resourcemanager.java

用java写的apple机上的小游戏aquamatic

package org.orchesta.mathena.aquamatic;

import java.awt.Canvas;
import java.awt.Color;
import java.awt.Graphics2D;
import java.awt.image.BufferedImage;
import java.io.BufferedReader;
import java.io.File;
import java.io.FileReader;
import java.io.IOException;
import java.util.Locale;

import javax.imageio.ImageIO;

public class ResourceManager {
	public  String getLocalizedMsg(Locale l) 
	{
		if (l.equals(Locale.CHINESE)) return ("已移动步骤："); 
		else return ("Moved: ");
		
	}
	int atomIndex = 100, rowNumber =0 , moleNumber, moleLength; 

	public GameMap getGameMap(String level) {
		GameMap gm = GameMap.getInstance();
		gm.setFail();
		try {
			File f = new File("resources/abilder.png");
    		BufferedImage bi = ImageIO.read(f);
    		
    		   		    		
			File f2 = new File("levels/level_"+level);
	 		BufferedReader br = new BufferedReader(new FileReader(f2));
	 		String line;
			int rowNumber =0 , moleNumber=0; 
			String[] mole = new String[15];
			
			gm.putAtom(GameMap.BLANK, bi.getSubimage(186,93,30,30));
			gm.putAtom(GameMap.WALL, bi.getSubimage(9*31, 31,30, 30));
			gm.putAtom(GameMap.UP, bi.getSubimage(8*31, 62, 30, 30)); //UP
			gm.putAtom(GameMap.DOWN, bi.getSubimage(8*31, 93, 30,30));	 //DOWN
			gm.putAtom(GameMap.LEFT, bi.getSubimage(7*31, 93, 30,30));  //LEFT
			gm.putAtom(GameMap.RIGHT, bi.getSubimage(9*31, 93, 30, 30));  //RIGHT
			
			while ((line = br.readLine())!=null)
			{
				//System.out.println(tag);
				char inital = line.charAt(0);
				switch (inital) {
				case '[': break; 	//[Level]
				case '#': break; 	//注释
					
					case 'N': {gm.setName(line.substring(5)); continue; } //Name
					
					case 'a': {//原子

						// Create an empty image;
						BufferedImage empty = new BufferedImage(30,30,BufferedImage.TYPE_4BYTE_ABGR);
			    		Graphics2D g = empty.createGraphics();
			    		Canvas c = new Canvas();
			    		// Find the right atom;
			    		BufferedImage atom;
			    		int id = (int)line.charAt(5);
			    		if (line.charAt(7)<='D'&&line.charAt(7)>='A') {
			    			int ai = (int)(line.charAt(7)-'A');
			    			atom = bi.getSubimage(62+31*ai, 93, 30,30);
			    		}
			    		
			    		else if (line.charAt(7)=='o') {
			    			//System.out.println("THIS IS AI:"+ai);
			    			atom = bi.getSubimage(31,31*3, 30, 30);	
			    		} else {
			    			int ai = (int)(line.charAt(7)-'1');
			    			//System.out.println("THIS IS AI IN number:"+ai);
			    			atom = bi.getSubimage(ai*31,0,30,30);
				    	}
						
						g.drawImage(atom, 0, 0, new Color(255,255,255, 0) , c);
						
						int i = line.length()-9;
						while (i>0) {
							char k = line.charAt(8+i);
							
							if (k>='a')
							{	
								int ki = (int) (k-'a');
								BufferedImage affix = bi.getSubimage(31*ki, 31, 30, 30);
			    				g.drawImage(affix, 0, 0, new Color(255,255,255, 0), c);
								i--;
							} else {
								int ki = (int) (k-'A');
								BufferedImage affix = bi.getSubimage(31*ki, 62, 30, 30);
			    				g.drawImage(affix, 0, 0, new Color(255,255,255, 0), c);
								i--;
							}
						}
						
						//System.out.println(id + "::::ID");
						gm.putAtom(id, empty.getSubimage(0,0,30,30));
						// PUT empty to a hashtable;
						continue;
					}	
				
					case 'f': {//布局
						for (int i =0; i<15; i++) 
							gm.setCell(rowNumber,i, line.charAt(i+8));
						
						rowNumber++;
						continue;
					}
					
					case 'm': {//答案
						mole[moleNumber] = line.substring(7);
						moleNumber++;
					}
				}
						
			}
			moleLength = mole[0].length();
			int[][] map = new int[moleNumber][moleLength]; 
			for (int k =0; k< moleNumber; k++)
				{
					for (int y = 0; y<mole[k].length(); y++)
					{
						map[k][y] = (int)mole[k].charAt(y);
						//System.out.print((char)map[k][y]);
					}
					//System.out.println("");
				}
						
			gm.setMolecular(map);
			// The goal;
			
		}catch(IOException e) {e.printStackTrace();}
		return gm;
	}
}
/*

[Level]
# 这个布局的名称
Name=Methane
#原子类型的构成，atom_1=1-c表示下图中1号原子为abilder.png第一行的第一列的原子，
#c表示其化学键，按顺序就是第二行的第三个小图
atom_1=1-c
atom_2=1-e
atom_3=2-aceg
atom_4=1-a
atom_5=1-g
#布局，.空白，#墙，数字球
feld_00=...............
feld_01=...............
feld_02=.#############.
feld_03=.#...#..#....#.
feld_04=.#.#..#5#....#.
feld_05=.#..#...#3.#.#.
feld_06=.#......####.#.
feld_07=.#.......#1#.#.
feld_08=.#.2..#..#...#.
feld_09=.#.#####....##.
feld_10=.#.4.#..#...#..
feld_11=.#####..#...#..
feld_12=........#####..
feld_13=...............
feld_14=...............
#答案
mole_0=.2.
mole_1=135
mole_2=.4.

 */