說明：

來源於一個很早的帖子，題目如下，有人做過相關的解答。

我也花了1個多小時寫了一個，不過這第二個輸出和5×5的地圖衝突，實際上第二個Rover會走出這個地圖。

以下是原題

A squad of robotic rovers are to be landed by NASA on a plateau on Mars.

This plateau, which is curiously rectangular, must be navigated by the rovers so that their on-board cameras can get a complete view of the surrounding terrain to send back to Earth.

A rover's position and location is represented by a combination of x and y co-ordinates and a letter representing one of the four cardinal compass points. The plateau is divided up into a grid to simplify navigation. An example position might be 0, 0, N, which means the rover is in the bottom left corner and facing North.

In order to control a rover, NASA sends a simple string of letters. The possible letters are 'L ', 'R ' and 'M '. 'L ' and 'R ' makes the rover spin 90 degrees left or right respectively, without moving from its current spot.

'M ' means move forward one grid point, and maintain the same heading.

Assume that the square directly North from (x, y) is (x, y+1).

INPUT:

The first line of input is the upper-right coordinates of the plateau, the lower-left coordinates are assumed to be 0,0.

The rest of the input is information pertaining to the rovers that have been deployed. Each rover has two lines of input. The first line gives the rover 's position, and the second line is a series of instructions telling the rover how to explore the plateau.

The position is made up of two integers and a letter separated by spaces, corresponding to the x and y co-ordinates and the rover 's orientation.

Each rover will be finished sequentially, which means that the second rover won 't start to move until the first one has finished moving.

OUTPUT

The output for each rover should be its final co-ordinates and heading.

INPUT AND OUTPUT

Test Input:

5 5

1 2 N

LMLMLMLMM

3 3 E

MMRMMRMRRM

火星探測器

一小隊機器人探測器將由NASA送上火星高原，探測器將在這個奇特的矩形高原上行駛。

用它們攜帶的照相機將周圍的全景地勢圖發回到地球。每個探測器的方向和位置將由一個x,y系座標圖和一個表示地理方向的字母表示出來。為了方便導航，平原將被劃分為網格狀。位置座標示例：0，0，N，表示探測器在座標圖的左下角，且面朝北方。為控制探測器，NASA會傳送一串簡單的字母。可能傳送的字母為： 'L ', 'R '和 'M '。 'L ',和 'R '分別表示使探測器向左、向右旋轉90度，但不離開他所在地點。 'M ' 表示向前開進一個網格的距離，且保持方向不變。假設以廣場（高原）的直北方向為y軸的指向。

輸入：首先輸入的line是座標圖的右上方，假定左下方頂點的座標為0，0。剩下的要輸入的是被分佈好的探測器的資訊。每個探測器需要輸入wo lines。第一條line 提供探測器的位置，第二條是關於這個探測器怎樣進行高原探測的一系列說明。位置是由兩個整數和一個區分方向的字母組成，對應了探測器的（x,y）座標和方向。每個探測器的移動將按序完成，即後一個探測器不能在前一個探測器完成移動之前開始移動。

輸出：每個探測器的輸出應該為它行進到的最終位置座標和方向。輸入和輸出 測試如下：

期待的輸入:

5 5

1 2 N

LMLMLMLMM

3 3 E

MMRMMRMRRM 期待的輸出

1 3 N

5 1 E

以下是我的答案

說明：

Java語言，OO思想，JDK1.4環境和4個Class：

• Lunch用於程式路口和收入輸出
• Controller用於解釋和指揮Rover的行為
• Area是Plateau的環境
• Rover就是探測器了

寫的匆忙，木有註釋，儘可能的發揮各位的想象力了，另外打算在東南西北的轉換上用迴圈雙向列表的，但是又偷懶了（用4取餘），導致這裡會有點難以理解，算是美中不足吧。

歡迎大家交流 :-)

Lunch.java

package mars;

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.*;
import java.util.ArrayList;

public class Lunch {
  private static Area area;
  private static Controller controller = new Controller();
  private static int lineNum = 0;
  private static ArrayList roverList = new ArrayList();
  private static ArrayList commandList = new ArrayList();

  public static void main(String[] args) {

    BufferedReader stdin = new BufferedReader(new InputStreamReader(System.in));
    while (true) {
      try {
        String input = stdin.readLine();
        if (lineNum == 0) {
          String[] initPoistion = input.split(" ");
          area = controller.createArea(Integer.parseInt(initPoistion[0])
                                       , Integer.parseInt(initPoistion[1]));

        }
        else {
          if (lineNum % 2 == 1) {
            String[] initPoistion = input.split(" ");
            Rover rover = controller
                .createRover(Integer.parseInt(initPoistion[0]),
                             Integer.parseInt(initPoistion[1])
                             , initPoistion[2]
                             , area);
            roverList.add(rover);
          }
          else {
            commandList.add(input);
          }
        }
        if (lineNum == 4)break;
        lineNum++;
      }
      catch (IOException ex) {
      }
    }
    for (int i = 0; i < roverList.size(); i++) {
      Rover rover = (Rover) roverList.get(i);
      String command = (String) commandList.get(i);
      for (int j = 0; j < command.length(); j++) {