回溯法解决01背包问题

最新推荐文章于 2024-01-20 10:33:35 发布

转载最新推荐文章于 2024-01-20 10:33:35 发布 · 2.7k 阅读

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#java #01背包 #回溯

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本文介绍了一种解决0/1背包问题的方法，通过回溯算法实现最优价值计算。使用了商品类（Goods）来存储物品的价值和重量，并在主函数中展示了如何初始化商品数组并创建求解对象以获取最大价值。

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原文参看：http://blog.youkuaiyun.com/ljmingcom304/article/details/50314839

package com.alo.offer;

import java.util.Arrays;
import java.util.Collections;

public class Recall {
	private Goods goods[] ;
	private int totalWeight;
	private int currentWeight;
	private int n;
	private int bestValue;
	private int currentValue;
	public Recall(Goods[] goods, int totalWeight, int n) {
		super();
		this.goods = goods;
		this.totalWeight = totalWeight;
		this.n = n;
		Arrays.sort(goods,Collections.reverseOrder());
	}
	
	public int solve(int i) {
		if(i>=n) {
			bestValue=currentValue;
			return bestValue;
		}
		if(currentWeight+goods[i].getWeight()<=totalWeight) {
			currentWeight+=goods[i].getWeight();
			currentValue+=goods[i].getValue();
			bestValue=solve(i+1);
			currentWeight-=goods[i].getWeight();
			currentValue-=goods[i].getValue();
		}
		if(currentValue+getSurplusValue(i+1)>bestValue) {
			bestValue=solve(i+1);
		}
		return bestValue;
	}

	private int getSurplusValue(int i) {
		// TODO Auto-generated method stub
		int surplusValue=0;
		for(int j=i;j<n;j++) {
			surplusValue+=goods[j].getValue();
		}
		return surplusValue;
	}
	public static void main(String[] args) {
	    Goods[] bags = new Goods[] { new Goods(2, 13),  
	            new Goods(1, 10), new Goods(3, 24), new Goods(2, 15),  
	            new Goods(4, 28), new Goods(5, 33), new Goods(3, 20),  
	            new Goods(1, 8) };  
	    int totalWeight = 12;  
	    Recall problem = new Recall(bags, totalWeight,bags.length);  
	  
	    System.out.println(problem.solve(0)); 
	}
}