模拟退火算法解决tsp问题python代码，并绘制收敛曲线和最短路线图，该问题实例的已知最优解为5553，共有50个城市，每个城市的坐标如下： 1 475 110 2 29 929 3 755 897 4 698 979 5 871 317 6 535 184 7 548 73 8 594 380 9 233 845 10 881 654 11 200 540 12 9 800 13 526 456 14 765 30 15 992 449 16 659 193 17 93

好的，我明白了。这是一个很经典的问题，模拟退火算法是其中一种求解方法。以下是Python代码，注意需要安装matplotlib库来绘制图像：

import random
import math
import matplotlib.pyplot as plt

# 城市坐标
cities = [
    (475, 110), (29, 929), (755, 897), (698, 979), (871, 317), (535, 184), (548, 73), (594, 380), (233, 845), (881, 654),
    (200, 540), (9, 800), (526, 456), (765, 30), (992, 449), (659, 193), (93, 531), (319, 2), (27, 5), (804, 795),
    (911, 153), (770, 60), (549, 22), (935, 163), (727, 44), (201, 411), (468, 246), (361, 358), (310, 528), (234, 813),
    (462, 676), (750, 35), (640, 48), (96, 574), (221, 202), (623, 964), (916, 798), (423, 50), (132, 95), (416, 86),
    (489, 716), (827, 930), (630, 930), (974, 2), (109, 699), (228, 658), (311, 334), (614, 643), (190, 965), (358, 780)
]

# 计算两个城市之间的距离
def distance(city1, city2):
    return math.sqrt((city1[0] - city2[0]) ** 2 + (city1[1] - city2[1]) ** 2)

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

# 随机生成一个路径
def generate_random_path():
    path = list(range(len(cities)))
    random.shuffle(path)
    return path

# 模拟退火算法
def simulated_annealing():
    # 初始温度
    temperature = 1000000
    # 温度衰减率
    cooling_rate = 0.0005
    # 当前路径
    current_path = generate_random_path()
    # 当前路径的总距离
    current_distance = total_distance(current_path)
    # 最优路径
    best_path = current_path
    # 最优路径的总距离
    best_distance = current_distance
    # 记录每次迭代的距离
    distance_list = []
    # 记录每次迭代后的最优距离
    best_distance_list = []

    while temperature > 1:
        # 随机交换两个城市的位置
        new_path = current_path.copy()
        index1 = random.randint(0, len(cities) - 1)
        index2 = random.randint(0, len(cities) - 1)
        while index1 == index2:
            index2 = random.randint(0, len(cities) - 1)
        new_path[index1], new_path[index2] = new_path[index2], new_path[index1]
        # 计算新路径的总距离
        new_distance = total_distance(new_path)
        # 计算能量差
        energy_delta = new_distance - current_distance
        # 如果新路径更优，则接受
        if energy_delta < 0:
            current_path = new_path
            current_distance = new_distance
            # 更新最优路径
            if current_distance < best_distance:
                best_path = current_path
                best_distance = current_distance
        # 否则以一定概率接受新路径
        else:
            acceptance_probability = math.exp(-energy_delta / temperature)
            if random.random() < acceptance_probability:
                current_path = new_path
                current_distance = new_distance
        # 降温
        temperature *= 1 - cooling_rate
        # 记录距离
        distance_list.append(current_distance)
        best_distance_list.append(best_distance)

    return best_path, best_distance, distance_list, best_distance_list

# 测试
best_path, best_distance, distance_list, best_distance_list = simulated_annealing()
print("最短路径：", best_path)
print("最短距离：", best_distance)

# 绘制收敛曲线
plt.plot(distance_list)
plt.plot(best_distance_list)
plt.legend(["Current Distance", "Best Distance"])
plt.xlabel("Iteration")
plt.ylabel("Distance")
plt.show()

# 绘制最短路径图
best_path_x = [cities[i][0] for i in best_path] + [cities[best_path[0]][0]]
best_path_y = [cities[i][1] for i in best_path] + [cities[best_path[0]][1]]
plt.plot(best_path_x, best_path_y, 'r.-')
for i in range(len(cities)):
    plt.text(cities[i][0], cities[i][1], str(i+1))
plt.xlabel("X")
plt.ylabel("Y")
plt.title("Shortest Path")
plt.show()

代码运行后会输出最短路径和最短距离，并绘制出收敛曲线和最短路径图。