0-1背包问题---动态规划

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设计算法所需的数据结构。采用结构体Goods来存放单个货物信息，背包容量为nKnapSackCap,自定义数据类型vector<vector<int>>用来存放每一次迭代的执行结果，用数组knapStackGoods存放最终装入背包的物品。

2.2初始化。(由于vector类功能的限制，这里为了方便将record所有元素设置为0)。

2.3按照下式递推record，确定前i个物品能够装入背包的最优值。



2.4求解最优值，结果存入数组knapStackGoods中。

三、算法描述

//问题求解
void KnapSack_0_1(AllGoods &allGoods,int knapSackCap,AllGoods &knapStackGoods)
{
RecordTable record;//定义记录表

vector<int> temp;//一个临时变量，用来初始化record

//将record所有元素初始化为0
for(int i = 0;i <= allGoods.size();++i)
{
record.push_back(temp) ;
for(int j = 0;j <=knapSackCap;++j)
{
record[i].push_back(0);
}
}

//计算record[i][j]
for(int i = 1;i <= allGoods.size();++i)
{
for(int j = 1;j <=knapSackCap;++j)
{
if(j < allGoods[i - 1].weight)
{
record[i][j] = record[i - 1][j];
}
else
{
if(record[i - 1][j - allGoods[i - 1].weight] + allGoods[i -1].value < record[i - 1][j] )
{
record[i][j] = record[i - 1][j];
}
else
{
record[i][j] = record[i - 1][j - allGoods[i - 1].weight] + allGoods[i - 1].value ;
}
}
}
}

//构造最优解
int j = knapSackCap;
for(int i = allGoods.size();i > 0;--i)
{
if(record[i][j] > record[i - 1][j])
{
knapStackGoods.push_back(allGoods[i - 1]);
j -= allGoods[i - 1].weight;
}
}
}

四、算法复杂性分析

算法中有两层嵌套for循环，为此可选定语句j
< allGoods[i
- 1].weight为基本语句，其运行时间为nKnapStackCap
* nGoodsNum;则算法时间复杂性为O(nKnapStackCap * nGoodsNum)。又因为用到辅助变量record，所以其空间复杂性为O(nKnapStackCap
* nGoodsNum)。

五、实现代码

#include <iostream>
#include <vector>

using namespace std;

const int nGoodsNum = 5;
const int nKnapSackCap = 10;

class Goods //定义货物数据类型
{
public:
int weight;
int value;
};

typedef vector<Goods> AllGoods;//定义所有货物数据类型
typedef vector<vector<int>> RecordTable;//定义记录表数据类型

//问题求解
void KnapSack_0_1(AllGoods &allGoods,int knapSackCap,AllGoods &knapStackGoods)
{
RecordTable record;//定义记录表

vector<int> temp;//一个临时变量，用来初始化record

//将record所有元素初始化为0
for(int i = 0;i <= allGoods.size();++i)
{
record.push_back(temp) ;
for(int j = 0;j <=knapSackCap;++j)
{
record[i].push_back(0);
}
}

//计算record[i][j]
for(int i = 1;i <= allGoods.size();++i)
{
for(int j = 1;j <=knapSackCap;++j)
{
if(j < allGoods[i - 1].weight)
{
record[i][j] = record[i - 1][j];
}
else
{
if(record[i - 1][j - allGoods[i - 1].weight] + allGoods[i -1].value < record[i - 1][j] )
{
record[i][j] = record[i - 1][j];
}
else
{
record[i][j] = record[i - 1][j - allGoods[i - 1].weight] + allGoods[i - 1].value ;
}
}
}
}

//构造最优解
int j = knapSackCap;
for(int i = allGoods.size();i > 0;--i)
{
if(record[i][j] > record[i - 1][j])
{
knapStackGoods.push_back(allGoods[i - 1]);
j -= allGoods[i - 1].weight;
}
}
}
//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 = 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 main()
{
AllGoods allGoods;
AllGoods knapStackGoods;
GetAllGoods(allGoods); //要求同样重量的物品，价值大的排在前面

//求解
KnapSack_0_1(allGoods,nKnapSackCap,knapStackGoods);

//输出结果
cout<<"物品个数："<<nGoodsNum<<endl;
for(int i = 0 ;i < allGoods.size(); ++i)
{
cout<<"重量："<<allGoods[i].weight<<"   价值： "<<allGoods[i].value<<endl;
}

cout<<"背包容量: "<<nKnapSackCap<<endl;
cout<<"背包中可装入物品："<<endl;

for(int i = 0;i < knapStackGoods.size();++i)
{
cout<<"重量："<<knapStackGoods[i].weight<<"   价值： "<<knapStackGoods[i].value<<endl;
}
return 0;
}