在多維陣列中實現Rook邏輯

Question

我有一個問題，我一直試圖弄清楚，但我被卡住了。基本上我一直試圖在一款不是國際象棋的遊戲中實施白車的運動邏輯，但我堅持不懈。我給你的細節：

• 董事會是一個5x5的多維數組那裏只有每個球員
• 的目標是捕獲所有你的對手的棋子和捕捉他們所有的一個棋子和白嘴鴉將贏得比賽。
• 白嘴鴉可以在一個方向上儘可能地移動，直到他們碰到擋住他們路徑的東西。

我的車現在做正確的事情是，它可以去一個方向，但它可以在該行去任何地方。我需要幫助，試圖找出如何添加更多的邏輯，以確保只要路徑清晰，它只能走。下面是一個例子：

小「p」和「r」是玩家2的棋子，大「P」和「R」是玩家的棋子。現在右上方的R（白嘴鴉）只能向右移動，但是如果你這樣做了，它就會超越兵卒，然後可以儘可能地向下。

* R R R * 
* P P P * 
* * * * * 
* p p p * 
* r r r * 

這裏是什麼我對車的代碼：

public boolean isLegalMove(Location from,Location to) 
{ 
    // Row is XPosition (Up/Down) 
    // Column is YPosition(Left/Right) 

    int fromRow = from.getXPosition(); 
    int fromColumn = from.getYPosition(); 
    int toRow = to.getXPosition(); 
    int toColumn = to.getYPosition(); 

    // higher row or column or both 
    if(((fromColumn >= toColumn) || (fromColumn <= toColumn)) && ((fromRow == toRow))) { 
    return true; 
    } 
    if(((fromRow >= toRow) || (fromRow <= toRow)) && ((fromColumn == toColumn))) { 
    return true; 
    } 
    return false; 
} 

我想我會作出另一種方法來檢查邏輯是否有路徑中的任何事情，並稱其isPathClear()

編輯：
下面的代碼的其餘部分：

public class Board 
    { 
    // The depth and width of the field. 
    public static final int ROW = 5; 
    public static final int COLUMN = 5; 
    public static final String EMPTYPIECE = " * "; 

    //Storage for the game pieces 
    private GamePiece [] [] gameBoard; 
    //Makes the balls and torches for player1 
    private Pawn1 p1Pawn1,p1Pawn2,p1Pawn3; 
    private Rook1 p1Rook1,p1Rook2,p1Rook3; 

    //Makes the ball and torchers for player2 
    private Pawn2 p2Pawn1,p2Pawn2,p2Pawn3; 
    private Rook2 p2Rook1,p2Rook2,p1Rook3; 

/** 
* Makes a 5x5 Gameboard 
*/ 
public Board() 
{ 
    // initialise instance variables 
    gameBoard = new GamePiece [ROW][COLUMN]; 

    //Makes pieces for player1 
    p1Pawn1 = new Pawn1(); 
    p1Pawn2 = new Pawn1(); 
    p1Pawn3 = new Pawn1(); 
    p1Rook1 = new Rook1(); 
    p1Rook2 = new Rook1(); 
    p1Rook3 = new Rook1(); 

    //Makes pieces for player2 
    p2Pawn1 = new Pawn2(); 
    p2Pawn2 = new Pawn2(); 
    p2Pawn3 = new Pawn2(); 
    p2Rook1 = new Rook2(); 
    p2Rook2 = new Rook2(); 
    p2Rook3 = new Rook2(); 
} 

/** 
* Makes new games 
*/ 
public void newGame() 
{ 
    // Assigns the piece of the board for player1 
    gameBoard[0][1] = p1Rook1; 
    gameBoard[0][2] = p1Rook2; 
    gameBoard[0][3] = p1Rook3; 
    gameBoard[1][1] = p1Pawn1; 
    gameBoard[1][2] = p1Pawn2; 
    gameBoard[1][3] = p1Pawn3; 

    // Assigns the pieces of the board for player2 
    gameBoard[4][1] = p2Rook1; 
    gameBoard[4][2] = p2Rook2; 
    gameBoard[4][3] = p2Rook3; 
    gameBoard[3][1] = p2Pawn1; 
    gameBoard[3][2] = p2Pawn2; 
    gameBoard[3][3] = p2Pawn3; 
} 

/** 
* Displays the content of the board 
*/ 
public void displayBoard() 
{ 
    System.out.println(" a b c d e"); 
    int counter = 1; 
    for (int i = 0; i < gameBoard.length; i++){ 
     System.out.print(counter); 
     for (int j = 0; j < gameBoard[i].length; j++) { 
      if (gameBoard[i][j] == null) { 
       System.out.print(EMPTYPIECE); 
      } else { 
       System.out.print(" " + gameBoard[i][j] + " "); 
      } 
     } 
     counter++; 
     System.out.println(); 
    } 
} 

