HDU 2602 Bone Collector【01背包】

最新推荐文章于 2023-07-08 15:07:45 发布

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最新推荐文章于 2023-07-08 15:07:45 发布

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分类专栏： # OJ 文章标签： oj

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本文介绍了一种经典的背包问题——骨收集者问题。该问题要求根据骨收集者的背包容量及不同骨头的价值和体积，计算出收集骨头的最大总价值。文章通过一个具体的样例输入输出，展示了如何使用动态规划算法来解决这个问题。

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Problem Description

Many years ago , in Teddy’s hometown there was a man who was called “Bone Collector”. This man like to collect varies of bones , such as dog’s , cow’s , also he went to the grave …
The bone collector had a big bag with a volume of V ,and along his trip of collecting there are a lot of bones , obviously , different bone has different value and different volume, now given the each bone’s value along his trip , can you calculate out the maximum of the total value the bone collector can get ?
这里写图片描述

Input

The first line contain a integer T , the number of cases.
Followed by T cases , each case three lines , the first line contain two integer N , V, (N <= 1000 , V <= 1000 )representing the number of bones and the volume of his bag. And the second line contain N integers representing the value of each bone. The third line contain N integers representing the volume of each bone.

Output

One integer per line representing the maximum of the total value (this number will be less than 231).

Sample Input

1
5 10
1 2 3 4 5
5 4 3 2 1

Sample Output

14

Code

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        sc.nextLine();
        while (N-- > 0) {
            String[] tmp = sc.nextLine().split(" ");
            int n = Integer.parseInt(tmp[0]);
            int w = Integer.parseInt(tmp[1]);
            int[] weight = new int[n];
            int[] value = new int[n];
            tmp = sc.nextLine().split(" ");
            for (int i = 0; i < n; i++)
                value[i] = Integer.parseInt(tmp[i]);
            tmp = sc.nextLine().split(" ");
            for (int i = 0; i < n; i++)
                weight[i] = Integer.parseInt(tmp[i]);

            int[][] dp = new int[n + 1][w + 1];
            for (int i = 0; i < n; i++)
                for (int j = 0; j <= w; j++) {
                    if (j < weight[i])
                        dp[i + 1][j] = dp[i][j];
                    else
                        dp[i + 1][j] = Math.max(dp[i][j], dp[i][j - weight[i]] + value[i]);
                }
            System.out.println(dp[n][w]);
        }
    }
}