说明:
来源于一个很早的帖子,题目如下,有人做过相关的解答。
我也花了1个多小时写了一个,不过这第二个输出和5×5的地图冲突,实际上第二个Rover会走出这个地图。
以下是原题A squad of robotic rovers are to be landed by NASA on a plateauon Mars.
This plateau, which is curiously rectangular, must be navigatedby the rovers so that their on-board cameras can get a completeview of the surrounding terrain to send back to Earth.
A rover\s position and location is represented by a combinationof x and y co-ordinates and a letter representing one of the fourcardinal compass points. The plateau is divided up into a grid tosimplify navigation. An example position might be 0, 0, N, whichmeans the rover is in the bottom left corner and facing North.
In order to control a rover, NASA sends a simple string ofletters. 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 sameheading.
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 theplateau, the lower-left coordinates are assumed to be 0,0.
The rest of the input is information pertaining to the roversthat have been deployed. Each rover has two lines of input. Thefirst line gives the rover \s position, and the second line is aseries of instructions telling the rover how to explore theplateau.
The position is made up of two integers and a letter separatedby spaces, corresponding to the x and y co-ordinates and the rover\s orientation.
Each rover will be finished sequentially, which means that thesecond rover won \t start to move until the first one has finishedmoving.
OUTpUT
The output for each rover should be its final co-ordinates andheading.
INpUT AND OUTpUT
Test Input:
5 5
1 2 N
LMLMLMLMM
3 3 E
MMRMMRMRRM
火星探测器
一小队机器人探测器将由NASA送上火星高原,探测器将在这个奇特的矩形高原上行驶。
用它们携带的照相机将周围的全景地势图发回到地球。每个探测器的方向和位置将由一个x,y系坐标图和一个表示地理方向的字母表示出来。为了方便导航,平原将被划分为网格状。位置坐标示例:0,0,N,表示探测器在坐标图的左下角,且面朝北方。为控制探测器,NASA会传送一串简单的字母。可能传送的字母为:\L \, \R \和 \M \。 \L \,和 \R \分别表示使探测器向左、向右旋转90度,但不离开他所在地点。 \M 表示向前开进一个网格的距离,且保持方向不变。假设以广场(高原)的直北方向为y轴的指向。
输入:首先输入的line是坐标图的右上方,假定左下方顶点的坐标为0,0。剩下的要输入的是被分布好的探测器的信息。每个探测器需要输入wolines。第一条line提供探测器的位置,第二条是关于这个探测器怎样进行高原探测的一系列说明。位置是由两个整数和一个区分方向的字母组成,对应了探测器的(x,y)坐标和方向。每个探测器的移动将按序完成,即后一个探测器不能在前一个探测器完成移动之前开始移动。
输出:每个探测器的输出应该为它行进到的最终位置坐标和方向。输入和输出 测试如下:
期待的输入:
5 5
1 2 N
LMLMLMLMM
3 3 E
MMRMMRMRRM 期待的输出
1 3 N
5 1 E
以下是我的答案说明:
Java语言,OO思想,JDK1.4环境和4个Class:
Lunch用于程序路口和收入输出Controller用于解释和指挥Rover的行为Area是plateau的环境Rover就是探测器了写的匆忙,木有注释,尽可能的发挥各位的想象力了,另外打算在东南西北的转换上用循环双向列表的,但是又偷懒了(用4取余),导致这里会有点难以理解,算是美中不足吧。
欢迎大家交流 :-)
Lunch.javapackage 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 = newController(); private static int lineNum = 0; private static ArrayList roverList = newArrayList(); private static ArrayList commandList = newArrayList();
public static void main(String[] args) {
BufferedReader stdin = new BufferedReader(newInputStreamReader(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++) {char c = command.charAt(j);if (c == Controller.TURN_LEFT) {controller.turnLeft(rover);}else if (c == Controller.TURN_RIGHT) {
controller.turnRight(rover);}else if (c == Controller.MOVE_FOWARD) {controller.moveFoward(rover, area);}else {System.out.println(\"commond error:\" + c);}}
controller.report(rover, area); }
}}
Controller.javapackage mars;
public class Controller { public static final char TURN_LEFT = \L\; public static final char TURN_RIGHT = \R\; public static final char MOVE_FOWARD = \M\;
public Controller() { }
public static void report(Rover rover, Areaarea) {
Rover[][]a = area.getArea(); for (int i =0; i < a.length; i++) {for (int j = 0; j < a[i].length; j++) {if (a[i][j] == rover) {System.out.println(i + \" \" + j + \" \" + rover.getDirection());break;}} } }
public Area createArea(int x, int y) { return newArea(x, y); }
public Rover createRover(int x, int y, Stringdirection, Area area) { Rover rover= new Rover(direction); Rover[][] a= area.getArea(); a[x][y] =rover; returnrover; }
public void turnLeft(Rover rover) {rover.turnLeft(); }
public void turnRight(Rover rover) {rover.turnRight(); }
public void moveFoward(Rover rover, Areaarea) { booleanneedBreak = false; Rover[][] a= area.getArea(); int x; int y; for (int i =0; i < a.length; i++) {for (int j = 0; j < a[i].length; j++) {if (a[i][j] == rover) {needBreak = true;x = i;y = j;a[x][y] = null;String direction = rover.getDirection();//System.out.println(\"X:\" + x + \" Y:\" + y+ \"direction:\"+direction);if (direction.equals(Rover.EAST_DIR)) {a[x + 1][y] = rover;}else if (direction.equals(Rover.SOUTH_DIR)) {a[x][y - 1] = rover;}else if (direction.equals(Rover.WEST_DIR)) {a[x - 1][y] = rover;}else if (direction.equals(Rover.NORTH_DIR)) {a[x][y + 1] = rover;}else {System.out.println(\"error direction:\" + direction);}if (needBreak)break;}if (needBreak)break;} } }
}
Area.javapackage mars;
public class Area {
Rover[][] area;
public Area(int x, int y) { area = newRover[x][y]; }
public Rover[][] getArea() { returnarea; }}
Rover.javapackage mars;
public class Rover {
public static final String NORTH_DIR =\"N\"; public static final String SOUTH_DIR =\"S\"; public static final String EAST_DIR = \"E\"; public static final String WEST_DIR = \"W\"; private static final String[] DIR = {EAST_DIR, SOUTH_DIR, WEST_DIR, NORTH_DIR}; private String direction;
public Rover(String direction) {this.direction = direction; }
public void turnRight() { for (int i =0; i < DIR.length; i++) {if (direction.equals(DIR[i])) {int tmep = i + 1;direction = DIR[tmep % 4];break;} } }
public void turnLeft() { for (int i =0; i < DIR.length; i++) {if (direction.equals(DIR[i])) {int tmep = i - 1;direction = DIR[ (tmep + 4) % 4];break;} } }
public String getDirection() { returndirection; }
}
查看全文
false