求解tsp问题，python实现，采用遗传算法，其中变异算子使用反转突变

很高兴回答你的问题。关于TSP问题的遗传算法实现，变异算子采用反转突变的思想，你可以参考以下的Python代码实现：

import random

# 遗传算法参数设置
POPULATION_SIZE = 100           # 种群数量
CROSSOVER_PROBABILITY = 0.9     # 交叉概率
MUTATION_PROBABILITY = 0.2      # 变异概率
GENERATION_NUM = 100            # 迭代次数

# 城市坐标（这里只列举了5个城市，你可以根据实际情况进行修改）
CITY_COORDINATES = [[1, 1], [2, 3], [5, 4], [7, 2], [4, 7]]

# 计算个体适应度函数（这里采用路径总长度作为适应度函数）
def fitness(individual):
    total_distance = 0
    for i in range(len(individual) - 1):
        # 计算两个城市之间的距离
        distance = ((CITY_COORDINATES[individual[i]][0] - CITY_COORDINATES[individual[i+1]][0])**2 +
                    (CITY_COORDINATES[individual[i]][1] - CITY_COORDINATES[individual[i+1]][1])**2)**0.5
        total_distance += distance
    return 1 / total_distance

# 初始化种群
def init_population():
    population = []
    for i in range(POPULATION_SIZE):
        individual = list(range(len(CITY_COORDINATES)))
        random.shuffle(individual)
        population.append(individual)
    return population

# 选择操作
def selection(population):
    fitness_sum = sum([fitness(individual) for individual in population])
    p = [fitness(individual) / fitness_sum for individual in population]
    cumulative_p = [sum(p[:i+1]) for i in range(len(p))]
    parents = []
    for i in range(len(population)):
        r = random.random()
        for j in range(len(cumulative_p)):
            if r < cumulative_p[j]:
                parents.append(population[j])
                break
    return parents

# 交叉操作
def crossover(parents):
    children = []
    for i in range(len(parents)):
        parent1, parent2 = random.sample(parents, 2)
        child = [-1] * len(parent1)
        start = random.randint(0, len(parent1)-1)
        end = random.randint(start, len(parent1)-1)
        for j in range(start, end+1):
            child[j] = parent1[j]
        for j in range(len(parent2)):
            if parent2[j] not in child:
                for k in range(len(child)):
                    if child[k] == -1:
                        child[k] = parent2[j]
                        break
        children.append(child)
    return children

# 变异操作
def mutation(children):
    for i in range(len(children)):
        r = random.random()
        if r < MUTATION_PROBABILITY:
            start = random.randint(0, len(children[i])-1)
            end = random.randint(start, len(children[i])-1)
            children[i][start:end+1] = reversed(children[i][start:end+1])
    return children

# 主函数
def tsp_ga():
    population = init_population()
    for i in range(GENERATION_NUM):
        parents = selection(population)
        children = crossover(parents)
        children = mutation(children)
        population = parents + children
        population = sorted(population, key=lambda x: fitness(x), reverse=True)
        population = population[:POPULATION_SIZE]
    # 返回最优个体
    return population[0], 1 / fitness(population[0])

# 测试代码
best_individual, best_fitness = tsp_ga()
print('Best individual:', best_individual)
print('Best fitness:', best_fitness)

需要注意的是，在上述代码中，并没有对变异算子进行“反转突变”的具体实现，而是利用了Python中的列表切片和反转函数进行实现。具体来说，变异操作中随机选择一个子序列，然后将该子序列进行反转处理即可。