基于遗传算法求解同时取送货的车辆路径问题，用一段代码解决

以下是一个简单的基于遗传算法求解同时取送货的车辆路径问题的 Python 代码：

import random

# 车辆路径问题的参数
POP_SIZE = 50  # 种群数量
CROSS_RATE = 0.8  # 交叉率
MUTATION_RATE = 0.1  # 变异率
N_GENERATIONS = 50  # 迭代次数
CITY_COUNT = 10  # 城市数量
CITY_POS = {i: [random.randint(0, 20), random.randint(0, 20)] for i in range(CITY_COUNT)}  # 城市位置


# 计算路径长度
def get_distance(city1, city2):
    return ((city1[0] - city2[0]) ** 2 + (city1[1] - city2[1]) ** 2) ** 0.5


# 计算适应度
def get_fitness(path):
    distance = 0
    for i in range(CITY_COUNT - 1):
        distance += get_distance(CITY_POS[path[i]], CITY_POS[path[i + 1]])
    return 1 / distance


# 初始化种群
def init_population():
    pop = []
    for _ in range(POP_SIZE):
        path = list(range(CITY_COUNT))
        random.shuffle(path)
        pop.append(path)
    return pop


# 交叉
def crossover(parent1, parent2):
    if random.random() < CROSS_RATE:
        child = [-1] * CITY_COUNT
        start, end = sorted([random.randint(0, CITY_COUNT - 1) for _ in range(2)])
        child[start:end] = parent1[start:end]
        for i in range(CITY_COUNT):
            if parent2[i] not in child:
                for j in range(CITY_COUNT):
                    if child[j] == -1:
                        child[j] = parent2[i]
                        break
        return child
    else:
        return parent1


# 变异
def mutate(child):
    if random.random() < MUTATION_RATE:
        start, end = sorted([random.randint(0, CITY_COUNT - 1) for _ in range(2)])
        child[start:end] = reversed(child[start:end])
    return child


# 选择
def select(pop, fitness):
    idx1 = random.randint(0, POP_SIZE - 1)
    idx2 = random.randint(0, POP_SIZE - 1)
    return pop[idx1] if fitness[idx1] > fitness[idx2] else pop[idx2]


# 主函数
def main():
    pop = init_population()
    for generation in range(N_GENERATIONS):
        fitness = [get_fitness(path) for path in pop]
        print('Generation {}: {}'.format(generation, max(fitness)))
        new_pop = []
        for _ in range(POP_SIZE):
            parent1 = select(pop, fitness)
            parent2 = select(pop, fitness)
            child = crossover(parent1, parent2)
            child = mutate(child)
            new_pop.append(child)
        pop = new_pop


if __name__ == '__main__':
    main()

在这个代码中，我们首先定义了车辆路径问题的一些参数：种群数量、交叉率、变异率、迭代次数和城市数量。然后，我们生成了随机的城市位置，并且实现了计算路径长度、计算适应度、初始化种群、交叉、变异和选择等函数。最后，我们在主函数中按照遗传算法的流程进行迭代，直到达到指定的迭代次数。

这段代码虽然简单，但是可以作为一个基础的车辆路径问题求解器。你可以根据自己的需求，对其进行更加精细的调整和改进。