Example analysis of how to implement knapsack algorithm in Java

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0-1 Knapsack problem

The Knapsack problem is an NP-complete problem of combinatorial optimization. The problem can be described as: given a set of items, each item has its own weight and price, within the limited total weight, how do we choose so that the total price of the items is the highest. The name of the problem comes from how to choose the most appropriate item to place in a given backpack.

This is the most basic backpack problem. Its characteristics are: there is only one item of each type, and you can choose to put it or not.

Use sub-problems to define the state: that is, f[i][v] represents the maximum value that can be obtained by putting the first i items into a backpack with a capacity of v. Then its state transition equation is:

f[i][v]=max{ f[i-1][v], f[i-1][v-w[i]]+v[i ] }.

public class Bag {

  static class Item {// 定义一个物品
    String id; // 物品id
    int size = 0;// 物品所占空间
    int value = 0;// 物品价值

    static Item newItem(String id, int size, int value) {
      Item item = new Item();
      item.id = id;
      item.size = size;
      item.value = value;
      return item;
    }

    public String toString() {
      return this.id;
    }
  }

  static class OkBag { // 定义一个打包方式
    List<Item> Items = new ArrayList<Item>();// 包里的物品集合

    OkBag() {
    }

    int getValue() {// 包中物品的总价值
      int value = 0;
      for (Item item : Items) {
        value += item.value;
      }
      return value;
    };

    int getSize() {// 包中物品的总大小
      int size = 0;
      for (Item item : Items) {
        size += item.size;
      }
      return size;
    };

    public String toString() {
      return String.valueOf(this.getValue()) + " ";
    }
  }

  // 可放入包中的备选物品
  static Item[] sourceItems = { Item.newItem("4号球", 4, 5), Item.newItem("5号球", 5, 6), Item.newItem("6号球", 6, 7) };
  static int bagSize = 10; // 包的空间
  static int itemCount = sourceItems.length; // 物品的数量

  // 保存各种情况下的最优打包方式 第一维度为物品数量从0到itemCount,第二维度为包裹大小从0到bagSize
  static OkBag[][] okBags = new OkBag[itemCount + 1][bagSize + 1];

  static void init() {
    for (int i = 0; i < bagSize + 1; i++) {
      okBags[0][i] = new OkBag();
    }

    for (int i = 0; i < itemCount + 1; i++) {
      okBags[i][0] = new OkBag();
    }
  }

  static void doBag() {
    init();
    for (int iItem = 1; iItem <= itemCount; iItem++) {
      for (int curBagSize = 1; curBagSize <= bagSize; curBagSize++) {
        okBags[iItem][curBagSize] = new OkBag();
        if (sourceItems[iItem - 1].size > curBagSize) {// 当前物品大于包空间.肯定不能放入包中.
          okBags[iItem][curBagSize].Items.addAll(okBags[iItem - 1][curBagSize].Items);
        } else {
          int notIncludeValue = okBags[iItem - 1][curBagSize].getValue();// 不放当前物品包的价值
          int freeSize = curBagSize - sourceItems[iItem - 1].size;// 放当前物品包剩余空间
          int includeValue = sourceItems[iItem - 1].value + okBags[iItem - 1][freeSize].getValue();// 当前物品价值+放了当前物品后剩余包空间能放物品的价值
          if (notIncludeValue < includeValue) {// 放了价值更大就放入.
            okBags[iItem][curBagSize].Items.addAll(okBags[iItem - 1][freeSize].Items);
            okBags[iItem][curBagSize].Items.add(sourceItems[iItem - 1]);
          } else {// 否则不放入当前物品
            okBags[iItem][curBagSize].Items.addAll(okBags[iItem - 1][curBagSize].Items);
          }
        }

      }
    }
  }

  public static void main(String[] args) {
    Bag.doBag();
    for (int i = 0; i < Bag.itemCount + 1; i++) {// 打印所有方案中包含的物品
      for (int j = 0; j < Bag.bagSize + 1; j++) {
        System.out.print(Bag.okBags[i][j].Items);
      }
      System.out.println("");
    }

    for (int i = 0; i < Bag.itemCount + 1; i++) {// 打印所有方案中包的总价值
      for (int j = 0; j < Bag.bagSize + 1; j++) {
        System.out.print(Bag.okBags[i][j]);
      }
      System.out.println("");
    }

    OkBag okBagResult = Bag.okBags[Bag.itemCount][Bag.bagSize];
    System.out.println("最终结果为:" + okBagResult.Items.toString() + okBagResult);

  }

}

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