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我希望能够将一个人的“整体”价值与另一个人进行比较。我不确定我是否正确存储它们，也不知道如何正确比较它们。我不知道如何访问任何一个人的“整体”价值观，这是我认为最困扰我的。

头文件

#ifndef Population_h
#define Population_h


class population
{
    friend class person;
private:
    int size;
    int generation;


public:
    void setsize(int x);    
    void tournament_selection(population x, int z);



};



class person
{
    friend class population;
private:
    float fit1;
    float fit2;
    float overall;


public:

    void generatefit();
    void setfit();
    void langerman();
    void printinfo();

};






#endif

人口.cpp

#include <iostream>
#include <iomanip>
#include <cstdlib>
#include <ctime>
#include <random>
#include <string>
#include <CMATH>
#include <vector>
#include "Population.h"
using namespace std;

void person ::generatefit()
{
    float randomnumb1;
    float randomnumb2;
    //float((rand() % 10)*0.1);

    randomnumb1 = static_cast <float> (rand()) / static_cast <float> (RAND_MAX);
    randomnumb2 = static_cast <float> (rand()) / static_cast <float> (RAND_MAX);

    fit1 = randomnumb1;
    fit2 = randomnumb2;


}

void person::setfit()
{   
    float x = fit1;
    float y = fit2;
}

void person::langerman()
{
    overall = 3 * pow(fit1, 2) + 2 * fit2 + 0.0252;
    for (overall; overall > 1; overall--);
}


void person::printinfo()
{
    cout << overall << " " << endl;
}



void population::setsize(int x)
{
    size = x;
}


void population::tournament_selection(population x, int total)
{
    float best = 0;
    for (int i = 0; i <= total; i++)
    {

    }


}

主文件

#include "Population.h"
#include <iostream>
#include <string>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <chrono>
#include <random>
#include <vector>
#include <stdlib.h>
using namespace std;

int main()
{
    cout << "Program is starting " << endl;

    srand(static_cast <unsigned> (time(0)));
    population pop;
    vector<person> popvector;
    vector<person> survivor;
    person *p1;

    int popsize = 500;
    pop.setsize(popsize);

    for (int i = 0; i <= popsize; i++)
    {
        p1 = new person;

        p1 ->generatefit();
        p1->setfit();
        p1->langerman();
        popvector.push_back(*p1);
        delete p1;
    }
    cout << "The fit values of the person are listed here: " << endl;
    vector<person> ::iterator it; //iterator to print everything in the vector 
    for (it = popvector.begin(); it != popvector.end(); ++it)
    {
        it->printinfo();
    }

    unsigned seed = std::chrono::system_clock::now().time_since_epoch().count(); // generate a seed for the shuffle process of the vector.

    cout << "Beggining selection process" << endl;

    shuffle(popvector.begin(), popvector.end(), std::default_random_engine(seed));

    //want to pick consecutive parents 

    int j = 0;



}

我希望能够比较人，将“获胜者”存储到“幸存者”向量中，然后继续使用“幸存者”向量通过 X 代的交叉和变异来创建新种群。

4

1 回答 1

1

您可以使用运算符重载来设置比较两个人的健康水平的自定义“规则”。是一个完美的例子”：通过而不是std::string直接进行相等操作，展示了运算符重载技术的优点。if(str1 == str2)if(!strcmp(str1, str2))

以下代码应满足您的需求：

class person {
    friend class population;
private:
    float fit1;
    float fit2;
    float overall;

public:
    void generatefit();
    void setfit();
    void langerman();
    void printinfo();

    //Overloading '<' operator
    bool operator(const person&);
};

//The following function defines
//rule of fitness in the jungle
bool person::operator < (const person& p2){
    //This is just an example, you can define your own rules
    return overall < p2.overall;
}

建立比较规则后，您可以按照该规则对总体进行排序：

//The "less fit" rule is defined so the population will be sorted
//in ascending order, if you want to sort them by descending order,
//just change the definition of your fitness rules accordingly.

sort(popvector, popvector + popsize);

或者，您可以首先使用有序容器来存储人口。这样的选择可以是set、map或priority_queue。有序容器中的元素将始终遵循您在声明此类容器对象时指定的确切顺序。

在您的情况下，我建议priority_queue您，因为您可以轻松地从最不合适的人中脱颖而出，如下所示：

#include<priority_queue>

//Still, the definition of "priority" is required beforehand
priority_queue<person> pqPerson;

person tmp;
for(int i = 0; i < popsize; ++i){
    tmp.setfit(fit1, fit2, overall);
    pqPerson.push(tmp);
}

for(int generation = 0; generation < X; +=generation){
    //The weakest group will perish
    for(int j = 0; j < tollSize; ++j){
        pqPerson.pop();
    }

    //Crossover and Mutation process
}
于 2016-04-11T04:09:13.297 回答