2015-04-17 158 views
0

我有這個任務,我應該在Java中創建一個迷宮求解器。我決定應用的算法按以下方式工作:它是一種遞歸方法,在每次找到路徑時再次調用自身。如果它運行到死衚衕,它會調用第二個遞歸方法「goBack」,它會一直返回,直到它找到一個新路徑。牆是0,路徑是1,走路是2,走過兩次的路徑是3s。這個想法很簡單,但我無法實現。 ArrayOutOfBounds異常一直顯示。有人對此有任何想法嗎?Java遞歸迷宮

public class Project5v2 { 

static String mazecsv = "/Users/amorimph/Documents/COMP 182/Project 5/mazeinput.csv"; 
static File solvedMaze = new File("/Users/amorimph/Documents/COMP 182/Project 5/solvedMaze.txt"); 
static int[][] maze = new int[50][50]; 
static int trigger = 0; 
static int mazeWidth; 
static int mazeHeight; 

public static void main(String[] args) { 

    readCSV(mazecsv); 
    start(maze); 
    mazeToString(maze); 

} 

public static void readCSV(String csvfile) { 

    BufferedReader br = null; 
    String line = ""; 
    String csvSplitBy = ","; 
    int x = 1; 
    int y = 0; 


    try { 

     br = new BufferedReader(new FileReader(csvfile)); 
     br.readLine(); 

      while ((line = br.readLine()) != null) { 

       String[] info = line.split(csvSplitBy); 

       for (x = 1; x < info.length; x++) {      
         maze[y][x] = Integer.parseInt(info[x]); 
       } 
       mazeWidth = info.length; 
       y++; 
       mazeHeight = y; 

      } 

    } catch (FileNotFoundException e) { 
     e.printStackTrace(); 
    } catch (IOException e) { 
     e.printStackTrace(); 
    } finally { 
     if (br != null) { 
      try { 
       br.close(); 
      } catch (IOException e) { 
       e.printStackTrace(); 
      } 
     } 
    } 

} 

public static void start(int[][] maze) { 

int i = 0; 
while(maze[0][i] != 1) { 
    i++; 
} 
System.out.println(i); 
move(maze,i,1); 

} 

public static void move(int[][] maze, int x, int y) { 


for (int i = 0; i < 4; i++) { 
    switch(i) { 

    case 0: if(maze[x][y-1] == 1) { 
     maze[x][y] = 2; 
     maze[x][y-1] = 2; 
     move(maze, x, y-1); 
     break; 
    } 

    case 1: if(maze[x-1][y] == 1) { 
     maze[x][y] = 2; 
     maze[x-1][y] = 2; 
     move(maze, x-1, y); 
     break; 
    } 

    case 2: if(maze[x+1][y] == 1) { 
     maze[x][y] = 2; 
     maze[x+1][y] = 2; 
     move(maze, x+1, y); 
     break; 
    } 

    case 3: if(maze[x][y+1] == 1) { 
     maze[x][y] = 2; 
     maze[x][y+1] = 2; 
     move(maze, x, y+1); 
     break; 
    } 

    //case 4: 
    // maze[x][y] = 2; 
    // goBack(maze, y, x); 
    // break; 

    } 
} 
} 

public static void goBack(int[][] maze, int x, int y) { 

for (int i = 0; i < 7; i++) { 
    switch(i) { 

    case 0: if(maze[x][y-1] == 1) { 
     maze[x][y] = 2; 
     maze[x][y-1] = 2; 
     move(maze, x, y-1); 
     break; 
    } 

    case 1: if(maze[x-1][y] == 1) { 
     maze[x][y] = 2; 
     maze[x-1][y] = 2; 
     move(maze, x-1, y); 
     break; 
    } 

    case 2: if(maze[x+1][y] == 1) { 
     maze[x][y] = 2; 
     maze[x+1][y] = 2; 
     move(maze, x+1, y); 
     break; 
    } 

    case 3: if(maze[x][y+1] == 1) { 
     maze[x][y] = 2; 
     maze[x][y+1] = 2; 
     move(maze, x, y+1); 
     break; 
    } 

    case 4: if(maze[x][y+1] == 2) { 
     maze[x][y] = 3; 
     goBack(maze, x, y+1); 
     break; 
    } 

    case 5: if(maze[x+1][y] == 2) { 
     maze[x][y] = 3; 
     goBack(maze, x+1, y); 
     break; 
    } 

    case 6: if(maze[x-1][y] == 2) { 
     maze[x][y] = 3; 
     goBack(maze, x-1, y); 
     break; 
    } 

    case 7: if(maze[x][y-1] == 2) { 
     maze[x][y] = 3; 
     goBack(maze, x, y-1); 
     break; 
    } 
    } 
} 
} 

public static void FWriter(String content, File file) { 

try {   
     if (!file.exists()) { 
     file.createNewFile(); 
     }  
     FileWriter fw = new FileWriter(file.getAbsoluteFile(), true); 
     BufferedWriter bw = new BufferedWriter(fw); 
     bw.write(content); 
     bw.close();   
}   
catch (IOException e) { 
     e.printStackTrace(); 
} 
} 

public static void mazeToString(int[][] maze) { 

    for(int i = 0; i < mazeHeight; i++) { 
     for(int j = 0; j < mazeWidth; j++) { 
      FWriter(Integer.toString(maze[i][j]), solvedMaze); 
     } 
     FWriter("\n", solvedMaze); 
    } 
} 

} 
+0

我知道迷宮是由我們的邊界的正方形組成的,如果邊界是有顏色的,那麼它是一堵牆,不能被穿過? – Alexey

回答

2

你的迷宮是50 * 50。看看你的移動方法,我看不到任何東西阻止你從迷宮中走出(即移動到大於49或小於0的索引)。

更一般地說,如果你想使用遞歸,並且你不想陷入麻煩,你必須檢查你什麼時候完成。你的移動方法永遠不會停止調用自己,所以想想移動方法應該終止的地方。