So I'm trying to generate an algorithm that will find the best combination of n items (in my case 4) that can only be placed in the knapsack once (0-1) with a maximum weight capacity. Summarized probably more effectively, I want to place no more than four unique items in my knapsack so that the their weights are less than some value W while maximizing their total value. My first attempt and assumption was to put a volume limit of 4 with all item volumes as 1 for a multidimensional knapsack problem. But I ran into the problem of it not being 0-1 (meaning either in the bag or not). Then I tried making an 0-1(bounded) knapsack code multidimensional but I was unable to add the volume limit as well as the 0-1 requirement. How do I code a 0-1 multidimensional knapsack problem? Or how do I adapt the code to only hold a volume of V with all item volumes as 1? The code doesnt have to be Java but that's what I have so far.
The Knapsack:
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
And the Item:
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
Here is a generic implementation to solve the knapsack 0-1 problem with 2 dimensions (size and volume). I used a matrix instead of a list of list because it is much easier. Here is the whole class with also the main method to test it.
To add dimensions just add new dimensions to the matrix and add inner cycles to check all conditions.
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);
}
}
来源：https://stackoverflow.com/questions/14420211/0-1-multidimensional-knapsack