0/1背包问题

0/1背包问题 - 动态规划（C++实现）

flyfish

以下代码在VC++2013下编译通过

#include "stdafx.h"#include <iostream>#include <algorithm>#include <vector>struct Item //物品定义{    int id, weight, value;//编号，重量，价值。编号为0的物品这里没有使用    Item(){}    Item(int i, int w, int v) :id(i), weight(w), value(v){}};const int n=4, C=10;//C背包所能承受的最大重量//物品个数nint f[n+1][C+1];std::vector<Item> allItems;//所有的物品std::vector<Item> selectedItems;//装入背包的物品int maxValue=0;//能够装入背包的最大价值void Result(){    int currentWeight = C;    for (int i = n; i > 0 && currentWeight > 0; i--) {        if (f[i][currentWeight] == f[i - 1][currentWeight - allItems[i].weight] + allItems[i].value){            selectedItems.push_back(allItems[i]);            currentWeight -= allItems[i].weight;        }    }}void KnapsackProblem_DynamicProgramming() {    for (int i = 1; i <= n; i++) {        for (int j = 0; j < allItems[i].weight; j++)            f[i][j] = f[i - 1][j];        for (int j = allItems[i].weight; j <= C; j++)            f[i][j] = std::max(f[i - 1][j], f[i - 1][j - allItems[i].weight] + allItems[i].value);    }    maxValue = f[n][C];    Result();}int _tmain(int argc, _TCHAR* argv[]){        memset(f, 0, sizeof(f));        allItems.push_back(Item(0, 0, 0));        allItems.push_back(Item(1, 3, 30));        allItems.push_back(Item(2, 5, 20));        allItems.push_back(Item(3, 4, 10));        allItems.push_back(Item(4, 2, 40));        KnapsackProblem_DynamicProgramming();        for (size_t i = 0; i < selectedItems.size(); i++)            std::cout << "物品编号：" << selectedItems[i].id             << "  重量：" << selectedItems[i].weight            << "  价值：" << selectedItems[i].value << std::endl;        std::cout << "背包最大价值：" << maxValue;    return 0;}

KnapsackProblem_DynamicProgramming 过程

0,30,50,50,90

result：后面的1表示放入，0表示不放入
0 + 30 =30 1
30 + 20 =50 1
50 + 0 =50 0
50 + 40 =90 1

输出