0-1多维背包(0-1 Multidimensional Knapsack)

2019-07-18 10:48发布

所以我想生成只能放置在背包一次(0-1),最大承重能力的算法,将发现的n项(在我的情况4)的最佳组合。 可能更有效地总结,我想将不超过四个独特的项目在我的背包,以使它们的权重小于一定值W同时最大限度地提高它们的总价值。 我第一次尝试和设想是把4音量限制所有项目作为卷1多维背包问题。 但我跑进它不是0-1(意味着无论是在袋或没有)的问题。 然后我试图做一个0-1(界)背包代码多维,但我无法添加音量限制以及0-1的要求。 如何编写一个0-1多维背包问题? 或者我怎么适应代码只持有Ⅴ与所有物品的体积为1卷? 该代码不必须是Java,但这是我到目前为止所。

背包:

package hu.pj.alg;

import hu.pj.obj.Item;
import java.util.*;

public class ZeroOneKnapsack {

    protected List<Item> itemList  = new ArrayList<Item>();
    protected int maxWeight        = 0;
    protected int solutionWeight   = 0;
    protected int profit           = 0;
    protected boolean calculated   = false;

    public ZeroOneKnapsack() {}

    public ZeroOneKnapsack(int _maxWeight) {
        setMaxWeight(_maxWeight);
    }

    public ZeroOneKnapsack(List<Item> _itemList) {
        setItemList(_itemList);
    }

    public ZeroOneKnapsack(List<Item> _itemList, int _maxWeight) {
        setItemList(_itemList);
        setMaxWeight(_maxWeight);
    }

    // calculte the solution of 0-1 knapsack problem with dynamic method:
    public List<Item> calcSolution() {
        int n = itemList.size();

        setInitialStateForCalculation();
        if (n > 0  &&  maxWeight > 0) {
            List< List<Integer> > c = new ArrayList< List<Integer> >();
            List<Integer> curr = new ArrayList<Integer>();

            c.add(curr);
            for (int j = 0; j <= maxWeight; j++)
                curr.add(0);
            for (int i = 1; i <= n; i++) {
                List<Integer> prev = curr;
                c.add(curr = new ArrayList<Integer>());
                for (int j = 0; j <= maxWeight; j++) {
                    if (j > 0) {
                        int wH = itemList.get(i-1).getWeight();
                        curr.add(
                            (wH > j)
                            ?
                            prev.get(j)
                            :
                            Math.max(
                                prev.get(j),
                                itemList.get(i-1).getValue() + prev.get(j-wH)
                            )
                        );
                    } else {
                        curr.add(0);
                    }
                } // for (j...)
            } // for (i...)
            profit = curr.get(maxWeight);

            for (int i = n, j = maxWeight; i > 0  &&  j >= 0; i--) {
                int tempI   = c.get(i).get(j);
                int tempI_1 = c.get(i-1).get(j);
                if (
                    (i == 0  &&  tempI > 0)
                    ||
                    (i > 0  &&  tempI != tempI_1)
                )
                {
                    Item iH = itemList.get(i-1);
                    int  wH = iH.getWeight();
                    iH.setInKnapsack(1);
                    j -= wH;
                    solutionWeight += wH;
                }
            } // for()
            calculated = true;
        } // if()
        return itemList;
    }

    // add an item to the item list
    public void add(String name, int weight, int value) {
        if (name.equals(""))
            name = "" + (itemList.size() + 1);
        itemList.add(new Item(name, weight, value));
        setInitialStateForCalculation();
    }

    // add an item to the item list
    public void add(int weight, int value) {
        add("", weight, value); // the name will be "itemList.size() + 1"!
    }

    // remove an item from the item list
    public void remove(String name) {
        for (Iterator<Item> it = itemList.iterator(); it.hasNext(); ) {
            if (name.equals(it.next().getName())) {
                it.remove();
            }
        }
        setInitialStateForCalculation();
    }

    // remove all items from the item list
    public void removeAllItems() {
        itemList.clear();
        setInitialStateForCalculation();
    }

    public int getProfit() {
        if (!calculated)
            calcSolution();
        return profit;
    }

    public int getSolutionWeight() {return solutionWeight;}
    public boolean isCalculated() {return calculated;}
    public int getMaxWeight() {return maxWeight;}

    public void setMaxWeight(int _maxWeight) {
        maxWeight = Math.max(_maxWeight, 0);
    }

    public void setItemList(List<Item> _itemList) {
        if (_itemList != null) {
            itemList = _itemList;
            for (Item item : _itemList) {
                item.checkMembers();
            }
        }
    }

    // set the member with name "inKnapsack" by all items:
    private void setInKnapsackByAll(int inKnapsack) {
        for (Item item : itemList)
            if (inKnapsack > 0)
                item.setInKnapsack(1);
            else
                item.setInKnapsack(0);
    }

    // set the data members of class in the state of starting the calculation:
    protected void setInitialStateForCalculation() {
        setInKnapsackByAll(0);
        calculated     = false;
        profit         = 0;
        solutionWeight = 0;
    }

