Google Code Jam 2011 Qualification Round ProblemA

Problem A. Bot Trust

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 "R_i P_i" where R_i is a robot color (always 'O' or 'B'), and P_i 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 ≤ P_i ≤ 100 for all i.

Small dataset

1 ≤ T ≤ 20.
1 ≤ N ≤ 10.

Large dataset

1 ≤ T ≤ 100.
1 ≤ N ≤ 100.

Sample

Input	Output
3 4 O 2 B 1 B 2 O 4 3 O 5 O 8 B 100 2 B 2 B 1	Case #1: 6 Case #2: 100 Case #3: 4

我的代码如下：

import java.io.BufferedReader; import java.io.BufferedWriter; import java.io.File; import java.io.FileReader; import java.io.FileWriter; import java.io.IOException; import java.util.ArrayList; class Button{ String hallway; int k; public Button(String hallway, int k){ this.hallway = hallway; this.k = k; } } class Sequence{ int N = 0; Button[] buttons; } public class ProblemA { boolean debug = false; int T = 0; int times = 0; Sequence[] cases; //解析输入序列 public void input(String inputPath){ File file = new File(inputPath); try { int i = 0; if(!file.exists()) file.createNewFile(); FileReader fileReader = new FileReader(file); BufferedReader bufReader = new BufferedReader(fileReader); String s = bufReader.readLine(); T = Integer.parseInt(s); cases = new Sequence[T]; while((s=bufReader.readLine())!=null){ cases[i] = new Sequence(); String[] splitedStr = s.split(" "); cases[i].N = Integer.parseInt(splitedStr[0]); if(debug){ System.out.print(cases[i].N+" "); } cases[i].buttons = new Button[cases[i].N]; for(int j = 0; j < splitedStr.length / 2; j++){ cases[i].buttons[j] = new Button(splitedStr[j*2+1], Integer.parseInt(splitedStr[j*2+2])); if(debug){ System.out.print(cases[i].buttons[j].hallway+" "+cases[i].buttons[j].k+" "); } } if(debug){ System.out.println(); } i++; } bufReader.close(); fileReader.close(); } catch (IOException e) { // TODO Auto-generated catch block e.printStackTrace(); } } //已知执行顺序，找出一个并行方案使时间最短 public String cal(Sequence cace, int caseNo){ //将序列分成O和B ArrayList<Integer> targetO = new ArrayList<Integer>(); ArrayList<Integer> targetB = new ArrayList<Integer>(); int i = 0; while(i < cace.buttons.length){ if(cace.buttons[i].hallway.equalsIgnoreCase("O")){ targetO.add(cace.buttons[i].k); }if(cace.buttons[i].hallway.equalsIgnoreCase("B")){ targetB.add(cace.buttons[i].k); } i++; } //开始计时 i = 0; Button currentTarget = cace.buttons[0]; times = 0; int curO = 0, curB = 0, curOPos = 1, curBPos = 1, distO = -1, distB = -1; if(!targetO.isEmpty()){ distO = Math.abs(targetO.get(curO)-curOPos); } if(!targetB.isEmpty()){ distB = Math.abs(targetB.get(curB)-curBPos); } while(i < cace.buttons.length){ times++; if(distO > -1){ distO--; }if(distB > -1){ distB--; }if(distO == -1&¤tTarget.hallway.equalsIgnoreCase("O")){ if(curO<targetO.size()-1){ curOPos = targetO.get(curO); curO++; distO = Math.abs(targetO.get(curO)-curOPos); } i++; if(i == cace.buttons.length) break; currentTarget = cace.buttons[i]; continue; }if(distB == -1&¤tTarget.hallway.equalsIgnoreCase("B")){ if(curB<targetB.size()-1){ curBPos = targetB.get(curB); curB++; distB = Math.abs(targetB.get(curB)-curBPos); } i++; if(i == cace.buttons.length) break; currentTarget = cace.buttons[i]; } } String str = new String("Case #"+(caseNo+1)+": "+times); return str; } //写文件 public void write(String outputPath, String[] content){ File file = new File(outputPath); try { if(!file.exists()) file.createNewFile(); FileWriter fileWriter = new FileWriter(file); BufferedWriter bufWriter = new BufferedWriter(fileWriter); for(int i=0; i<content.length;i++){ bufWriter.write(content[i]); bufWriter.newLine(); } bufWriter.close(); fileWriter.close(); }catch (IOException e) { // TODO Auto-generated catch block e.printStackTrace(); } } public static void main(String[] args){ String[] result; ProblemA problemA = new ProblemA(); problemA.input("E:/A-large.in"); result = new String[problemA.T]; for(int i = 0; i < problemA.T; i++){ result[i] = new String(problemA.cal(problemA.cases[i], i)); } problemA.write("E:/A-large.out", result); } }