanf("%d%d%d",&w,&h,&m),w||h||m)
        printf("%d\n",dfs(m-1,w,h));
    return 0;
}