GCJ 2011 - Magicka

Below is the description of a problem called 'Magicka', from 'Google Code Jam - 2011':


Introduction
------------

Magicka™ is an action-adventure game developed by Arrowhead Game Studios. In Magicka you play a wizard, invoking and combining elements to create Magicks. This problem has a similar idea, but it does not assume that you have played Magicka.

Note: "invoke" means "call on." For this problem, it is a technical term and you don't need to know its normal English meaning.

Problem
-------

As a wizard, you can invoke eight elements, which are the "base" elements. Each base element is a single character from {Q, W, E, R, A, S, D, F}. When you invoke an element, it gets appended to your element list. For example: if you invoke W and then invoke A, (we'll call that "invoking WA" for short) then your element list will be [W, A].

We will specify pairs of base elements that combine to form non-base elements (the other 18 capital letters). For example, Q and F might combine to form T. If the two elements from a pair appear at the end of the element list, then both elements of the pair will be immediately removed, and they will be replaced by the element they form. In the example above, if the element list looks like [A, Q, F] or [A, F, Q] at any point, it will become [A, T].

We will specify pairs of base elements that are opposed to each other. After you invoke an element, if it isn't immediately combined to form another element, and it is opposed to something in your element list, then your whole element list will be cleared.

For example, suppose Q and F combine to make T. R and F are opposed to each other. Then invoking the following things (in order, from left to right) will have the following results:

QF → [T] (Q and F combine to form T)
QEF → [Q, E, F] (Q and F can't combine because they were never at the end of the element list together)
RFE → [E] (F and R are opposed, so the list is cleared; then E is invoked)
REF → [] (F and R are opposed, so the list is cleared)
RQF → [R, T] (QF combine to make T, so the list is not cleared)
RFQ → [Q] (F and R are opposed, so the list is cleared)

Given a list of elements to invoke, what will be in the element list when you're done?

Input
-----

The first line of the input gives the number of test cases, T. T test cases follow. Each test case consists of a single line, containing the following space-separated elements in order:

First an integer C, followed by C strings, each containing three characters: two base elements followed by a non-base element. This indicates that the two base elements combine to form the non-base element. Next will come an integer D, followed by D strings, each containing two characters: two base elements that are opposed to each other. Finally there will be an integer N, followed by a single string containing N characters: the series of base elements you are to invoke. You will invoke them in the order they appear in the string (leftmost character first, and so on), one at a time.

Output
------

For each test case, output one line containing "Case #x: y", where x is the case number (starting from 1) and y is a list in the format "[e0, e1, ...]" where ei is the ith element of the final element list. Please see the sample output for examples.

Limits
------

1 ≤ T ≤ 100.
Each pair of base elements may only appear together in one combination, though they may appear in a combination and also be opposed to each other.
No base element may be opposed to itself.
Unlike in the computer game Magicka, there is no limit to the length of the element list.

Small dataset
-------------
0 ≤ C ≤ 1.
0 ≤ D ≤ 1.
1 ≤ N ≤ 10.

Large dataset
-------------
0 ≤ C ≤ 36.
0 ≤ D ≤ 28.
1 ≤ N ≤ 100.

Sample Input
------------
5
0 0 2 EA
1 QRI 0 4 RRQR
1 QFT 1 QF 7 FAQFDFQ
1 EEZ 1 QE 7 QEEEERA
0 1 QW 2 QW

Output
------
Case #1: [E, A]
Case #2: [R, I, R]
Case #3: [F, D, T]
Case #4: [Z, E, R, A]
Case #5: []


This problem can be easily coded, if you can use some efficient data structures to look up, which elements when invoked merge and which elements oppose each other. The problem requirements should also be carefully read and implemented, for example: "After you invoke an element, if it isn't IMMEDIATELY combined to form another element, and it is opposed to something in your element list, then your whole element list will be cleared."

