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| public class Board {
public static final int NUMBEROFCELLS = 50;
public static final int NUMBEROFCELLSINROW = 5;
private BoardCell[] boardCells = new BoardCell[NUMBEROFCELLS];
public Board() {
for (int i = 0; i < NUMBEROFCELLS; i++) {
boardCells = new BoardCell();
}
for (int i = 0; i < NUMBEROFCELLS; i++) {
initializeBoardCell(boardCells, i);
}
}
/**
* Initialize the neighbours of the given boardCell at the given
* index on the board
*/
private void initializeBoardCell(BoardCell boardCell, int index) {
int row = index/NUMBEROFCELLSINROW;
// Check if cell is in last or first column
boolean isFirst = (index%NUMBEROFCELLSINROW == 0);
boolean isLast = ((index+1)%NUMBEROFCELLSINROW == 0);
if (row%2 == 0) { // Even rows
if (row != 0) {
// Northern neighbours
if (!isFirst) {
boardCell.setNeighbour(Cell.NORTHWEST, boardCells[index-6]);
}
boardCell.setNeighbour(Cell.NORTHEAST, boardCells[index-5]);
}
if (row != ((NUMBEROFCELLS/NUMBEROFCELLSINROW)-1)) {
// Southern neighbours
if (!isFirst) {
boardCell.setNeighbour(Cell.SOUTHWEST, boardCells[index+4]);
}
boardCell.setNeighbour(Cell.SOUTHEAST, boardCells[index+5]);
}
}
else { // Uneven rows
// Northern neighbours
if (!isLast) {
boardCell.setNeighbour(Cell.NORTHEAST, boardCells[index-4]);
}
boardCell.setNeighbour(Cell.NORTHWEST, boardCells[index-5]);
// Southern neighbours
if (row != ((NUMBEROFCELLS/NUMBEROFCELLSINROW)-1)) {
if (!isLast) {
boardCell.setNeighbour(Cell.SOUTHEAST, boardCells[index+6]);
}
boardCell.setNeighbour(Cell.SOUTHWEST, boardCells[index+5]);
}
}
// Set the east and west neighbours
if (!isFirst) {
boardCell.setNeighbour(Cell.WEST, boardCells[index-1]);
}
if (!isLast) {
boardCell.setNeighbour(Cell.EAST, boardCells[index+1]);
}
}
public void findOccupiedBoardCells(
ArrayList occupiedCells, PieceCell pieceCell, BoardCell boardCell) {
if (pieceCell != null && boardCell != null && !pieceCell.isProcessed()) {
occupiedCells.add(boardCell);
/* Neighbouring cells can form loops, which would lead to an
infinite recursion. Avoid this by marking the processed
cells. */
pieceCell.setProcessed(true);
// Repeat for each neighbour of the piece cell
for (int i = 0; i < Cell.NUMBEROFSIDES; i++) {
findOccupiedBoardCells(occupiedCells,
(PieceCell)pieceCell.getNeighbour(i),
(BoardCell)boardCell.getNeighbour(i));
}
}
}
public boolean placePiece(Piece piece, int boardCellIdx) {
// We will manipulate the piece using its first cell
return placePiece(piece, 0, boardCellIdx);
}
public boolean
placePiece(Piece piece, int pieceCellIdx, int boardCellIdx) {
// We're going to process the piece
piece.resetProcessed();
// Get all the boardCells that this piece would occupy
ArrayList occupiedBoardCells = new ArrayList();
findOccupiedBoardCells(occupiedBoardCells,
piece.getPieceCell(pieceCellIdx),
boardCells[boardCellIdx]);
if (occupiedBoardCells.size() != Piece.NUMBEROFCELLS) {
// Some cells of the piece don't fall on the board
return false;
}
for (int i = 0; i < occupiedBoardCells.size(); i++) {
if (((BoardCell)occupiedBoardCells.get(i)).getPiece() != null)
// The board cell is already occupied by another piece
return false;
}
// Occupy the board cells with the piece
for (int i = 0; i < occupiedBoardCells.size(); i++) {
((BoardCell)occupiedBoardCells.get(i)).setPiece(piece);
}
return true; // The piece fits on the board
}
public void removePiece(Piece piece) {
for (int i = 0; i < NUMBEROFCELLS; i++) {
// Piece objects are unique, so use reference equality
if (boardCells.getPiece() == piece) {
boardCells.setPiece(null);
}
}
}
}
|
既然我们已经实现了图板的数据结构，则转向下一个问题：编写将块放入图板的 placePiece() 方法。编写该方法的最难部分是确定块是否适合图板上的给定位置。确定块是否适合的一种方法是：首先，找出所有这样的图板单元，如果将块放入图板，则这些图板单元就可能会被块单元占用。获得这组图板单元后，我们可以容易地确定新的块是否合适：所有相应的块单元都需要是空的并且块要完全适合图板。这个过程是由显示在清单 6 中的 findOccupiedBoardCells() 方法和 placePiece() 方法实现的。注：我们使用 PieceCell 对象的 processed 字段以避免 findOccupiedBoardCells() 方法中的无限递归。