题目介绍

本题的目标是找出两个人访问N个城市所需的总时间，并且尽量减少这个时间，以保证他们不会见面。

此题是一个典型的旅行商问题 (Traveling Salesman Problem)，属于 NP 完全问题，因此不存在时间复杂度短又能完美解决的算法。

常见的解决方法包括暴力枚举、遗传算法、模拟退火算法等。

在实际应用中，通常采用近似算法来解决该问题。

解决方案

1. 暴力枚举

暴力枚举是一种简单粗暴的方法，能够找到最短路径，但随着城市数量的增加，时间复杂度呈指数级增长，因此只适用于小规模的数据。

时间复杂度: $O(n!)$

def dfs(path, visited, current, cost, min_cost):
    """
    path: 已经访问过的城市序列
    visited: 标记某个城市是否被访问过
    current: 当前访问的城市编号
    cost: 从起点到当前城市的时间花费
    min_cost: 最小时间花费
    """
    if len(path) == n:  # 所有城市都访问过
        min_cost = min(min_cost, cost)  # 更新最小时间花费
        return min_cost

    for i in range(n):
        if not visited[i]:
            visited[i] = True
            path.append(i)
            min_cost = dfs(path, visited, i, cost + times[current][i], min_cost)
            path.pop()
            visited[i] = False
    return min_cost

n = 5  # 城市数量
times = [[0, 2, 4, 5, 1],
         [2, 0, 6, 3, 2],
         [4, 6, 0, 7, 5],
         [5, 3, 7, 0, 3],
         [1, 2, 5, 3, 0]]  # 城市之间的时间花费
visited = [False] * n
visited[0] = True
path = [0]  # 起点为0号城市
current = 0
cost = 0
min_cost = float('inf')

min_cost = dfs(path, visited, current, cost, min_cost)
print(min_cost)

2. 遗传算法

遗传算法是通过模拟进化过程来求解问题的一种通用优化方法，具有全局寻优能力和并行处理能力，适用于解决复杂的优化问题。

时间复杂度: $O(n^2)$

import random

n = 5  # 城市数量
times = [[0, 2, 4, 5, 1],
         [2, 0, 6, 3, 2],
         [4, 6, 0, 7, 5],
         [5, 3, 7, 0, 3],
         [1, 2, 5, 3, 0]]  # 城市之间的时间花费

def get_path_distance(path):
    """
    计算路径长度
    """
    distance = 0
    for i in range(n - 1):
        distance += times[path[i]][path[i+1]]
    return distance

def generate_random_population(size):
    """
    生成随机种群
    """
    population = []
    for i in range(size):
        path = list(range(n))
        random.shuffle(path)
        population.append(path)
    return population

def crossover(parent1, parent2):
    """
    交叉操作
    """
    child1 = [-1] * n
    child2 = [-1] * n
    begin = random.randint(0, n-1)
    end = random.randint(begin+1, n)
    for i in range(begin, end):
        child1[i] = parent1[i]
        child2[i] = parent2[i]
    for i in range(n):
        if parent2[i] not in child1:
            for j in range(n):
                if child1[j] == -1:
                    child1[j] = parent2[i]
                    break
        if parent1[i] not in child2:
            for j in range(n):
                if child2[j] == -1:
                    child2[j] = parent1[i]
                    break
    return child1, child2

def mutate(path):
    """
    变异操作
    """
    i = random.randint(0, n-1)
    j = random.randint(0, n-1)
    path[i], path[j] = path[j], path[i]
    return path

def select_fittest(population, num_parents):
    """
    选择适应度高的个体
    """
    population_fitness = [(path, get_path_distance(path)) for path in population]
    sorted_fitness = sorted(population_fitness, key=lambda x: x[1])
    return [path for path, _ in sorted_fitness[:num_parents]]

def run_genetic_algorithm(population_size, num_generations):
    population = generate_random_population(population_size)
    for i in range(num_generations):
        # 选择适应度好的个体
        fittest = select_fittest(population, 10)

        new_population = []
        new_population.extend(fittest)

        # 交叉操作
        for i in range(population_size - len(fittest)):
            parent1, parent2 = random.sample(fittest, k=2)
            child1, child2 = crossover(parent1, parent2)
            new_population.append(mutate(child1))
            new_population.append(mutate(child2))

        population = new_population

    # 返回适应度最好的路径
    return select_fittest(population, 1)[0]

print(get_path_distance(run_genetic_algorithm(50, 100)))

3. 模拟退火算法

模拟退火算法是通过模拟固体物理退火过程来求解优化问题的一种通用算法，具有全局寻优能力和随机性，适用于解决连续优化问题。

时间复杂度: $O(n)$

import math
import random

n = 5  # 城市数量
times = [[0, 2, 4, 5, 1],
         [2, 0, 6, 3, 2],
         [4, 6, 0, 7, 5],
         [5, 3, 7, 0, 3],
         [1, 2, 5, 3, 0]]  # 城市之间的时间花费

def get_path_distance(path):
    """
    计算路径长度
    """
    distance = 0
    for i in range(n - 1):
        distance += times[path[i]][path[i+1]]
    return distance

def get_initial_solution():
    """
    生成初始解
    """
    path = list(range(n))
    random.shuffle(path)
    return path

def get_neighbor(path):
    """
    生成邻居解
    """
    neighbor = path.copy()
    i = random.randint(0, n-1)
    j = random.randint(0, n-1)
    neighbor[i], neighbor[j] = neighbor[j], neighbor[i]
    return neighbor

def acceptance_probability(old_cost, new_cost, T):
    """
    计算接受概率
    """
    if new_cost < old_cost:
        return 1.0
    else:
        return math.exp((old_cost - new_cost) / T)

def run_simulated_annealing(T):
    solution = get_initial_solution()
    current_cost = get_path_distance(solution)
    for i in range(1000):
        neighbor = get_neighbor(solution)
        new_cost = get_path_distance(neighbor)
        ap = acceptance_probability(current_cost, new_cost, T)
        if random.random() < ap:
            solution = neighbor
            current_cost = new_cost
        T *= 0.99  # 降温
    return get_path_distance(solution)

print(run_simulated_annealing(1000))

总结

以上是三种解决旅行商问题的方法，其中暴力枚举能找到最优解，但时间复杂度很高，只适用于小规模的数据；遗传算法和模拟退火算法虽不能保证找到最优解，但能在较短的时间内找到较优解，适用于大规模数据的处理。

在实际应用中，可根据问题的实际情况选择不同的解决方法。