算法分析棋盘覆盖问题java
时间: 2023-11-10 14:05:01 浏览: 64
棋盘覆盖问题是指在一个大小为2^n * 2^n的棋盘上,恰好有一个方格是缺失的,现在要用L型骨牌(类似于俄罗斯方块中的形状)覆盖整个棋盘,求解覆盖方案。
算法分析:
采用分治算法,将棋盘分成四个大小相等的子棋盘,然后递归地对每个子棋盘进行覆盖。具体实现时,先找到缺失方格所在的子棋盘,然后将该子棋盘中除缺失方格外的其它方格用L型骨牌覆盖,最后递归地对其它三个子棋盘进行同样的操作。
Java代码实现:
```
public class ChessboardCover {
private int[][] board;
private int tile = 1;
public void cover(int tr, int tc, int dr, int dc, int size) {
if (size == 1) {
return;
}
int t = tile++;
int s = size / 2;
// 覆盖左上角子棋盘
if (dr < tr + s && dc < tc + s) {
cover(tr, tc, dr, dc, s);
} else {
board[tr + s - 1][tc + s - 1] = t;
cover(tr, tc, tr + s - 1, tc + s - 1, s);
}
// 覆盖右上角子棋盘
if (dr < tr + s && dc >= tc + s) {
cover(tr, tc + s, dr, dc, s);
} else {
board[tr + s - 1][tc + s] = t;
cover(tr, tc + s, tr + s - 1, tc + s, s);
}
// 覆盖左下角子棋盘
if (dr >= tr + s && dc < tc + s) {
cover(tr + s, tc, dr, dc, s);
} else {
board[tr + s][tc + s - 1] = t;
cover(tr + s, tc, tr + s, tc + s - 1, s);
}
// 覆盖右下角子棋盘
if (dr >= tr + s && dc >= tc + s) {
cover(tr + s, tc + s, dr, dc, s);
} else {
board[tr + s][tc + s] = t;
cover(tr + s, tc + s, tr + s, tc + s, s);
}
}
public void printBoard() {
for (int i = 0; i < board.length; i++) {
for (int j = 0; j < board.length; j++) {
System.out.print(board[i][j] + "\t");
}
System.out.println();
}
}
public static void main(String[] args) {
int size = 8;
int missingRow = 3;
int missingCol = 5;
ChessboardCover cover = new ChessboardCover();
cover.board = new int[size][size];
cover.board[missingRow][missingCol] = -1;
cover.cover(0, 0, missingRow, missingCol, size);
cover.printBoard();
}
}
```