【编程题】0-1背包问题

0-1背包问题

1、问题描述

给定n中物品和一背包。物品i的重量是wi，其价值为vi，背包的容量为c。问应如何选择装入背包中的物品，使得装入背包中的物品的总价值最大？

2、我的代码

class Knapsack {

    public static void main(String[] args) throws Exception {
        int val[] = {10, 40, 30, 50};
        int wt[] = {5, 4, 6, 3};
        int W = 10;

        System.out.println(knapsack(val, wt, W));
    }

    public static int knapsack(int val[], int wt[], int W) {

        //Get the total number of items. 
        //Could be wt.length or val.length. Doesn't matter
        int N = wt.length; 

        //Create a matrix. 
        //Items are in rows and weight at in columns +1 on each side
        int[][] V = new int[N + 1][W + 1]; 


        //What if the knapsack's capacity is 0 - Set
        //all columns at row 0 to be 0
        for (int col = 0; col <= W; col++) {
            V[0][col] = 0;
        }

        //What if there are no items at home.  
        //Fill the first row with 0
        for (int row = 0; row <= N; row++) {
            V[row][0] = 0;
        }

        for (int item=1;item<=N;item++){

            //Let's fill the values row by row
            for (int weight=1;weight<=W;weight++){

                //Is the current items weight less
                //than or equal to running weight
                if (wt[item-1]<=weight){

//Given a weight, check if the value of the current 
//item + value of the item that we could afford 
//with the remaining weight is greater than the value
//without the current item itself
                    V[item][weight]=Math.max (val[item-1]+V[item-1][weight-wt[item-1]], V[item-1][weight]);
                }
                else {
//If the current item's weight is more than the
//running weight, just carry forward the value
//without the current item
                    V[item][weight]=V[item-1][weight];
                }
            }

        }

        //Printing the matrix
        for (int[] rows : V) {
            for (int col : rows) {

                System.out.format("%5d", col);
            }
            System.out.println();
        }

        return V[N][W];

    }

}