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0for(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0for(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allGoodsvoid 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;}