First off, when parsing the input, I'm using Java's HashMap data structures to build a convenient data structure to look up, which elements oppose and which elements combine to form an other element. The code to build these data structures, is as below:

String[] splitString = line.split(" ");
int index = 0;
int num1 = Integer.parseInt(splitString[index]);
HashMap<String, String> mergeMap = new HashMap<String, String>();
          
for(int i=0;i<num1;i++) {
  index++;
  String str = splitString[index];
  String sub1 = str.substring(0,2);
  String sub2 = str.substring(2);

  mergeMap.put(sub1, sub2);
  mergeMap.put(sub1.charAt(1) + "" + sub1.charAt(0), sub2);
}
          
index++;
int num2 = Integer.parseInt(splitString[index]);
HashMap<String, ArrayList<String>> opposingMap = new HashMap<String, ArrayList<String>>();
          
for(int i=0;i<num2;i++) {
  index++;
  String str = splitString[index];
  String sub1 = "" + str.charAt(0);
  String sub2 = "" + str.charAt(1);
            
  if(opposingMap.containsKey(sub1)) {
     opposingMap.get(sub1).add(sub2);
  } else {
     ArrayList<String> keyList = new ArrayList<String>();
     keyList.add(sub2);

     opposingMap.put(sub1, keyList);
  }

  if(opposingMap.containsKey(sub2)) {
     opposingMap.get(sub2).add(sub1);
  } else {
     ArrayList<String> keyList = new ArrayList<String>();
     keyList.add(sub1);

     opposingMap.put(sub2, keyList);
  }
}

index++;
int invokeStringLength = Integer.parseInt(splitString[index]); 
          
index++;
String invokeString = splitString[index];

And...with the above input, I'm using the below code, for solving the problem:

GCJ 2011 - Bot Trust

Below is the description of a problem called 'Bot Trust', from 'Google Code Jam - 2011':

Problem
-------

Blue and Orange are friendly robots. An evil computer mastermind has locked them up in separate hallways to test them, and then possibly give them cake.

Each hallway contains 100 buttons labeled with the positive integers {1, 2, ..., 100}. Button k is always k meters from the start of the hallway, and the robots both begin at button 1. Over the period of one second, a robot can walk one meter in either direction, or it can press the button at its position once, or it can stay at its position and not press the button. To complete the test, the robots need to push a certain sequence of buttons in a certain order. Both robots know the full sequence in advance. How fast can they complete it?

For example, let's consider the following button sequence:

O 2, B 1, B 2, O 4

Here, O 2 means button 2 in Orange's hallway, B 1 means button 1 in Blue's hallway, and so on. The robots can push this sequence of buttons in 6 seconds using the strategy shown below:

Time | Orange | Blue
-----+------------------+-----------------
1 | Move to button 2 | Stay at button 1
2 | Push button 2 | Stay at button 1
3 | Move to button 3 | Push button 1
4 | Move to button 4 | Move to button 2
5 | Stay at button 4 | Push button 2
6 | Push button 4 | Stay at button 2

Note that Blue has to wait until Orange has completely finished pushing O 2 before it can start pushing B 1.

Input
-----

The first line of the input gives the number of test cases, T. T test cases follow.

Each test case consists of a single line beginning with a positive integer N, representing the number of buttons that need to be pressed. This is followed by N terms of the form "Ri Pi" where Ri is a robot color (always 'O' or 'B'), and Pi is a button position.

Output
------

For each test case, output one line containing "Case #x: y", where x is the case number (starting from 1) and y is the minimum number of seconds required for the robots to push the given buttons, in order.

Limits
------

1 ≤ Pi ≤ 100 for all i.

Small dataset
-------------

1 ≤ T ≤ 20.
1 ≤ N ≤ 10.

Large dataset
-------------

1 ≤ T ≤ 100.
1 ≤ N ≤ 100.

Sample Input
------------
3
4 O 2 B 1 B 2 O 4
3 O 5 O 8 B 100
2 B 2 B 1

Output
------

Case #1: 6
Case #2: 100
Case #3: 4


The solution to the problem can be easily coded, after understanding the fact that the robot (which was not scheduled to press the switch) can freely move to it's destination (next switch), so as to make the time required to solve the problem, as minimal as possible. Both the robots start at 1. So, for example, if one robot has to press the switch at 10 (first move), it takes 9 seconds (10 - 1) to move and 1 to press the switch...So, it takes a total of 10 seconds...In these 10 seconds, the other robot can move as long as possible and stay-put there, waiting for its turn. I used Java datastructures 'Queue' and 'Deque' to solve the problem.

Below is my solution, in Java: