用于AI游戏的Java代码

Question

我目前正在开发一个类似于跳棋的Java游戏。

有两个玩家，player X和player Y。我不想为这个游戏的图形用户界面，只是下面的输出。在游戏中，玩家X从[0,0]开始，玩家Y从[7,7]开始。

我的第一个问题是我不确定如何进入[3,6]，让播放器X移动到[3,6]。

当player X从[0,0]移动时，[0,0]应该被标记为不可用，因此，任何一个玩家都不能进入那个空间或跳过它，这是我的第二个问题，因为我不确定如何做到这一点。

这个游戏的目的实际上是让其中一个玩家处于不能移动的位置。我们打算将不同的AI添加到其中，但如果我得到了基础工作，我可以尝试开发AI的。

我不是最强的程序员，因此，如果我的代码可以编辑在任何方式使事情变得更容易，任何帮助将不胜感激。谢谢

我的代码如下：

public static AIProject {

    public static String [][] board = new String[9][9];

    public static void addPiece(int x, int y, String r) {
        board [x][y] = r;
    }

    public static void showBoard() {
        for(int row = 0; row < board.length; row++) {
            System.out.println(" ");
            System.out.println("-------------------");

            for (int col = 0; col < board[row].length; col++) {
                System.out.print("I");
                if(board[col][row] == null) {
                    System.out.print(" ");
                } else {
                    System.out.print(board[col][row]);
                }
            }
        }

        System.out.println(" ");
        System.out.println("-------------------");
    }   

    public static void mainn(String[] args) {
        System.out.println(board.length);

        addPiece(0,0,"0");
        addPiece(0,1,"1");
        addPiece(0,2,"2");
        addPiece(0,3,"3");
        addPiece(0,4,"4");
        addPiece(0,5,"5");
        addPiece(0,6,"6");
        addPiece(0,7,"7");
        addPiece(0,8,"8");
        addPiece(1,0,"1");
        addPiece(2,0,"2");
        addPiece(3,0,"3");
        addPiece(4,0,"4");
        addPiece(5,0,"5");
        addPiece(6,0,"6");
        addPiece(7,0,"7");
        addPiece(8,0,"8");

        addPiece(1,1,"X");
        addPiece(8,8,"Y");

        showBoard();
     }
}

Answer 1

我的第一个问题是我不确定如何进入，即3,6，让玩家X移动到3,6。

你可以使用一些有意义的坐标-例如，如果用户想要转到3,6，然后输入36，等等。坐标的标签应该在船上可见。

当玩家X从0,0移动时，0,0应该被标记为不可用，因此，任何一个玩家都不能进入那个空间或跳过它，这是我的第二个问题，因为我不确定如何做到这一点。

您可以创建新的类，例如BoardItem -它将包含isOccupied和label字段- label将显示玩家标签，如果该字段将被占用或该字段未被占用的坐标。

我还建议从1开始计算数组，而不是从0开始计算。这对用户来说更有意义-“当我想要转到1行1列时，我必须输入11”。

我提出的解决方案：

public static AIProject {

private static int BOARD_SIZE = 8;

static BoardItem[][] board = new BoardItem[BOARD_SIZE][BOARD_SIZE];

// available players
public static String[] players = { "X", "Y" };

// x,y are the coordinates, label is the string, it can take coordinates or
// player name if the field is occupied
public static void addPiece(int x, int y, String label, boolean isOccupied)    {
// here we are assigning to a specific board piece a BoardItem - label      and if it
// is occupied
    board[x][y] = new BoardItem(label, isOccupied);
}

public static void showBoard() {
    for (int row = 0; row < BOARD_SIZE; row++) {
        for (int col = 0; col < BOARD_SIZE; col++) {
            if (board[col][row] == null) {
                System.out.print("\t");
            } else {
                System.out.print(board[col][row] + "\t");
            }
        }
        System.out.println("\n_____________________________________________________________");
    }
}

public static void main(String[] args) {
    // board creating
    for (int x = 1; x < BOARD_SIZE ; x++) {
        for (int y = 1; y < BOARD_SIZE ; y++) {
          // as you can see in method definition we are passing coordinates 
         // as integers in 2 first arguments, next coordinates as string
         // this "label" would be displayed on board
         // the last parameter is indicating that the field are not occupied
            addPiece(x, y, x + "" + y, false);
        }
    }
    // however, you want to point that 1,1 is occupied by player 'X'...
    addPiece(1, 1, players[0], true);
    // and 7,7 is occupied by player 'Y'
    addPiece(7, 7, players[1], true);

    // 'X' is active player now
    String activePlayer = players[0];
    Scanner keyboard = new Scanner(System.in);

    // infinite loop - in the loop condition you should check
    // if someone lose/win to end the game
    while (true) {
        // clearing output
        clearScreen();
        // showing board            
        showBoard();
        // moving player - waiting for input from player and then 
        // mark chosen coordinates as occupied by active player
        playerMove(activePlayer, keyboard);
        // changing active player. First X, then Y, then X and so on.
        activePlayer = changeActivePlayer(activePlayer);
    }

}

private static void playerMove(String player, Scanner keyboard) {

    BoardItem boardItem = getBoardItemForPosition(keyboard.next());
    // it should be isNotValid
    while (boardItem.isOccupied()) {
        System.out.println("Chosen position is occupied! Please choose another one");
        boardItem = getBoardItemForPosition(keyboard.next());
    }
    boardItem.setOccupied(true);
    boardItem.setLabel(player);
}

private static BoardItem getBoardItemForPosition(String position) {
    int x = Integer.parseInt(position.substring(0, 1));
    int y = Integer.parseInt(position.substring(1, 2));
    return board[x][y];
}

private static String changeActivePlayer(String activePlayer) {
    if (activePlayer.equals(players[0])) {
        return players[1];
    }
    return players[0];
}

private static void clearScreen() {
    for (int i = 0; i < 50; ++i)
        System.out.println();
}
}

和BoardItem类：

public class BoardItem {

private String label;

private boolean isOccupied;

public BoardItem(String label, boolean isOccupied) {
    this.label = label;
    this.isOccupied = isOccupied;
}

public String getLabel() {
    return label;
}

public void setLabel(String label) {
    this.label = label;
}

public boolean isOccupied() {
    return isOccupied;
}

public void setOccupied(boolean isOccupied) {
    this.isOccupied = isOccupied;
}

@Override
public String toString() {
    return label;
}

}

另一个小贴士：

使用main方法的

• 类应该只启动应用程序，整个逻辑应该在其他地方；
• 应该为用户输入编写验证，并检查用户是否在适当的范围内输入正确的数字，字段是否被占用，以及用户是否可以移动到指定的字段；
• 我会将所有数组更改为新的类，但这只是一个建议:)一个类应该负责一件事:启动应用程序，创建和打印视图，使用适当的验证进行玩家移动。