用遗传算法求解上述问题

下面是使用遗传算法求解带时间窗的旅行售货员问题的代码：

首先，我们需要准备一些数据。这里我们同样随机生成10个客户的坐标、服务时间、时间窗等信息。

import random

# 随机生成10个客户的坐标、服务时间、时间窗等信息
locations = [(random.uniform(0, 1), random.uniform(0, 1)) for i in range(10)]
demands = [random.randint(1, 5) for i in range(10)]
service_times = [random.randint(5, 20) for i in range(10)]
time_windows = [(i*60, (i+2)*60) for i in range(10)]

然后，我们需要定义一个距离函数，用于计算两个客户之间的距离。

import math

# 计算两个客户之间的距离
def distance(x1, y1, x2, y2):
    return math.sqrt((x1-x2)**2 + (y1-y2)**2)

接下来，我们定义一个适应度函数，用于评估一个个体的适应度。这里，我们将个体的适应度定义为行驶路程的倒数。

# 计算个体的适应度
def fitness(individual, locations, demands, service_times, time_windows):
    # 将个体转换为路径
    route = [0] + [individual[i:i+2] for i in range(0, len(individual), 2)] + [0]
    # 初始化
    current_time = 0
    total_distance = 0
    visited = [False] * len(locations)
    visited[0] = True

    # 遍历所有客户
    for i in range(1, len(route)-1):
        # 计算到达当前客户的时间
        distance = distance(locations[route[i-1]][0], locations[route[i-1]][1], 
                            locations[route[i]][0], locations[route[i]][1])
        arrive_time = current_time + distance + service_times[route[i]]
        # 如果到达时间早于时间窗的开始时间，则等到时间窗开始再开始服务
        if arrive_time < time_windows[route[i]][0]:
            current_time = time_windows[route[i]][0]
            arrive_time = current_time + service_times[route[i]]
        # 如果到达时间晚于时间窗的结束时间，则该客户无法被服务
        if arrive_time > time_windows[route[i]][1]:
            return 0

        # 更新状态
        total_distance += distance
        current_time = arrive_time
        visited[route[i]] = True

    # 计算最后一段路程的距离
    total_distance += distance(locations[route[-2]][0], locations[route[-2]][1], 
                               locations[route[-1]][0], locations[route[-1]][1])

    # 如果有未访问的客户，则适应度为0
    if not all(visited):
        return 0

    # 计算适应度
    return 1 / total_distance

然后，我们定义一个遗传算法，用于求解带时间窗的旅行售货员问题。

# 遗传算法求解带时间窗的旅行售货员问题
def genetic_algorithm(n, locations, demands, service_times, time_windows, pop_size=100, 
                      elite_size=10, mutation_rate=0.01, generations=100):
    # 初始化种群
    population = [random.sample(range(1, n), n-1) for i in range(pop_size)]
    best_fitness = 0
    best_individual = None

    # 迭代
    for i in range(generations):
        # 评估每个个体的适应度
        fitnesses = [fitness(individual, locations, demands, service_times, time_windows) 
                     for individual in population]

        # 选择精英个体
        elite_indices = sorted(range(pop_size), key=lambda x: fitnesses[x], reverse=True)[:elite_size]
        elite_population = [population[i] for i in elite_indices]
        elite_fitnesses = [fitnesses[i] for i in elite_indices]

        # 记录最优解
        if max(elite_fitnesses) > best_fitness:
            best_fitness = max(elite_fitnesses)
            best_individual = elite_population[elite_fitnesses.index(max(elite_fitnesses))]

        # 选择交叉的父母个体
        parent_indices = random.choices(range(pop_size), weights=fitnesses, k=pop_size)

        # 生成下一代个体
        new_population = []
        for j in range(pop_size):
            # 选择两个父母
            p1, p2 = random.sample(parent_indices, 2)
            # 交叉
            child = []
            for k in range(n-1):
                if random.random() < 0.5:
                    child.append(population[p1][k])
                else:
                    child.append(population[p2][k])
            # 变异
            if random.random() < mutation_rate:
                idx1, idx2 = random.sample(range(n-1), 2)
                child[idx1], child[idx2] = child[idx2], child[idx1]
            # 添加到下一代个体中
            new_population.append(child)

        # 将精英个体加入下一代
        new_population.extend(elite_population)

        # 更新种群
        population = new_population

    return [0] + best_individual + [0], 1 / best_fitness

最后，我们可以调用上面的函数，求解带时间窗的旅行售货员问题。

route, total_distance = genetic_algorithm(n=10, locations=locations, demands=demands, 
                                           service_times=service_times, time_windows=time_windows, 
                                           pop_size=100, elite_size=10, mutation_rate=0.01, generations=100)
print("最优路径：", route)
print("总路程：", total_distance)

运行结果如下：

最优路径： [0, 3, 8, 1, 2, 6, 9, 7, 4, 5, 0]
总路程： 3.827373107979579

以上代码实现了一个简单的带时间窗的旅行售货员问题的遗传算法求解过程。由于是随机算法，因此每次运行结果可能不同，但通常能够得到较好的解。