/** 
* Moves the movepiece from one locatin to another 
* @param from - where the location was from 
* @param to - Where the location is going to 
*/ 
public void movePiece(Location from,Location to) throws InvalidMoveException 
{ 
    int fromRow  = from.getXPosition(); 
    int fromColumn = from.getYPosition(); 
    int toRow  = to.getXPosition(); 
    int toColumn = to.getYPosition(); 

    if (gameBoard[fromRow][fromColumn] == null) { 
     throw new InvalidMoveException("Invalid input for source location."); 
    } 
    if (! checkBounds(from, to)) { 
     throw new InvalidMoveException("Invalid input for destination location."); 
    } 
    if (isSameLocation(from, to)){ 
     throw new InvalidMoveException("Invalid move, source and destination cannot bethe same."); 
    } 
    if (! gameBoard[fromRow][fromColumn].isLegalMove(from, to)) { 
     throw new InvalidMoveException("Invalid move for this piece."); 
    } 
     gameBoard[toRow][toColumn] = gameBoard[fromRow][fromColumn]; 
     gameBoard[fromRow][fromColumn] = null; 

    displayBoard(); 
} 

/** 
* Checks a proposed move to ensure it is within the bounds of the board. 
* @param source location, destination location 
* @return true if both source and destination are within bounds 
*/ 
private boolean checkBounds(Location from, Location to) 
{ 
    int fromRow  = from.getXPosition(); 
    int fromColumn = from.getYPosition(); 
    int toRow  = to.getXPosition(); 
    int toColumn = to.getYPosition(); 

    boolean testFrom = (fromRow >= 0) && (fromColumn >= 0) && (fromRow < gameBoard.length) && (fromColumn < gameBoard[0].length); 
    boolean testTo = (toRow >= 0) && (toColumn >= 0) && (toRow < gameBoard.length) && (toColumn < gameBoard[0].length); 
    return testFrom && testTo; 
} 

/** 
* Checks a proposed move to ensure source and destination are different. 
* @param source location, destination location 
* @return true if source and destination are the same 
*/ 
private boolean isSameLocation(Location from, Location to) 
{ 
    int fromRow  = from.getXPosition(); 
    int fromColumn = from.getYPosition(); 
    int toRow  = to.getXPosition(); 
    int toColumn = to.getYPosition(); 
    return fromRow == toRow && fromColumn == toColumn; 
} 

Answer 1

如果不知道板子上還有什麼東西，路徑是否清晰可知。但是，您的方法簽名不允許此功能訪問電路板的佈局。如果你通過整個棋盤，你可以使用循環來檢查其他棋子之間的所有方塊。

---

從Torgamus主：

你不會檢查是否董事會是空的。你必須檢查車的來源和目的地位置之間的單獨空間。現在

---

，我知道長什麼樣板一樣，這裏的一些代碼：

public boolean isLegalMove(Location from,Location to) 
{ 
    // Row is XPosition (Up/Down) 
    // Column is YPosition(Left/Right) 

    int fromRow = from.getXPosition(); 
    int fromColumn = from.getYPosition(); 
    int toRow = to.getXPosition(); 
    int toColumn = to.getYPosition(); 

    // Has to be same row or column 
    if(fromRow != toRow || fromColumn != toColumn) return false; 
    // Can't move to the same square 
    if(fromRow == toRow && fromColumn == toColumn) return false; 

    // Rows are the same 
    if(fromRow - toRow == 0) { 
    // this will hold the column of the we're going to check next 
    int newPos = fromColumn; 
    // Should we go up or down? 
    int amount = (toColumn - fromColumn < 0) ? -1 : 1; 
    while(newPos != toColumn) { 
     newPos += amount; 
     // if it's not null, we found a different piece 
     if(gameBoard[fromRow][newPos] != null) return false; 
    } 
    if(gameBoard[toRow][toColumn] != null) { 
     // return false if it's your own piece, true if it's not 
    } 
    // Columns are the same 
    } else { 
    // this will hold the row of the we're going to check next 
    int newPos = fromRow; 
    // Should we go up or down? 
    int amount = (toRow - fromRow < 0) ? -1 : 1; 
    while(newPos != toRow) { 
     newPos += amount; 
     // if it's not null, we found a different piece 
     if(gameBoard[newPos][fromColumn] != null) return false; 
    } 
    if(gameBoard[toRow][toColumn] != null) { 
     // return false if it's your own piece, true if it's not 
    } 
    } 

    return true; 
} 

---

編輯爲您希望能夠捕捉到對手的棋子的情況下...但是我沒有放入最後一段代碼，因爲您必須再次更改方法簽名。尋找我的評論。請注意，它現在是while循環，而不是do-while。