在给定旅行预算（以分钟为单位）的情况下，如何使用旅行时间矩阵找到我可以访问的最大城市数量

Posted 2023-02-19

【中文标题】在给定旅行预算（以分钟为单位）的情况下，如何使用旅行时间矩阵找到我可以访问的最大城市数量

【英文标题】：How can I find the maximum number of cities that I can visit given a travel budget (in minutes) using a travel time matrix

【发布时间】：2021-11-23 11:42:55

【问题描述】：

我有一个 12 个相互连接的城市的列表，无一例外。唯一担心的是旅行时间。每个城市的名称是here。城市对之间的距离矩阵（以分钟为单位表示旅行时间）为here。

在给定的旅行预算（例如 800 分钟）的情况下，我如何才能知道我可以从原籍城市（可以是 12 个城市中的任何一个）访问多少个城市。

您在旅途中不能两次访问同一个城市，也不必担心返回原籍。我不能超出我的旅行预算。

【问题讨论】：

除了蛮力尝试所有方法之外，没有任何神奇的方法。您需要使用递归。

*** 旨在帮助解决具体的技术问题，而不是开放式的代码或建议请求。

【参考方案1】：

import numpy as np
from scipy.spatial import distance_matrix
from sklearn.cluster import AgglomerativeClustering


def find_cities(dist, budget):     # dist: a 12x12 matrix of travel time in minutes between city pairs; budget: max travel time allowed for the trip (in mins)

    assert len(dist) == 12        # there are 12 cities to visit and each one has a pairwise cost with all other 11 citis 

    clusters = []                 # list of cluster labels from 1..n where n is no of cities to be visited  

    dists = [0] + [row[1:] for row in dist]       # start-to-start costs have been excluded from the distances array which only contains intercity distances     
    linkage = 'complete'             # complete linkage used here because we want an optimal solution i.e., finding minimum number of clusters required  

    ac = AgglomerativeClustering(affinity='precomputed', linkage=linkage, compute_full_tree=True)          # affinity must be precomputed or function otherwise it will use euclidean distance by default !!! 
                        # compute full tree ensures that I get all possible clustesr even if they don't make up entire population! This is needed so that I can determine how many clusters need to be created given my budget constraints below  

    Z = ac.fit_predict(dists).tolist()   # AgglomerativeClustering.fit_predict returns list of cluster labels for each city 

    while budget >= min(dists):          # while my budget is greater than the minimum intercity travel cost, i.e., I can still visit another city
        if len(set(Z)) > 1:              # at least 2 clusters are needed to form a valid tour so continue only when there are more than one cluster left in Z
            c1 = np.argmax([sum([i==j for j in Z]) for i in set(Z)])      # find which clustes has max no of cities associated with it and that will be the next destination (cities within this cluster have same label as their parent cluster!)          # numpy argmax returns index of maximum value along an axis; here we want to know which group has most elements! 
            c2 = [j for j,val in enumerate(Z) if val == Z[c1]][0]         # retrieve first element from the group whose parent is 'cluster' returned by previous line      

            clusters += [c2 + 1]                             ## add new destination found into our trip plan/list "clusters" after converting its city id back into integer starting from 1 instead of 0 like array indices do!!              
            dists += [dist[c1][c2]]                           ## update total distance travelled so far based on newly added destination ... note: distances between two adjacent cities always equals zero because they fall under same cluster
            budget -= dists[-1]                               ## update travel budget by subtracting the cost of newly added destination from our total budget 

        else: break   # when there is only one city left in Z, then stop! it's either a single city or two cities are connected which means cost between them will always be zero!              

    return clusters   # this is a list of integers where each integer represents the id of city that needs to be visited next! 
    
def main():

    with open('uk12_dist.txt','r') as f:      ## read travel time matrix between cities from file           ## note: 'with' keyword ensures file will be closed automatically after reading or writing operation done within its block!!! 
        dist = [[int(num) for num in line.split()] for line in f]       ## convert text data into array/list of lists using list comprehension; also ensure all data are converted into int before use!        

    with open('uk12_name.txt','r') as f:      ## read names of 12 cities from file              ## note: 'with' keyword ensures file will be closed automatically after reading or writing operation done within its block!!! 
        name = [line[:-1].lower().replace(" ","") for line in f]          ## remove newline character and any leading/trailing spaces, then convert all characters to lowercase; also make sure there's no space between first and last name (which results in empty string!) otherwise won't match later when searching distances!!

    budget = 800                             # max travel budget allowed (in mins) i.e., 8 hrs travelling at 60 mins per km which means I can cover about 800 kms on a full tank!                   

    print(find_cities(dist,budget), "\n")     ## print(out list of city ids to visit next!
    print("Total distance travelled: ", sum(dist[i][j] for i, j in enumerate([0]+find_cities(dist,budget))), "\n" )     # calculate total cost/distance travelled so far by adding up all distances between cities visited so far - note index '0' has been added at start because 0-2 is same as 2-0 and it's not included in find_cities() output above !

    while True:
        try:                                ## this ensures that correct input from user will be obtained only when required!!              
            budget = int(raw_input("\nEnter your travel budget (in minutes): "))   # get new travel budget from user and convert into integer before use!!! 
            
            if budget <= 800: break          ## stop asking for valid input only when the value entered by user isn't greater than 800 mins or 8 hrs !!                              

        except ValueError:                   ## catch exception raised due to invalid data type; continue asking until a valid number is given by user!!                     
            pass

    print(name[find_cities(dist,budget)[1]],"->",name[find_cities(dist,budget)[2]],"-> ...",name[find_cities(dist,budget)[-1]] )## print out the city names of cities to visit next!

    return None

if __name__ == '__main__': main()

【讨论】：

我得到一个 ValueError: Expected 2D array, got 1D array instead: array=[0 list([300, 352, 466, 217, 238, 431, 336, 451, 47, 415, 515]) list([0, 638, 180, 595, 190, 138, 271, 229, 236, 214, 393]) - 我也可以添加始发城市名称吗？

待机。检查代码，很快就会回复您。如果你有时间给我发电子邮件。我有一个问题要问你，这将引发对话。 （在 pm.me 上进行评论）