0-1背包问题---回溯法

一、问题描述

             0-1背包问题可描述为:n个物体和一个背包。对物体i,其价值为value,重量为weight,背包的容量为W。如何选取物品装入背包,使背包中所装入的物品总价值最大?


二、算法设计

  2.1 用到的数据结构

class Goods //定义货物数据类型
{
public:
	int weight;//货物重量
	int value;//货物价值
	friend ostream& operator<<(ostream &os, const Goods &out);
};

class Knapsack//背包
{
private:
	int capacity;//背包容量
	int nGoodsNum;//物品数
	vector<Goods> goods; //所有货物
	int nMaxValue;//之前背包中装入的最大价值物品
	int nCurrentWeight;//当前背包中装入物品的数量
	int nCurrentValue;//当前背包中物品的价值
	vector<bool> bestResult;//之前背包中物品最大价值时的物品
	vector<bool> currentResult;//当前背包中的物品
}

 

  2.2 算法步骤 


  1)定义解空间。(X0 , X1X2X3…..Xn,Xi的值为truefalse

     (i = 0,1,2,3….n)

    2)确定解空间。问题的解空间描述了2^n种可能的解,采用一个二叉满树组织,解

        空间的深度为问题的规模。

    3)搜索解空间

       a.约束条件。背包中物品重量小于背包容量。

     b.限界条件。nCurrentValue为当前背包中物品价值,nMaxValue之前背包中装

        入的最大价值的物品。nP = bound(i + 1),第 i个物品之后的所有物品可

        装入背包的最大价值。要求:nP + nCurrentValue > nMaxValue.

     c.以深入优先的方式进行搜索.首先以根节点即第一个物品开始搜索.



三、算法描述


 int bound(int i)//限界函数
	{
		int nLeftCapacity = capacity - nCurrentWeight;
		int tempMaxValue = nCurrentValue;
		
		while (i < nGoodsNum && goods[i].weight <= nLeftCapacity)
		{
			nLeftCapacity     -= goods[i].weight;
			tempMaxValue   += goods[i].value;
		}

		if (i < nGoodsNum)
		{
			tempMaxValue += (float)(goods[i].value) / goods[i].weight * nLeftCapacity;
		}

		return tempMaxValue;
	}

void backTrack(int t)//递归回溯
	{
		if (t >= nGoodsNum)
		{
			for (int i = 0; i < nGoodsNum; ++i)
			{
				bestResult[i] = currentResult[i];
			}
			nMaxValue = nCurrentValue;
			return;
		}

		if (nCurrentWeight + goods[t].weight <= capacity)
		{
			currentResult[t] = true;

			nCurrentWeight += goods[t].weight;
			nCurrentValue    += goods[t].value;

			backTrack(t + 1);

			nCurrentWeight -= goods[t].weight;
			nCurrentValue -= goods[t].value;
		}

		if (bound(t + 1) > nMaxValue)
		{
			currentResult[t] = false;
			backTrack(t + 1);
		}
	}

//寻找最优结果
int BacktrackingKnapsack0_1(AllGoods &allGoods, int nKnapSackCap)
{
	Knapsack knap(allGoods,nKnapSackCap);

	knap.printGoods();
	knap.sortByUintValue();
	cout << "sort" << endl;
	knap.printGoods();

	knap.backTrack(0);
	knap.printResult();
	return 0;
}

 

四、算法复杂性分析

   时间复杂度:判断约束函数需O(1),最坏情况下有2^n – 1个左孩子,约束函数耗时最坏为O(2^n).计算上界函数需O(n),在最坏情况下有2^n – 1个右孩子,限界函数耗时最坏为O(n2^n)。则背包问题最坏的时间复杂度为O(n2^n)

空间复杂度:创建Knapsack类对象,其中有三个长度为n的数组,而后调用backTrack()函数递归深度为n。所以,此算法的空间复杂度为O(n).


