背包问题填表法

假设背包容量为W,物体数目为n，填表法得到的时间复杂度和空间复杂度均为O(nW)。

#include <iostream>
#include <ctime>
#include <string>
#include <vector>
#include <fstream>
#include <sstream>
using namespace std;
//迭代的方法，时间复杂度(nW)
/***************物体属性******************/
class item
{
public:
    long value;
    int weight;
    item() : value(0), weight(0){}
};

class Knapsack
{
public:
/***************构造函数读取数据*********************/
    Knapsack()
    {
        ifstream fin("4.txt");
        string line;
        stringstream stream;
        if(getline(fin, line))
        {
            stream.clear();
            stream << line;
            stream >> knapsack_size;
            stream >> num_of_items;
        }
        int cnt = 1;
        init_items();
        while(getline(fin, line))
        {
            long _value;
            int _weight;
            stream.clear();
            stream << line;
            stream >> _value;
            stream >> _weight;
            items[cnt].value = _value;
            items[cnt].weight = _weight;
            cnt++;
        }
    }

    void init_items()
    {
        items.resize(num_of_items+1);
        Avalue.resize(num_of_items+1);
        for(int i = 0; i <= num_of_items; i++)
        {    
            Avalue[i].resize(knapsack_size+1);
        }
    }
    /***************返回最大值**************/
    long max(long a, long b)
    {
        return a > b ? a : b;        
    }
/***************遍历填表*****************/
    void sack()
    {    
        for(int k = 0; k <= knapsack_size; k++)        
        {
            Avalue[0][k] = 0;  //初始化边界
            
        }
        for(int i = 1; i <= num_of_items; i++)  //填表
        {
            for(int x = 0; x <= knapsack_size; x++)
            {
                if(x-items[i].weight >= 0)
                {
                    Avalue[i][x] = max(Avalue[i-1][x], Avalue[i-1][x-items[i].weight] + items[i].value);
                } else {
                    Avalue[i][x] = Avalue[i-1][x];
                }
            }
        }
    }
/****************根据填好的表重构最优解的组成元素*************/
    void reconstruct()
    {
        int temp = Avalue[num_of_items][knapsack_size];
        int i = num_of_items;
        int x = knapsack_size;
        while( i > 0)
        {
//            temp = Avalue[i][x];
            if( (temp > Avalue[i-1][x]) )
            {
                cout << "(" << x << "," << i << ")" << temp <<  endl;
                set.push_back(i);
                x = x - items[i].weight;
                i = i-1;
            } else  {
                //set.push_back(i-1);
                i = i - 1;
            }
            temp = Avalue[i][x]; 
            cout << "(" << x << "," << i << ")" << temp  << endl;
        }
        
    }

/**********打印看输出*********/
    void print()
    {
        ofstream fout;
        fout.open("out1.txt");
        for(int j = knapsack_size; j >= 0; j--)
        {
            for(int i = 0; i <= num_of_items; i++)        
            {
                fout << "[w,i] [" << j << "," << i << "]" << Avalue[i][j] << " ;";
            }
            fout << endl;
        }
    }
    
    void write()
    {
        ofstream fout;
        fout.open("out2.txt");
        for(int i = 0; i < set.size(); i++)
        {
            fout << set[i] << " ";    
        }
        cout << endl;
    }
    
public:
    int knapsack_size;   //背包容量
    int num_of_items;    //物体数目
    vector<item> items;  //物体集合
    vector< vector<long> > Avalue; //表的数据
    vector<int> set; //存储最优解
};

int main()
{
    clock_t start, end;
    start = clock();
    Knapsack knapsack;
    knapsack.sack();
    knapsack.print();
    knapsack.reconstruct();
    knapsack.print();
    knapsack.write();
    end = clock();
    cout << "running time:" << (double)(end-start)/CLOCKS_PER_SEC << "s" << endl;
    return 0;
}