改进的背包算法

Question

我有不同类别的物品。每个项目都有value 和weight。

例如：

ClassA: [A1, A2, A3]

ClassB: [B1, B2, B3]

ClassC: [C1, C2, C3]

我应该如何修改经典的 0-1 背包问题，以便算法优化解决方案以最大化整体价值，考虑类中的所有项目但允许从一个类中最多选择一项？

package knapsack;

import java.util.ArrayList;
import java.util.List;

public class Knapsack {

    private int totalNumberOfItems;
    private int maxmimumKnapsackCapacityUnits;

    private double[][] optimum;
    private boolean[][] solution;

    private double [] value;
    private int [] weight;

    public Knapsack(int knapsackCapacityUnits, List<KnapsackItem> items){

        this.totalNumberOfItems = items.size();
        this.maxmimumKnapsackCapacityUnits = knapsackCapacityUnits;

        this.optimum = new double[items.size() + 1][knapsackCapacityUnits + 1];
        this.solution = new boolean[items.size() + 1][knapsackCapacityUnits + 1];

        this.value = new double[items.size() + 1];
        this.weight = new int[items.size() + 1];

        int index=1;
        for(KnapsackItem item : items){
            value[index] = item.value;
            weight[index] = item.weight;
            index++;
        }


}

public List<KnapsackItem> optimize(){

    for(int currentItem = 1; currentItem <= totalNumberOfItems; currentItem++){
        for(int currentWeightUnit = 1; currentWeightUnit <= maxmimumKnapsackCapacityUnits; currentWeightUnit++){

            double option1 = optimum[currentItem - 1][currentWeightUnit];

            double option2 = Integer.MIN_VALUE;

            if(weight[currentItem] <= currentWeightUnit){
                option2 = value[currentItem] + optimum[currentItem-1][currentWeightUnit - weight[currentItem]];
            }

            optimum[currentItem][currentWeightUnit] = Math.max(option1, option2);
            solution[currentItem][currentWeightUnit] = (option2 > option1);
        }
    }

    boolean take [] = new boolean[totalNumberOfItems + 1];
    for(int currentItem = totalNumberOfItems,
            currentWeightUnit = maxmimumKnapsackCapacityUnits; 
            currentItem > 0; currentItem --){
        if(solution[currentItem][currentWeightUnit]){
            take[currentItem] = true;
            currentWeightUnit = currentWeightUnit - weight[currentItem];
        }
        else{
            take[currentItem] = false;
        }
    }

    List<KnapsackItem> items = new ArrayList<KnapsackItem>();
    for(int i = 0; i < take.length; i++){
        KnapsackItem newItem = new KnapsackItem();
        newItem.value = value[i];
        newItem.weight = weight[i];
        newItem.isTaken = take[i];
        items.add(newItem);
    }

    return items;
}
}

谢谢！

Answer 1

经典算法是这样的：

for i in items:
    for w in possible total weights (downwards):
        if w is achievable with maximum value v:
            (w + weight[i]) is also achievable with at least value (v + value[i])

这里的方法将是一个轻微的变化：

for c in classes:
    for w in possible total weights (downwards):
        if w is achievable with maximum value v:
            for i in items of class c:
                (w + weight[i]) is also achievable with at least value (v + value[i])

---

使用您的代码，更改将如下：

1. 也许您会想要为每个类制作一个单独的项目列表。根据目前所做的，我希望 value 和 weight 成为列表列表，并且一些变量和数组命名为 numberOfClasses 和 numberOfClassItems 来监视新列表的长度。
   例如，假设两个 1 类项目是 (w=2,v=3) 和 (w=3,v=5)，三个 2 类项目是 (w=1,v=1), (w=4, v=1) 和 (w=1,v=4)。然后我们将有：
   totalNumberOfItems = 5,
   numberOfClasses = 2,
   numberOfClassItems = [2, 3],
   value = [[3, 5], [1, 1, 4]] 和
   weight = [[2, 3], [1, 4, 1]]。
   也就是说，如果您从0 索引。像您一样从1 索引将在每个列表的开头留下未使用的零或空列表。

2. for (currentItem) 循环将变为for (currentClass) 循环。数组optimum 和solution 将被currentClass 索引而不是currentItem。

3. option2 的值实际上将被计算为几个选项中的最佳值，每个类项一个： double option2 = Integer.MIN_VALUE; for (currentItem = 1; currentItem <= numberOfClassItems[currentClass]; currentItem++){ if(weight[currentClass][currentItem] <= currentWeightUnit){ option2 = Math.max (option2, value[currentClass][currentItem] + optimum[currentClass - 1][currentWeightUnit - weight[currentClass][currentItem]]); } }

1234563带上option1，不要使用此类的任何物品。

Answer 2

解决此问题的方法是将类 A,B,C 视为项目本身，然后继续在各个类中进行选择以从中选出最好的。

number of items = number of classes = N
Knapsack capacity  = W
Let item[i][k] be kth item of ith class.

简单的 DP 解决方案通过对背包解决方案的简单修改：-

int DP[n][W+1]={0};

//base condition for item = 0

for(int i=0;i<=W;i++) {

   for(int j=0;j<item[0].size();j++) {
         if(i>=item[0][j].weight) {
              DP[0][i] = max(DP[0][i],item[0][j].value);
         }
   }

} 

//Actual DP 

for(int k=0;k<=W;k++) {

  for(int i=0;i<n;i++) {  

    for(int j=0;j<item[i].size();j++) {
         if(k>=item[i][j].weight) {
              DP[i][k] = max(DP[i][k],DP[i-1][k-item[i][j].weight]+item[i][j].value);
         }
     }

     DP[i][k] = max(DP[i][k],DP[i-1][k]);

  }

}

print("max value : ",DP[n-1][W]);