分支限界法求解货郎问题伪码

下面是使用分支限界法求解货郎问题的伪码：

class City:
    def __init__(self, name, distance):
        self.name = name
        self.distance = distance

def tsp(cities):
    n = len(cities)
    path = [0] * (n + 1)
    visited = [False] * n
    best_path = [0] * (n + 1)
    best_distance = float('inf')
    
    def calculate_distance(path):
        total_distance = 0
        for i in range(n):
            total_distance += cities[path[i]].distance[path[i+1]]
        return total_distance
    
    def branch_and_bound(curr_city, curr_distance, depth):
        nonlocal best_distance, best_path
        
        if depth == n:
            curr_distance += cities[curr_city].distance[0]
            if curr_distance < best_distance:
                best_distance = curr_distance
                best_path = path.copy()
        
        for i in range(1, n):
            if not visited[i]:
                path[depth] = i
                visited[i] = True
                new_distance = curr_distance + cities[curr_city].distance[i]
                
                if new_distance < best_distance:
                    branch_and_bound(i, new_distance, depth + 1)
                
                visited[i] = False
    
    branch_and_bound(0, 0, 0)
    
    return best_path, best_distance

这段伪码实现了一个`City`类来表示每个城市，其中包含了城市的名称和与其他城市的距离。`tsp`函数使用分支限界法来解决货郎问题。它通过递归地在每个节点处进行分支和限界来搜索最优路径。`path`列表用于保存当前的路径，`visited`列表用于标记已经访问过的城市。`best_path`和`best_distance`分别用于保存最优路径和最小距离。`calculate_distance`函数用于计算给定路径的总距离。`branch_and_bound`函数是核心的分支限界算法实现，它通过遍历每个未访问过的城市进行分支，并在每个节点处进行限界剪枝判断。最终，函数返回最优路径和最小距离。

请注意，这只是一个简化的伪码示例，实际的实现可能需要更多的优化和边界条件处理。