生成求解柔性调度问题代码

柔性车间调度问题是一个复杂的组合优化问题，可以采用各种算法进行求解，包括遗传算法、模拟退火、蚁群算法等。下面是一个简单的 Python 代码示例，用遗传算法求解柔性车间调度问题：

import random

# 柔性车间调度问题的参数
mutation_rate = 0.1  # 变异率
population_size = 100  # 种群大小
max_gen = 1000  # 最大迭代次数
num_jobs = 10  # 作业数量
num_machines = 3  # 机器数量
job_list = [i for i in range(num_jobs)]  # 作业列表

# 遗传算法的基本操作
def crossover(parent1, parent2):
    # 交叉操作，生成新的后代
    child = []
    for i in range(num_jobs):
        if random.random() < 0.5:
            child.append(parent1[i])
        else:
            child.append(parent2[i])
    return child

def mutate(individual):
    # 变异操作，对个体进行微调
    if random.random() < mutation_rate:
        i = random.randint(0, num_jobs - 1)
        j = random.randint(0, num_jobs - 1)
        individual[i], individual[j] = individual[j], individual[i]

def evaluate(individual):
    # 评估个体适应度
    makespan = [0] * num_machines
    for job in individual:
        machine = makespan.index(min(makespan))
        makespan[machine] += job
    return max(makespan)

# 初始化种群
population = []
for i in range(population_size):
    individual = random.sample(job_list, num_jobs)
    population.append(individual)

# 进行遗传算法迭代
for gen in range(max_gen):
    # 评估当前种群
    fitness = [evaluate(individual) for individual in population]
    best_fitness = min(fitness)
    best_individual = population[fitness.index(best_fitness)]
    print("Generation:", gen, "Best Fitness:", best_fitness, "Best Individual:", best_individual)
    
    # 选择下一代
    next_population = []
    for i in range(population_size):
        parent1 = random.choice(population)
        parent2 = random.choice(population)
        child = crossover(parent1, parent2)
        mutate(child)
        next_population.append(child)
    population = next_population

这个代码示例使用遗传算法对柔性车间调度问题进行求解，其中包括种群初始化、交叉、变异、评估和选择等步骤。这只是一个简单的示例，实际应用中需要根据具体情况进行调整和优化。