五、算法实现与测试

  5.1 实现代码

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;
const int NKNAPSACKCAP = 10;
class Goods //定义货物数据类型
{
public:
	int weight;
	int value;
	friend ostream& operator<<(ostream &os, const Goods &out);
};

ostream& operator<<(ostream &os, const Goods &out)
{
	os << "重量:" << out.weight << " 价值: " << out.value;
	return os;
}
typedef vector<Goods> AllGoods;//定义所有货物数据类型

class Knapsack
{
private:
	int capacity;//背包容量
	int nGoodsNum;//物品数
	vector<Goods> goods;
	int nMaxValue;
	int nCurrentWeight;
	int nCurrentValue;
	vector<bool> bestResult;
	vector<bool> currentResult;

	int bound(int i)
	{
		int nLeftCapacity = capacity - nCurrentWeight;
		int tempMaxValue = nCurrentValue;
		
		while (i < nGoodsNum && goods[i].weight <= nLeftCapacity)
		{
			nLeftCapacity     -= goods[i].weight;
			tempMaxValue   += goods[i].value;
		}

		if (i < nGoodsNum)
		{
			tempMaxValue += (float)(goods[i].value) / goods[i].weight * nLeftCapacity;
		}

		return tempMaxValue;
	}
public:
	Knapsack(AllGoods &AllGoods, int nKnapsackCap)
	{
		nGoodsNum         = AllGoods.size();
		capacity                 = nKnapsackCap;
		nCurrentWeight  = 0;
		nCurrentValue     = 0;
		nMaxValue            = 0;

		for (int i = 0; i < nGoodsNum; ++i)
		{
			goods.push_back(AllGoods[i]);
			bestResult.push_back(false);
			currentResult.push_back(false);
		}
	}

	void sortByUintValue()
	{
		stable_sort(goods.begin(), goods.end(), [](const Goods& left, const Goods& right)
		{return (left.value * right.weight > left.weight * right.value); });
	}

	void printGoods()
	{
		for (size_t i = 0; i < goods.size(); ++i)
		{
			cout << goods[i] << endl;
		}
	}

	void printResult()
	{
		cout << "MAX VALUE: " << nMaxValue << endl;
		for (int i = 0; i < nGoodsNum; ++i)
		{
			if (bestResult[i])
			{
				cout << goods[i] << endl;
			}
		}
	}

	void backTrack(int t)
	{
		if (t >= nGoodsNum)
		{
			for (int i = 0; i < nGoodsNum; ++i)
			{
				bestResult[i] = currentResult[i];
			}
			nMaxValue = nCurrentValue;
			return;
		}

		if (nCurrentWeight + goods[t].weight <= capacity)
		{
			currentResult[t] = true;

			nCurrentWeight += goods[t].weight;
			nCurrentValue    += goods[t].value;

			backTrack(t + 1);

			nCurrentWeight -= goods[t].weight;
			nCurrentValue -= goods[t].value;
		}

		if (bound(t + 1) > nMaxValue)
		{
			currentResult[t] = false;
			backTrack(t + 1);
		}
	}
};


//获取物品信息,此处只是将书上例子输入allGoods
void GetAllGoods(AllGoods &allGoods)
{
	Goods goods;

	goods.weight = 2;
	goods.value = 6;
	allGoods.push_back(goods);


	goods.weight = 2;
	goods.value = 3;
	allGoods.push_back(goods);

	goods.weight = 2;
	goods.value = 8;
	allGoods.push_back(goods);

	goods.weight = 6;
	goods.value = 5;
	allGoods.push_back(goods);

	goods.weight = 5;
	goods.value = 4;
	allGoods.push_back(goods);

	goods.weight = 4;
	goods.value = 6;
	allGoods.push_back(goods);
	
}

int BacktrackingKnapsack0_1(AllGoods &allGoods, int nKnapSackCap)
{
	Knapsack knap(allGoods,nKnapSackCap);
	knap.printGoods();
	knap.sortByUintValue();
	cout << "sort" << endl;
	knap.printGoods();
	knap.backTrack(0);
	knap.printResult();
	return 0;
}
int main()
{
	AllGoods allGoods;
	GetAllGoods(allGoods); //要求按照单位物品价值由大到小排序
	AllGoods::iterator it;
	BacktrackingKnapsack0_1(allGoods, NKNAPSACKCAP);
	return 0;
}

  5.2 运行结果


《0-1背包问题---回溯法》

  注意:代码中可能会用到C++11新特性,请在支持C++11标准的环境下编译运行

         (如VS2013)

 

    原文作者:回溯法
    原文地址: https://blog.csdn.net/u013507368/article/details/40862905
    本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。
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