Howdy! This time, problems started getting more and more complex, and that’s good, because I like the challenge! The next problem is Get to Work:

You work for a company that has E employees working in town T. There are N towns in the area where the employees live. You want to ensure that everyone will be able to make it to work. Some of the employees are drivers and can drive P passengers. A capacity of P == 1 indicates that the driver can only transport themselves to work. You want to ensure that everyone will be able to make it to work and you would like to minimize the number of cars on the road.

You want to calculate the number of cars on the road, with these requirements:

• Every employee can get to town T.
• The only way an employee may travel between towns is in a car belonging to an employee.
• Employees can only take rides from other employees that live in the same town.
• The minimum number of cars is used.

Find whether it is possible for everyone to make it to work, and if it is, how many cars will end up driving to the office.

Whoa, that’s way more complex than the other problems, but still, not that hard! What we need to do is: check who are drivers and passengers for each town (since they can only help same town buddies), if the total capacity of cars from a town is inferior to the number of passengers, the problem cannot be solved and so, reported as “IMPOSSIBLE”. If we can carry everyone, start filling the cars from the “minivan” (can carry more passengers) to the “mini cooper”. 😀 That way we can make sure there are fewer cars on the road.

Some rough, but commented, and validated (!) python code follows:

for case in range(1,int(raw_input())+1):                  # For all test cases
    town_len, office = map(int, raw_input().split(" "))   # Get amount of towns and office

    carriers = [[] for x in xrange(town_len)]             # Create new lists for every town
    capacities = town_len * [0]                           # Initialize "capacity of transportation"
    payload = town_len * [0]                              # Initialize "cargo" for each town

    for i in range(int(raw_input())):                     # Get and go through employee list
        town, capacity = map(int, raw_input().split(" ")) # Get the current employee info
        if town == office:                                # If employee is from office town
            continue                                      #   No need to commute, ignore

        town -= 1                                         # Adjust "town" to zero-based index
        payload[town] += 1                                # Employee needs to commute, count

        if capacity != 0:                                 # If employee has a car
            capacities[town] += capacity                  # Add to our capacities for the town
            carriers[town].append(capacity)               # Append its car capacity

    resultStr = ""                                        # Create our result string

    for i in range(town_len):                             # For all towns
        if capacities[i] < payload[i]:                    # If we can't carry everyone
            resultStr = "IMPOSSIBLE "                     #   declare it impossible, give up
            break;

        cars = 0                                          # We can carry, start with 0 cars
        carriers[i].sort()                                # Sort and reverse the car list
        carriers[i].reverse()
        for car in carriers[i]:                           # So we can start from the bigger ones
            if (payload[i] <= 0):                         # If we're carrying everybody already
                break                                     #  we're done for this town

            payload[i] -= car                             # Else, fill a car, report as being used
            cars += 1

        resultStr += str(cars)+" "                        # Add this town filled vehicles

    print "Case #%d: %s" % (case, resultStr[:-1])         # Report results, remove last space

As the last problem, the same can be told about this code. There are many ways to improve its performance, but this way it can already solve the large inputs in an instant. Bitbangers apart, it solves the problem and everybody gets happy. Also, I think it is quite clear to understand, even more with the comments. 🙂 Bye!

Hey there! So, another day, another problem! This time, it is called Odd Man Out, and it is quite simple to “get”:

You are hosting a party with G guests and notice that there is an odd number of guests! When planning the party you deliberately invited only couples and gave each couple a unique number C on their invitation. You would like to single out whoever came alone by asking all of the guests for their invitation numbers.

Well, just get the number that’s not duplicated! There are many ways to do this, be it counting the occurrences of every member of the list and finding the “odd man”, or the way I did, which is basically some push and pop into a new list. Python code follows:

for case in range(1,int(raw_input())+1):        # For all test cases
    raw_input()                                 # Discard the guest list size
    guests = raw_input().split(" ")             # Get the list of guest IDs

    guestList = []                              # Create an empty guest list

    for guest in guests:                        # For all guests
        if guestList.count(guest) > 0:          # If it's already on the final list
            guestList.remove(guest)             #    if is a couple, remove
        else:                                   # If it's not in the list yet
            guestList.append(guest)             #    Add its ID

    print "Case #%d: %s" % (case, guestList[0]) # Reports results

Needless to say, I bet there are many ways to improve this algorithm, but for this problem it performs decently, so no worries (unless you’re a hardcore bit banger). 🙂 See you tomorrow!

Whoa! Keeping up with the Africa 2010 contest, enter the Qualification C problem: T9 Spelling! It was a very fun problem and says:

The Latin alphabet contains 26 characters and telephones only have ten digits on the keypad. We would like to make it easier to write a message to your friend using a sequence of keypresses to indicate the desired characters. The letters are mapped onto the digits as shown below. To insert the character B for instance, the program would press 22. In order to insert two characters in sequence from the same key, the user must pause before pressing the key a second time. The space character ' ' should be printed to indicate a pause. For example, 2 2 indicates AA whereas 22 indicates B.

Reminds me of those days of texting through that old indestructible Nokias! Anyway, it’s not hard to do, but requires some data disposition. I chose to create a map to lists containing keys and the amount of keystrokes for every letter (including space, which would be 0), in order to simply match letters and print the sequence right along. Also, to know when to pause, I would check the last printed character before, so I can print a space and then the sequence.

Python code followsas usual:

# Build a map with the number to write and its repetition count
t9 = {
        "a" : ["2", 1], "b" : ["2", 2], "c": ["2", 3],
        "d" : ["3", 1], "e" : ["3", 2], "f": ["3", 3],
        "g" : ["4", 1], "h" : ["4", 2], "i": ["4", 3],
        "j" : ["5", 1], "k" : ["5", 2], "l": ["5", 3],
        "m" : ["6", 1], "n" : ["6", 2], "o": ["6", 3],
        "p" : ["7", 1], "q" : ["7", 2], "r": ["7", 3], "s": ["7", 4],
        "t" : ["8", 1], "u" : ["8", 2], "v": ["8", 3],
        "w" : ["9", 1], "x" : ["9", 2], "y": ["9", 3], "z": ["9", 4],
        " " : ["0", 1],
}

for case in range(1,int(raw_input())+1): # For all test cases
    message = raw_input()                # Get the list of words

    attr = t9[message[0]]                # Start with the first character
    t9str = attr[0]*attr[1]              # Initialize t9str with first T9 "word"

    for i in range(1,len(message)):      # From the second character on
        attr = t9[message[i]]            # Get the map values
        if t9str[-1] == attr[0]:         # If it's the same number as the last
            t9str += " "                 # Add a space

        t9str = t9str+attr[0]*attr[1]    # Append the new T9 "word"

    print "Case #%d: %s" % (case, t9str) # Reports results

It was the hardest of the Qualification problems, but still very easy to develop. 🙂 Fun! See ya tomorrow!