【算法】【动态规划】【背包九讲】Knapsack Problem

本文详细解析了01背包、完全背包及多重背包问题的算法实现,包括朴素二维数组、一维优化、时间优化等方法,并提供了Java代码示例。

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本文章用 AcWing 题目举例

一、01背包

题目:01背包问题

d p i , j dp_{i,j} dpi,j 代表 i 件物品 j 容量背包可装的最大价值。

则其可以由两种结果递推得到

  • 第i件物品不装到背包里,前i-1件物品用了j容量的包。 d p [ i + 1 ] [ j ] = d p [ i ] [ j ] dp[i+1][j] = dp[i][j] dp[i+1][j]=dp[i][j]
  • 第i件物品装到背包里,前i-1件物品用了 j − w i j-w_{i} jwi容量的包。 d p [ i + 1 ] [ j ] = d p [ i ] [ j − v [ i + 1 ] ] + w [ i + 1 ] dp[i+1][j] =dp[i][j-v[i+1]] + w[i+1] dp[i+1][j]=dp[i][jv[i+1]]+w[i+1]

取两种递推结果的最大值即可。

1.1 朴素二维数组

import java.util.Scanner;
public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        while (scanner.hasNextInt()) {
            int n = scanner.nextInt();
            int total = scanner.nextInt();
            int[][] dp = new int[n+1][total + 1];
            for (int i = 0; i < n; i++) {
                int v = scanner.nextInt();
                int w = scanner.nextInt();
                for (int j = 0; j <= total; j++) {
                    if (j < v) {
                        dp[i+1][j] = dp[i][j];
                    } else {
                        dp[i+1][j] = Math.max(dp[i][j-v] + w,dp[i][j]);
                    }
                }
            }
            System.out.println(dp[n][total]);
        }
    }
}

1.2 一维优化

二维转一维的方法是画出二维表格图,看在一维滚动的情况下,如何使用旧数据。

import java.util.Scanner;
public class Main {
     public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        while (scanner.hasNextInt()) {
            int n = scanner.nextInt();
            int total = scanner.nextInt();
            int[] dp = new int[total + 1];
            for (int i = 0; i < n; i++) {
                int v = scanner.nextInt();
                int w = scanner.nextInt();
                for (int j = total; j >= v; j--) {
                    dp[j] = Math.max(dp[j-v] + w,dp[j]);
                }
            }
            System.out.println(dp[total]);
        }
    }
}

二、完全背包

题目:完全背包问题

第i件物品可以选0次、1次、2次……,递推时判断所有选法的最大值

2.1 暴力推导

可以直观地得到公式:

d p [ i + 1 ] [ j ] = M a t h . m a x ( d p [ i ] [ j − v ∗ k ] + w ∗ k , d p [ i + 1 ] [ j ] ) dp[i+1][j] = Math.max(dp[i][j-v*k] + w* k,dp[i+1][j]) dp[i+1][j]=Math.max(dp[i][jvk]+wk,dp[i+1][j])

时间复杂度为 O ( n v s ) O(nvs) O(nvs)


import java.util.Scanner;

public class Main {
    /**
     * dp[i][j] = dp[i-1][j],dp[i-1][j-v[i]]+w[i],...
     */
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        while (scanner.hasNextInt()) {
            int n = scanner.nextInt();
            int total = scanner.nextInt();
            int[][] dp = new int[n+1][total+1];
            for (int i = 0; i < n; i++) {
                int v = scanner.nextInt();
                int w = scanner.nextInt();
                for (int j = 0; j <= total; j++) {
                    for (int k = 0; k * v <= j; k++) {
                        dp[i+1][j] = Math.max(dp[i][j-v*k] + w* k,dp[i+1][j]);
                    }
                }
            }
            System.out.println(dp[n][total]);
        }
    }
}

2.2 时间优化

第i件物品取0次、1次、2次……
d p [ i + 1 ] [ j ] = M a t h . m a x ( d p [ i ] [ j − v [ i + 1 ] ∗ k ] + w [ i + 1 ] ∗ k ) , k = 0 , 1 , 2 , 3... dp[i+1][j] = Math.max(dp[i][j-v[i+1]*k] + w[i+1]* k), k=0,1,2,3... dp[i+1][j]=Math.max(dp[i][jv[i+1]k]+w[i+1]k),k=0,1,2,3...
可以转化为:第i件物品取0次和至少取1次
d p [ i + 1 ] [ j ] = M a t h . m a x ( d p [ i ] [ j ] , d p [ i + 1 ] [ j − v [ i + 1 ] ] + w [ i + 1 ] ) dp[i+1][j] = Math.max( dp[i][j],dp[i+1][j-v[i+1]] + w[i+1]) dp[i+1][j]=Math.max(dp[i][j],dp[i+1][jv[i+1]]+w[i+1])

import java.util.Scanner;

/**
 * 完全背包
 */
public class KnapsackFull {
    /**
     * dp[i][j] = dp[i-1][j],dp[i][j]+w[i],...
     */
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        while (scanner.hasNextInt()) {
            int n = scanner.nextInt();
            int total = scanner.nextInt();
            int[][] dp = new int[n+1][total+1];
            for (int i = 0; i < n; i++) {
                int v = scanner.nextInt();
                int w = scanner.nextInt();
                for (int j = 0; j <= total; j++) {
                    if (j >= v) {
                        dp[i+1][j] = Math.max(dp[i+1][j-v] + w,dp[i][j]);
                    } else {
                        dp[i+1][j] = dp[i][j];
                    }
                }
            }
            System.out.println(dp[n][total]);
        }
    }
}

2.3 一维优化


/**
 * 完全背包
 */
public class KnapsackFull {
    /**
     * dp[i][j] = dp[i-1][j],dp[i][j]+w[i],...
     */
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        while (scanner.hasNextInt()) {
            int n = scanner.nextInt();
            int total = scanner.nextInt();
            int[] dp = new int[total+1];
            for (int i = 0; i < n; i++) {
                int v = scanner.nextInt();
                int w = scanner.nextInt();
                for (int j = 0; j <= total; j++) {
                    if (j >= v) {
                        dp[j] = Math.max(dp[j-v] + w,dp[j]);
                    } else {
                        dp[j] = dp[j];
                    }
                }
            }
            System.out.println(dp[total]);
        }
    }
}

三、多重背包

题目:多重背包问题 I

多重背包是有限制的完全背包。

3.1 暴力求解

可以在2.1节上稍加修改,增加限制


import java.util.Scanner;

/**
 * 多重背包
 */
public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        while (scanner.hasNextInt()) {
            int n = scanner.nextInt();
            int total = scanner.nextInt();
            int[][] dp = new int[n+1][total+1];
            for (int i = 0; i < n; i++) {
                int v = scanner.nextInt();
                int w = scanner.nextInt();
                int s = scanner.nextInt();
                for (int j = 0; j <= total; j++) {
                    for (int k = 0; k <= s && k * v <= j; k++) {
                        dp[i+1][j] = Math.max(dp[i][j-v*k] + w* k,dp[i+1][j]);
                    }
                }
            }
            System.out.println(dp[n][total]);
        }
    }
}

3.2 二进制优化

3.3 单调队列优化法

参考 https://blog.youkuaiyun.com/weixin_43693379/article/details/89432283

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