python用分支限界法实现旅行商问题

以下是使用Python实现旅行商问题的分支限界算法的示例代码：

import numpy as np

class Node:
    def __init__(self, level, path, bound):
        self.level = level
        self.path = path
        self.bound = bound

def tsp_branch_bound(cities):
    n = len(cities)
    # 初始化起始节点
    start_node = Node(-1, [], 0)
    start_node.bound = bound(start_node, cities)
    # 初始化最优路径和最优距离
    best_path = []
    best_distance = np.inf
    # 初始化队列
    queue = [start_node]
    while queue:
        # 取出队列中的第一个节点
        node = queue.pop(0)
        # 如果该节点的界限小于等于当前最优距离，则不再扩展该节点
        if node.bound < best_distance:
            # 扩展该节点的子节点
            for i in range(n):
                if i not in node.path:
                    new_path = node.path + [i]
                    new_bound = bound(Node(node.level+1, new_path, 0), cities)
                    if new_bound < best_distance:
                        new_node = Node(node.level+1, new_path, new_bound)
                        if new_node.level == n-2:
                            # 如果新节点是倒数第二个节点，则直接计算最终距离
                            last_city = list(set(range(n)) - set(new_node.path))[0]
                            new_path = new_node.path + [last_city]
                            new_distance = distance(new_path, cities)
                            if new_distance < best_distance:
                                best_path = new_path
                                best_distance = new_distance
                        else:
                            # 否则将新节点加入队列
                            queue.append(new_node)
    return best_path, best_distance

def bound(node, cities):
    # 计算当前节点的界限
    n = len(cities)
    path = node.path
    # 计算已经走过的路径长度
    distance = 0
    for i in range(len(path)-1):
        distance += cities[path[i], path[i+1]]
    # 计算当前节点到每个未走过的城市的最小距离和次小距离
    min_distances = []
    for i in range(n):
        if i not in path:
            distances = list(cities[i, j] for j in range(n) if j not in path)
            min_distances.append(min(distances))
    # 计算界限
    bound = distance + sum(sorted(min_distances)[:2])
    return bound

def distance(path, cities):
    # 计算路径长度
    distance = 0
    for i in range(len(path)-1):
        distance += cities[path[i], path[i+1]]
    return distance

# 示例
cities = np.array([[0, 10, 15, 20], [10, 0, 35, 25], [15, 35, 0, 30], [20, 25, 30, 0]])
best_path, best_distance = tsp_branch_bound(cities)
print("最优路径：", best_path)
print("最优距离：", best_distance)