给我两个不同的方法来解决时间旅行商问题的代码

下面是两种不同的解决旅行商问题的方法及其代码实现：

方法一：遗传算法

遗传算法是一种基于进化论的优化算法，它模拟自然界中生物进化的过程，通过交叉、变异等操作来产生新的解，并逐步优化解的质量。下面是使用遗传算法解决旅行商问题的 Python 代码实现：

import random

# 旅行商问题函数
def tsp(city_list, pop_size=50, elite_size=10, mutation_rate=0.01, generations=100):
    # 初始化种群
    def init_population(pop_size, city_list):
        population = []
        for i in range(pop_size):
            individual = random.sample(city_list, len(city_list))
            population.append(individual)
        return population

    # 计算适应度函数
    def fitness(individual):
        dist = 0
        for i in range(len(individual)-1):
            dist += distance(individual[i], individual[i+1])
        dist += distance(individual[-1], individual[0])
        return 1/dist

    # 计算两点之间的距离
    def distance(city1, city2):
        return ((city1[0]-city2[0])**2 + (city1[1]-city2[1])**2)**0.5

    # 选择
    def selection(population, elite_size):
        fitness_scores = [(individual, fitness(individual)) for individual in population]
        fitness_scores.sort(key=lambda x: x[1], reverse=True)
        elite = [x[0] for x in fitness_scores[:elite_size]]
        selection_probs = [(x[0], x[1]/sum([y[1] for y in fitness_scores])) for x in fitness_scores]
        selection_probs.sort(key=lambda x: x[1], reverse=True)
        selection = [x[0] for x in selection_probs[:len(population)-elite_size]]
        return elite + selection

    # 交叉
    def crossover(parent1, parent2):
        gene1 = random.randint(0, len(parent1)-1)
        gene2 = random.randint(0, len(parent1)-1)
        start = min(gene1, gene2)
        end = max(gene1, gene2)
        child = [-1] * len(parent1)
        for i in range(start, end+1):
            child[i] = parent1[i]
        for i in range(len(parent2)):
            if parent2[i] not in child:
                for j in range(len(child)):
                    if child[j] == -1:
                        child[j] = parent2[i]
                        break
        return child

    # 变异
    def mutation(individual, mutation_rate):
        for i in range(len(individual)):
            if random.random() < mutation_rate:
                j = random.randint(0, len(individual)-1)
                individual[i], individual[j] = individual[j], individual[i]
        return individual

    # 进化
    def evolve(population, elite_size, mutation_rate):
        elite = selection(population, elite_size)
        children = []
        while len(children) < len(population) - elite_size:
            parent1 = random.choice(elite)
            parent2 = random.choice(elite)
            child = crossover(parent1, parent2)
            child = mutation(child, mutation_rate)
            children.append(child)
        return elite + children

    # 主函数
    population = init_population(pop_size, city_list)
    for i in range(generations):
        population = evolve(population, elite_size, mutation_rate)
    best_route = max([(individual, fitness(individual)) for individual in population], key=lambda x: x[1])[0]
    return best_route

方法二：动态规划

动态规划是一种基于状态转移的优化算法，它通常用于求解一些具有重叠子问题和最优子结构的问题。对于旅行商问题，可以使用动态规划来求解最短路径。下面是使用动态规划解决旅行商问题的 Python 代码实现：

import itertools

# 旅行商问题函数
def tsp(city_list):
    # 初始化距离矩阵
    dist = [[0]*len(city_list) for _ in range(len(city_list))]
    for i in range(len(city_list)):
        for j in range(len(city_list)):
            dist[i][j] = ((city_list[i][0]-city_list[j][0])**2 + (city_list[i][1]-city_list[j][1])**2)**0.5

    # 动态规划求解最短路径
    n = len(city_list)
    dp = [[float('inf')]*n for _ in range(1<<n)]
    dp[1][0] = 0
    for S in range(1<<n):
        for i in range(n):
            if S&(1<<i):
                for j in range(n):
                    if j != i and S&(1<<j):
                        dp[S][i] = min(dp[S][i], dp[S^(1<<i)][j] + dist[j][i])
    return dp[(1<<n)-1][0]

这两种方法各有优缺点，遗传算法的优点是可以处理大规模问题，但需要进行参数调整；动态规划的优点是精度高，但对于大规模问题可能会出现计算时间过长的问题。因此，在实际应用中需要根据问题规模和实际需求选择合适的算法。