背包问题——动态规划——Java实现

背包问题

动态规划实现

动态规划：本质上是枚举（不重复的暴力枚举），因此其复杂度很好分析，有多少个状态就要被计算多少次，复杂度就为多少。

public class Knapsack {
    public static void main(String[] args) {
        int n = 5;
        int c = 10;
        int[] w = new int[]{2, 2, 6, 5, 4};
        int[] v = new int[]{6, 3, 5, 4, 6};
        Knapsack knapsack = new Knapsack();
        int[][] res = knapsack.bag(n, c, w, v);
        for (int i = 1; i < n + 1; i++) {
            for (int j = 1; j < c + 1; j++) {
                System.out.print(res[i][j] + " ");
            }
            System.out.println();
        }
        knapsack.show(n, c, w, res);
    }

    public int[][] bag(int n, int c, int[] w, int[] v) {
        int[][] res = new int[n + 1][c + 1];
        for (int i = 1; i < n + 1; i++) {
            for (int j = 1; j < c + 1; j++) {
                res[i][j] = res[i - 1][j];
                if (j >= w[i - 1] && res[i - 1][j - w[i - 1]] + v[i - 1] > res[i][j]) {
                    res[i][j] = res[i - 1][j - w[i - 1]] + v[i - 1];
                }
            }
        }
        return res;
    }

    public void show(int n, int c, int[] w, int[][] res) {
        System.out.println("最大价值为：" + res[n][c]);
        boolean[] x = new boolean[n];
        int j = c;
        for (int i = n; i > 0; i--) {
            if (res[i][j] > res[i - 1][j]) {
                x[i - 1] = true;
                j -= w[i - 1];
            }
        }
        System.out.println("选择的物品为：");
        for (int i = 0; i < n; i++) {
            if (x[i]) {
                System.out.println("第 " + i + " 个");
            }
        }
        System.out.println();
    }
}