代码给多一点，邮局选址问题，用分治法

对于邮局选址问题，直接使用分治法并不常见，因为它通常是动态规划或贪婪算法的应用场景。不过，如果你想要模拟分治的思想去寻找解决方案，可以尝试将大问题拆分为若干子问题，每个子问题都是选择在某点设置邮局，然后遍历剩余城市来评估总成本。下面是一个简单的Python实现，仅作为教学演示：

def postal_office_location(locations, max_cost):
    def choose_post_office(locations, cost=0, best_index=None):
        n = len(locations)
        if n == 0 or cost > max_cost:
            return locations, float('inf')  # 如果没有城市或者超过预算，返回无解
        min_avg_dist = float('inf')
        for i in range(n):
            new_locations, avg_dist = choose_post_office(locations[i+1:], cost + locations[i], locations[i])
            if avg_dist < min_avg_dist:
                min_avg_dist = avg_dist
                best_index = i
        return new_locations, min_avg_dist

    # 初始化
    best_locations, best_cost = choose_post_office(locations)

    return best_locations, best_cost

# 示例
locations = [1, 2, 3, 4, 5]  # 假设这是城市之间的距离列表
max_cost = 10  # 预算限制
best_postal_locations, total_cost = postal_office_location(locations, max_cost)
print(f"最佳邮局地点: {best_postal_locations}, 最低总成本: {total_cost}")