给出符合下列要求的代码，1.初始化种群：第一个解决方案：由NEH2规则产生，选择完工时间最小的位置。第二个解决方案：对于每个作业𝑗，将其插入到所有工厂中所有可能的位置，然后选择总能耗最小的位置。第三个解决方案：对于一个作业𝑗，将其插入到所有工厂中的所有可能的位置，并选择具有最小总能耗和最小最大完工时间的位置。其余的解决方案随机产生，以保持种群的多样性。编码规则如下：有一组列表𝝅={𝝅1，...，𝝅𝑓，…，𝝅𝐹}，其中一个列表𝝅𝑓表示分配给工厂𝑓的部分作业，该列表的作业序列表示第一阶段的作业处理序列。解码机制如下：对于第一阶段的列表𝝅，每个列表𝝅𝑓中的作业被分配给工厂𝑓中的第一台可用机器；在下一阶段，将根据上一阶段的完成时间重新排序作业；另外，如果某些作业在当前阶段有相同的开始时间，那么它们将按照与前一个阶段相同的顺序排列。

以下是符合要求的 Python 代码：

import random

# 初始化种群
def initialize_population(num_individuals, num_jobs, num_factories):
    population = []
    # 第一个解决方案：使用 NEH2 规则产生
    neh_sol = neh2(num_jobs, num_factories)
    neh_fitness = calculate_fitness(neh_sol)
    population.append((neh_sol, neh_fitness))
    # 第二个解决方案：对于每个作业𝑗，将其插入到所有工厂中所有可能的位置，然后选择总能耗最小的位置
    for i in range(num_individuals//3):
        solution = []
        for job in range(num_jobs):
            costs = []
            for factory in range(num_factories):
                for pos in range(len(solution[factory])+1):
                    new_solution = solution.copy()
                    new_solution[factory].insert(pos, job)
                    cost = calculate_fitness(new_solution)
                    costs.append((cost, factory, pos))
            min_cost, min_factory, min_pos = min(costs)
            solution[min_factory].insert(min_pos, job)
        fitness = calculate_fitness(solution)
        population.append((solution, fitness))
    # 第三个解决方案：对于一个作业𝑗，将其插入到所有工厂中的所有可能的位置，并选择具有最小总能耗和最小最大完工时间的位置
    for i in range(num_individuals//3):
        solution = []
        for job in range(num_jobs):
            costs = []
            for factory in range(num_factories):
                for pos in range(len(solution[factory])+1):
                    new_solution = solution.copy()
                    new_solution[factory].insert(pos, job)
                    cost = calculate_fitness(new_solution)
                    makespan = calculate_makespan(new_solution)
                    costs.append((cost, makespan, factory, pos))
            min_cost, min_makespan, min_factory, min_pos = min(costs)
            solution[min_factory].insert(min_pos, job)
        fitness = calculate_fitness(solution)
        population.append((solution, fitness))
    # 其余的解决方案随机产生，以保持种群的多样性
    for i in range(num_individuals - len(population)):
        solution = []
        for job in range(num_jobs):
            factory = random.randint(0, num_factories-1)
            pos = random.randint(0, len(solution[factory]))
            solution[factory].insert(pos, job)
        fitness = calculate_fitness(solution)
        population.append((solution, fitness))
    return population

# NEH2 规则
def neh2(num_jobs, num_factories):
    jobs = list(range(num_jobs))
    job_times = [random.randint(1, 10) for _ in range(num_jobs)]
    makespans = []
    for i in range(num_jobs):
        job_order = [jobs[i]] + makespans
        factory_times = [sum(job_times[j] for j in job_order[:k+1]) for k in range(num_factories)]
        total_time = factory_times[-1]
        makespans.append((job_order, total_time))
    neh_order = [job for job, _ in sorted(makespans, key=lambda x: x[1])]
    solution = [[] for _ in range(num_factories)]
    for i, job in enumerate(neh_order):
        solution[i % num_factories].append(job)
    return solution

# 计算适应度函数值
def calculate_fitness(solution):
    total_cost = 0
    for factory in solution:
        factory_cost = 0
        machine_times = [0] * len(factory)
        for job in factory:
            job_time = job_times[job]
            machine_times = [max(machine_times[i], machine_times[i-1]) + job_time for i in range(1, len(machine_times))]
            machine_times[0] += job_time
            factory_cost += sum(machine_times)
        total_cost += factory_cost
    return total_cost

# 计算最大完工时间
def calculate_makespan(solution):
    makespans = []
    for factory in solution:
        machine_times = [0] * len(factory)
        for job in factory:
            job_time = job_times[job]
            machine_times = [max(machine_times[i], machine_times[i-1]) + job_time for i in range(1, len(machine_times))]
            machine_times[0] += job_time
        makespans.append(max(machine_times))
    return max(makespans)

其中 `num_individuals` 表示种群大小，`num_jobs` 表示作业数量，`num_factories` 表示工厂数量，`job_times` 表示每个作业的处理时间。