货物运输系统，我们不确定如何显示任务的最后一部分

如何解决货物运输系统，我们不确定如何显示任务的最后一部分

这是我们即将完成的任务的代码，只是我们停留在的最后一部分 “最快：3次旅行（一辆面包车，3辆小型货车，645美元）” 我们不确定如何在方括号中显示值，我们只能显示3次行程。 是否也可以在括号内显示值？ 我们使用

int min = *min_element(vTrips.begin(),vTrips.end());
cout << "Fastest: " << min << " trips" << endl; 

但这只会显示3次旅行。

#include <iostream>
#include <vector>
#include <iterator>
#include <fstream>
#include<algorithm>
using namespace std;



class CTS //cargo transport system
{
    int i;
    int cargo,lorryprice,vanprice,lorrysize,vansize,allOps;
    
public:
    void set_cargo(int);
    void set_lorryprice(int);
    void set_vanprice(int);
    void set_lorrysize(int);
    void set_vansize(int);

};
void CTS::set_cargo(int total_cargo) {
    cargo = total_cargo;
}
void CTS::set_lorryprice(int lorryP) {
    lorryprice = lorryP;
}

void CTS::set_vanprice(int vanP) {
    vanprice = vanP;
}
void CTS::set_lorrysize(int lorryS) {
    lorrysize = lorryS;
}
void CTS::set_vansize(int vanS)
{
    vansize = vanS;
}

int main()
{
    int cargo,options,i,no_lorry,no_van,cost,trips;

    ifstream infile;
    infile.open("size.txt");
    if (infile.is_open()) {
        infile >> cargo;
        infile >> lorryprice;
        infile >> vanprice;
        infile >> lorrysize;
        infile >> vansize;
    }
    CTS run;
    run.set_cargo(cargo);
    run.set_lorryprice(lorryprice);
    run.set_vanprice(vanprice);
    run.set_lorrysize(lorrysize);
    run.set_vansize(vansize);

    infile.close();
    options = (cargo / lorrysize) + 1;
    no_lorry = (cargo / lorrysize);
    no_van = (cargo / vansize) + 3;


    if (cargo % lorrysize == 0) {
        no_van = -3;
    }

    if (cargo % lorrysize != 0) {
        no_van = ((cargo % lorrysize) / 10) - 3;
    }


    /*it = numbervan.begin();
    for (auto ir = numbervan.rbegin(); ir != numbervan.rend(); ++ir) {

        cout << *ir << endl;
    }*/
    vector<int> vCost,vVan,vTrips,vLorry;
    vector <int>::iterator it;

    for (i = 1; i < options + 1; i++)
    {

        int numberlorry = no_lorry;
        cout << "Option " << i << ":" << endl;

        cout << "Number of Mini-Lorries : " << no_lorry-- << endl;

        if (no_van >= -3) {
            no_van += 3;

        }
        cout << "Number of Vans : " << no_van << endl;
        int numbervan = no_van;

        

        if (numberlorry > numbervan) {
            trips = numberlorry;
        }
        else {
            trips = numbervan;
        }
        cout << "Trips Needed : " << trips << endl;
        cost = (numberlorry * lorryprice) + (no_van * vanprice);
        cout << "Total Cost : $" << cost << endl;
        vCost.push_back(cost);
        vLorry.push_back(numberlorry);
        vVan.push_back(numbervan);
        vTrips.push_back(trips);


    }
    int counter = vCost.size() - 1;

    //std::vector<int>::reverse_iterator ir = vCost.rbegin();
    for (i = 1; i < 4; i++) {



        //cout << "Lowest #" << i << ": "<<cost<<endl;
        cout << "Lowest #" << i << ": $" << vCost[counter] << "(" << vVan[counter] << " Vans," << vLorry[counter] << " Mini-Lorry," << vTrips[counter] << " Trips)" << endl;
        counter--;
    }
    



    int min = *min_element(vTrips.begin(),vTrips.end()); // this line of code we figured out how to                                                          
    cout << "Fastest: " << min << " trips" << endl;       //display the number of trips using algorithm

    return 0;
}

解决方法

您的设计很尴尬；您创建了CTS run;的实例，并且从不使用它。

假设您正确进行了计算，则需要知道在min找到了什么索引。如果存储min_element()返回的迭代器，则可以通过从中减去vTrips.begin()来获得索引。然后，vCost，vLorry和vVan向量中的相应元素将包含所需的数据。

但是，如果您定义一个包含预先计算的值的结构，并将其推入某个向量中，将会更加容易。在这种情况下，所有相关数据都保存在一起。