C++巧克力拼图

C++ Chocolate Puzzle

本文关键字：拼图 巧克力 C++      更新时间：2023-10-16

假设我有N块巧克力，它们必须按照到货顺序准确地装进p个盒子。每个巧克力也有一定数量的卡路里X，每个盒子的容量K必须小于或等于3*sum（x1，x2，…，xn）+max。

在这项任务中，我得到了每种巧克力的N、p和X，我必须找出尽可能低的K。有人能帮我吗？

示例：N＝8，P＝3，X＝{1，4，5，6，3，2，5，3}

K for first three chocolates = 3*(1+4+5) + 5^2 - 1^2 = 54
K for next two chocolates = 3*(6+3) + 6^2 - 3^2 = 54
K for last three chocolates = 3*(2+5+3) + 5^2 - 2^2  = 51
Lowest possible K = 54

因此，目标是使用K 最低的p盒来找到最佳组合

谢谢！

以下是我在Java中解决此问题的方法：

import java.util.HashMap;
import java.util.Map;
import java.util.Random;
public class ChocolatePuzzle {
    private static final Map <String, Integer> solutions =
        new HashMap <String, Integer> ();
    private static final Map <String, Integer> bestMoves =
            new HashMap <String, Integer> ();
    private static int [] x;
    private static int k (int from, int to)
    {
        int sum = x [from];
        int max = x [from];
        int min = x [from];
        for (int i = from + 1; i < to; i++)
        {
            sum += x [i];
            max = Math.max (max, x [i]);
            min = Math.min (min, x [i]);
        }
        return sum * 3 + max * max - min * min;
    }
    public static int solve (int n, int p)
    {
        String signature = n + "," + p;
        Integer solution = solutions.get (signature);
        if (solution == null)
        {
            solution = Integer.valueOf (doSolve (n, p, signature));
            solutions.put (signature, solution);
        }
        return solution.intValue ();
    }
    public static int doSolve (int n, int p, String signature)
    {
        if (p == 1)
        {
            bestMoves.put (signature, Integer.valueOf (x.length - n));
            return k (n, x.length);
        }
        else
        {
            int result = Integer.MAX_VALUE;
            int bestMove = 0;
            int maxI = x.length - n - p + 1;
            for (int i = 1; i <= maxI; i++)
            {
                int k = Math.max (k (n, n + i), solve (n + i, p - 1));
                if (k < result)
                {
                    result = k;
                    bestMove = i;
                }
            }
            bestMoves.put (signature, Integer.valueOf (bestMove));
            return result;
        }
    }
    public static void main(String[] args) {
        int n = 20;
        int p = 5;
        x = new int [n];
        Random r = new Random ();
        for (int i = 0; i < n; i++)
            x [i] = r.nextInt (9) + 1;
        System.out.println("N: " + n);
        System.out.println("P: " + p);
        System.out.print("X: {");
        for (int i = 0; i < n; i++)
        {
            if (i > 0) System.out.print (", ");
            System.out.print (x [i]);
        }
        System.out.println("}");
        System.out.println();
        int k = solve (0, p);
        int o = 0;
        for (int i = p; i > 0; i--)
        {
            int m = bestMoves.get (o + "," + i);
            System.out.print ("{");
            for (int j = 0; j < m; j++)
            {
                if (j > 0)
                    System.out.print (", ");
                System.out.print (x [o + j]);
            }
            System.out.print ("} (k: ");
            System.out.print(k (o, o + m));
            System.out.println (")");
            o += m;
        }
        System.out.println("min(k): " + k);
    }
}

也许您可以在这段代码中找到一些有用的提示。

样本输入：

N: 20
P: 5
X: {1, 7, 6, 6, 5, 5, 7, 9, 1, 3, 9, 5, 3, 7, 9, 1, 4, 2, 4, 8}

样本输出：

{1, 7, 6, 6} (k: 108)
{5, 5, 7, 9} (k: 134)
{1, 3, 9, 5} (k: 134)
{3, 7, 9} (k: 129)
{1, 4, 2, 4, 8} (k: 120)
min(k): 134
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