动态规划-01背包问题

public class Main {
    public static void main(String[] args) {
        int[] w = {1,4,3};//物品的重量
        int[] val = {1500,3000,2000};//物品的价格
        int n = 4;//背包的最大容量

        //dp[i][j]代表最大容量为i时的最大总价值
        int[][] dp = new int[w.length+1][n+1];
        //初始化
        //没有物品的行
        for (int i = 0; i < n+1; i++) {
            dp[0][i] = 0;
        }
        //最大容量为0的列
        for (int i = 0; i < w.length+1; i++) {
            dp[i][0] = 0;
        }
        for (int i = 0; i < dp.length; i++) {
            for (int j = 0; j < dp[i].length; j++) {
                System.out.print(dp[i][j]+" ");
            }
            System.out.println();
        }
        for (int i = 1; i < dp.length; i++) {
            for (int j = 1; j < dp[i].length; j++) {
                if (w[i-1] > j) {
                    //物品的重量大于背包的最大容量
                    dp[i][j] = dp[i-1][j];
                } else {
                    dp[i][j] = Math.max(dp[i-1][j],val[i-1]+dp[i-1][j-w[i-1]]);
                }
            }
        }
        System.out.println();
        for (int i = 0; i < dp.length; i++) {
            for (int j = 0; j < dp[i].length; j++) {
                System.out.print(dp[i][j]+" ");
            }
            System.out.println();
        }
        System.out.println(dp[dp.length-1][dp[0].length-1]);
    }
}

如何拿到最大价值组合的商品：

//记录放入商品的情况
int[][] path = new int[w.length+1][n+1];

//                    dp[i][j] = Math.max(dp[i-1][j],val[i-1]+dp[i-1][j-w[i-1]]);
                    if (dp[i-1][j] < val[i-1]+dp[i-1][j-w[i-1]) {
                        dp[i][j] = val[i-1]+dp[i-1][j-w[i-1];
                        path[i][j] = 1;
                    } else {
                        dp[i][j] = dp[i-1][j];

        int i = path.length-1;
        int j = path[0].length-1;
        while (i >= 0 && j >= 0) {
            if (path[i][j] == 1) {
                System.out.printf("第%d个商品放入背包n",i);
                j -= w[i-1];
            }
            i--;
        }