} // class

和项目:

package hu.pj.obj;

public class Item {

    protected String name    = "";
    protected int weight     = 0;
    protected int value      = 0;
    protected int bounding   = 1; // the maximal limit of item's pieces
    protected int inKnapsack = 0; // the pieces of item in solution

    public Item() {}

    public Item(Item item) {
        setName(item.name);
        setWeight(item.weight);
        setValue(item.value);
        setBounding(item.bounding);
    }

    public Item(int _weight, int _value) {
        setWeight(_weight);
        setValue(_value);
    }

    public Item(int _weight, int _value, int _bounding) {
        setWeight(_weight);
        setValue(_value);
        setBounding(_bounding);
    }

    public Item(String _name, int _weight, int _value) {
        setName(_name);
        setWeight(_weight);
        setValue(_value);
    }

    public Item(String _name, int _weight, int _value, int _bounding) {
        setName(_name);
        setWeight(_weight);
        setValue(_value);
        setBounding(_bounding);
    }

    public void setName(String _name) {name = _name;}
    public void setWeight(int _weight) {weight = Math.max(_weight, 0);}
    public void setValue(int _value) {value = Math.max(_value, 0);}

    public void setInKnapsack(int _inKnapsack) {
        inKnapsack = Math.min(getBounding(), Math.max(_inKnapsack, 0));
    }

    public void setBounding(int _bounding) {
        bounding = Math.max(_bounding, 0);
        if (bounding == 0)
            inKnapsack = 0;
    }

    public void checkMembers() {
        setWeight(weight);
        setValue(value);
        setBounding(bounding);
        setInKnapsack(inKnapsack);
    }

    public String getName() {return name;}
    public int getWeight() {return weight;}
    public int getValue() {return value;}
    public int getInKnapsack() {return inKnapsack;}
    public int getBounding() {return bounding;}

} // class

Answer 1:

这里是一个通用的实施2点的尺寸(大小和体积),以解决该背包0-1问题。 我用一个矩阵,而不是列表的列表,因为它要容易得多。 下面是也来测试它的主要方法全班。
要添加维度只是增加新的维度矩阵,并添加内循环,检查所有条件。

public class MultidimensionalKnapsack {

    /** The size of the knapsack */
    private static int size;
    /** The volume of the knapsack */
    private static int vol;

    private static class Item {
        public int value;
        public int size;
        public int volume;
        public Item(int v, int w, int vol) {
            value = v;
            size = w;
            volume = vol;
        }
    }

    // Knapsack 0/1 without repetition
    // Row: problem having only the first i items
    // Col: problem having a knapsack of size j
    // Third dimension: problem having a knapsack of volume h
    private static int[][][] dynNoRep;

    private static void noRep(Item[] items) {
        dynNoRep = new int[items.length + 1][size + 1][vol + 1];
        for(int j = 0; j <= size; j++) {
            dynNoRep[0][j][0] = 0;
        }
        for(int i = 0; i <= vol; i++) {
            dynNoRep[0][0][i] = 0;
        }
        for(int i = 0; i <= items.length; i++) {
            dynNoRep[i][0][0] = 0;
        }
        for(int i = 1; i <= items.length; i++)
            for(int j = 0; j <= size; j++) {
                for(int h = 0; h <= vol; h++) {
                    if(items[i - 1].size > j)
                        // If the item i is too big, I  can't put it and the solution is the same of the problem with i - 1 items
                        dynNoRep[i][j][h] = dynNoRep[i - 1][j][h];  
                else {
                    if(items[i - 1].volume > h)
                        // If the item i is too voluminous, I can't put it and the solution is the same of the problem with i - 1 items
                        dynNoRep[i][j][h] = dynNoRep[i - 1][j][h];
                    else {
                        // The item i could be useless and the solution is the same of the problem with i - 1 items, or it could be 
                        // useful and the solution is "(solution of knapsack of size j - item[i].size and volume h - item[i].volume) + item[i].value" 
                        dynNoRep[i][j][h] = Math.max(dynNoRep[i - 1][j][h], dynNoRep[i - 1][j - items[i - 1].size][h - items[i - 1].volume] + items[i - 1].value);
                    }
                }
            }
        }
    }

    public static void main(String[] args) {
        size = 15;
        vol = 12;
        Item[] items = {new Item(2, 4, 1), new Item(1, 5, 4), new Item(6, 3, 9), 
            new Item(3, 3, 19), new Item(7, 2, 7), new Item(1, 2, 6), new Item(2, 1, 2),
            new Item(10, 9, 12), new Item(9, 10, 2), new Item(24, 23, 11)};
        System.out.print("We have the following " + items.length + " items (value, size, volume): ");
        for(int i = 0; i < items.length; i++)
            System.out.print("(" + items[i].value + ", " + items[i].size + ", " + items[i].volume + ") ");
        System.out.println();
        System.out.println("And a knapsack of size " + size + " and volume " + vol);

        noRep(items);
        System.out.println();
        // Print the solution
        int j = size, h = vol, finalSize = 0, finalValue = 0, finalVolume = 0;
        System.out.print("Items picked (value, size, volume) for 0/1 problems without repetitions: ");
        for(int i = items.length; i > 0; i--) {
            if(dynNoRep[i][j][h] != dynNoRep[i - 1][j][h]) {
                System.out.print("(" + items[i - 1].value + ", " + items[i - 1].size + ", " + items[i - 1].volume + ") ");
                finalSize += items[i - 1].size;
                finalValue += items[i - 1].value;
                finalVolume += items[i - 1].volume;
                j -= items[i - 1].size;
                h -= items[i - 1].volume;
            }
        }
        System.out.println();
        System.out.println(" Final size: " + finalSize);
        System.out.println(" Final volume: " + finalVolume);
        System.out.println(" Final value: " + finalValue);
    }

}



文章来源: 0-1 Multidimensional Knapsack