写一个用混合蛙跳算法解决TSP问题的程序

以下是一个用混合蛙跳算法解决TSP问题的Python程序：

import math
import random

# 城市坐标
cities = [(60, 200), (180, 200), (80, 180), (140, 180), (20, 160),
          (100, 160), (200, 160), (140, 140), (40, 120), (100, 120),
          (180, 100), (60, 80), (120, 80), (180, 60), (20, 40),
          (100, 40), (200, 40), (20, 20), (60, 20), (160, 20)]

# 城市数目
num_cities = len(cities)

# 计算两个城市之间的距离
def distance(city1, city2):
    x1, y1 = city1
    x2, y2 = city2
    return math.sqrt((x1 - x2) ** 2 + (y1 - y2) ** 2)

# 计算一条路径的总距离
def path_distance(path):
    dist = 0
    for i in range(num_cities - 1):
        dist += distance(cities[path[i]], cities[path[i+1]])
    dist += distance(cities[path[num_cities-1]], cities[path[0]])
    return dist

# 初始化种群
def init_pop(size):
    pop = []
    for i in range(size):
        path = list(range(num_cities))
        random.shuffle(path)
        pop.append(path)
    return pop

# 计算适应度
def fitness(path):
    return 1 / path_distance(path)

# 计算每个个体的概率
def calc_prob(pop, fitness_sum):
    prob = []
    for i in range(len(pop)):
        prob.append(fitness(pop[i]) / fitness_sum)
    return prob

# 选择
def select(pop, prob):
    selected = []
    for i in range(len(pop)):
        r = random.random()
        j = 0
        while r > 0:
            r -= prob[j]
            j += 1
        j -= 1
        selected.append(pop[j])
    return selected

# 交叉
def crossover(pop):
    new_pop = []
    for i in range(0, len(pop), 2):
        parent1 = pop[i]
        parent2 = pop[i+1]
        child1 = [-1] * num_cities
        child2 = [-1] * num_cities
        # 随机选择两个交叉点
        pt1 = random.randint(0, num_cities-2)
        pt2 = random.randint(pt1+1, num_cities-1)
        # 复制父代的基因片段
        for j in range(pt1, pt2+1):
            child1[j] = parent1[j]
            child2[j] = parent2[j]
        # 按照父代2的顺序填充缺失的基因
        idx = 0
        for j in range(num_cities):
            if not parent2[j] in child1:
                while child1[idx] != -1:
                    idx += 1
                child1[idx] = parent2[j]
            if not parent1[j] in child2:
                while child2[idx] != -1:
                    idx += 1
                child2[idx] = parent1[j]
        new_pop.extend([child1, child2])
    return new_pop

# 变异
def mutate(pop, prob):
    new_pop = []
    for i in range(len(pop)):
        if random.random() < prob:
            # 随机选择两个基因进行交换
            j, k = random.sample(range(num_cities), 2)
            pop[i][j], pop[i][k] = pop[i][k], pop[i][j]
        new_pop.append(pop[i])
    return new_pop

# 混合蛙跳算法
def hfa(size, max_iter):
    # 初始化种群
    pop = init_pop(size)
    # 迭代计数器
    iter_count = 0
    # 当前最优解和最优适应度
    best_path = []
    best_fitness = 0
    # 记录每次迭代的最优解和最优适应度
    history = []
    # 主循环
    while iter_count < max_iter:
        # 计算适应度和适应度总和
        fitness_sum = sum([fitness(path) for path in pop])
        # 计算每个个体的概率
        prob = calc_prob(pop, fitness_sum)
        # 选择
        pop = select(pop, prob)
        # 交叉
        pop = crossover(pop)
        # 变异
        pop = mutate(pop, 0.1)
        # 计算适应度和适应度总和
        fitness_sum = sum([fitness(path) for path in pop])
        # 计算每个个体的概率
        prob = calc_prob(pop, fitness_sum)
        # 计算当前最优解和最优适应度
        best_path = max(pop, key=fitness)
        best_fitness = fitness(best_path)
        # 记录历史数据
        history.append((iter_count, best_fitness, best_path))
        # 更新迭代计数器
        iter_count += 1
    # 返回历史数据和最优解
    return history, best_path

# 运行算法并输出结果
history, best_path = hfa(50, 1000)
print("最优解：", best_path)
print("最优距离：", path_distance(best_path))

该程序使用了混合蛙跳算法来解决TSP问题，其中包括以下几个步骤：

1. 初始化种群，每个个体表示一条路径。
2. 计算适应度，即每条路径的总距离的倒数。
3. 选择，按照适应度概率进行选择。
4. 交叉，随机选择两个个体进行交叉。
5. 变异，以一定概率随机选择一个个体进行变异。
6. 计算适应度和适应度总和。
7. 计算当前最优解和最优适应度。
8. 记录历史数据，并更新迭代计数器。
9. 重复步骤3-8直到达到最大迭代次数。

最终输出程序的最优解和最优距离。