changetime = 10 # A connection is a list of four items: # [, , , ] # departure and arrival time , , # are expressed in whole numbers of minutes, not strings # The problem statement is at # https://cw.fel.cvut.cz/wiki/courses/be5b33pge/practices/examexample1 # and also at the end of this file as a comment. def tominstr( t ): s1 = str( t // 60) s2 = str( t % 60) if len(s1) == 1: s1 = "0" + s1 if len(s2) == 1: s2 = "0" + s2 return s1 + ":" +s2 def tomin( HM ): # "06:08" --> whole minutes --> 368 (min) hm = HM.split(':') return int(hm[0])*60 + int(hm[1]) def time( t1, t2, t3, t4 ): global changetime if t3 - t2 < 10: return 10**11, 10**11 return t4-t1, t4-t3 + t2-t1 # , , , ] conn1 A to B # , , , ] conn2 B to C # 0 1 2 3 def sol( src, trg, allconn): bestTwhole = 10**10 # T stands for time, initial value impossibly big bestTdrive = 10**10 bestConn1, bestConn2 = None, None for conn1 in allconn: if conn1[0] != src: continue for conn2 in allconn: if conn2[0] != conn1[3] or conn2[3] != trg: continue Twhole, Tdrive = time( conn1[1], conn1[2], conn2[1], conn2[2] ) if Twhole < bestTwhole: bestTwhole, bestTdrive = Twhole, Tdrive bestConn1, bestConn2 = conn1, conn2 # printSol(conn1, conn2, Twhole, Tdrive ) continue if Twhole == bestTwhole: if Tdrive < bestTdrive: bestTdrive = Tdrive bestConn1, bestConn2 = conn1, conn2 # printSol(conn1, conn2, Twhole, Tdrive ) return bestConn1, bestConn2, bestTwhole, bestTdrive def printSol(con1, con2, t1min, t2min): print( "plain result" , con1, con2, t1min, t2min ) print( con1[0], tominstr(con1[1]), tominstr(con1[2]), con1[3]) print( con2[0], tominstr(con2[1]), tominstr(con2[2]), con2[3]) print( tominstr(t1min) ) print( tominstr(t2min) ) # g = [, , , ] def getallConn(): allConn = [] N = int(input()) src = input() trg = input() for i in range(N): s = input().strip() #print(s) ss = s.split(":") # ["Mossley West 06", "28 08", "22 Tara Street"] pos0 = ss[0].rfind(" ") # reverse find starts at the end of string pos2 = ss[2].find(" ") pos1 = ss[1].find(" ") g = [] g.append( ss[0][:pos0] ) g.append( tomin( ss[0][pos0:]+":"+ss[1][:pos1] )) # "06:28" --> whole minutes --> 388 min g.append( tomin( ss[1][pos1:]+":"+ss[2][:pos2] )) g.append( ss[2][pos2+1:] ) allConn.append(g) return src, trg, allConn # --------------------------------------------------------------------------- # M A I N # --------------------------------------------------------------------------- # load data src, trg, allConn = getallConn() print(allConn) # debug feature # solve bestConn1, bestConn2, bestTwhole, bestTdrive = sol(src, trg, allConn) # debug printing: #print(bestConn1) #print(bestConn2) #print(bestTwhole) #print(bestTdrive) # final print printSol(bestConn1, bestConn2, bestTwhole, bestTdrive) # --------------------------------------------------------------------------- # P R O B L E M S T A T E M E N T # with examples # --------------------------------------------------------------------------- ''' Hugo is going to travel by train from station A to station B. Unfortunately, there is no direct train connection from A to B. Hugo will have to go by train to another station C, and from there take another train to B. This means that the journey will consist of exactly two consecutive train connections. It does not matter which station will be chosen as a transfer station C. Hugo has a list of all train connections between the stations in the region. He wants to complete the journey in a single day and he wants the duration of the journey to be as short a possible. However, he is limited by the transfer time. He decided he will consider only those pair of connections in which the second connection departs at least 10 minutes after the arrival of the first connection. The duration of a journey is the difference between the arrival time of the second connection and the departure time of the first connection. The transfer duration of a journey is the difference between the departure time of the second connection and the arrival time of the first connection. A journey X is considered better than a journey Y if and only if either the duration of X is smaller than the duration of Y or the duration of X and Y is the same and the transfer duration of X is bigger than the transfer duration of Y. Write a program that will automate the search for the best connection. Input The first input line contains a positive integer N, the number of connections in the list. The second and the third line contain the name of the start and the end station of the journey, respectively. The list of connections occupies next N lines, each line describes one connection. A connection description consists of four items. The first and the fourth items are different station names. The second item is the departure time from the station specified in the first item. The third item is the arrival time to the station specified in the fourth item. Each time item consists of hour and minute specification, both are two digits long and are separated by a colon. All items on a line are separated by spaces. The value of N is between 2 and 1000. The station names do not contain digits or colons. Output The output consists of two lines. The first line contains the duration of the best journey. The the second line contains the duration of the shortest journey decreased by the transfer duration of that journey. Both durations are presented in the HH:MM format which is the same as the format of times in the input. It is guaranteed that there is always at least one best journey. Example 1 Input 5 A B A 01:00 02:30 C # in the program [A, 60, 150, c] A 01:00 02:00 C C 03:00 04:30 B C 02:30 03:30 B C 03:00 04:00 B Output 02:30 02:00 Comment: The connections in the best journey are A 01:00 02:00 C C 02:30 03:30 B Example 2 Input 7 Mossley West Ashtown Mossley West 15:29 16:18 Titanic Quarter/Bridge End Tara Street 08:33 09:54 Ashtown Mossley West 06:28 08:22 Tara Street Tara Street 12:40 14:34 Ashtown Mossley West 09:55 12:27 Tara Street Titanic Quarter/Bridge End 14:20 17:11 Tara Street Mossley West 06:38 07:41 Tara Street Output 03:16 02:24 Comment: The connections in the best journey are Mossley West 06:38 07:41 Tara Street Tara Street 08:33 09:54 Ashtown Example 3 Input 24 Mossley West Newry Longford 04:10 06:10 Newry Newry 15:09 17:21 Fota Fota 17:48 18:15 Mossley West Newry 00:22 03:01 Glenageary Longford 08:27 09:23 Newry Mossley West 06:58 09:06 Fota Newry 08:34 11:08 Glenageary Longford 18:37 21:57 Glenageary Longford 03:27 06:28 Fota Longford 16:09 18:04 Newry Glenageary 00:25 02:35 Mossley West Glenageary 14:31 17:52 Mossley West Mossley West 03:45 07:43 Glenageary Mossley West 14:18 18:08 Fota Glenageary 18:39 20:51 Newry Fota 11:33 11:55 Longford Glenageary 15:15 16:14 Newry Newry 10:29 12:00 Fota Mossley West 01:17 05:10 Longford Newry 02:57 05:00 Longford Glenageary 08:01 10:31 Mossley West Mossley West 04:05 07:56 Glenageary Glenageary 10:03 12:36 Mossley West Output 08:06 04:49 Comment: The connections in the best journey are Mossley West 01:17 05:10 Longford Longford 08:27 09:23 Newry '''