题目描述：

使用分治法求解棋盘覆盖问题。

棋盘覆盖问题的描述：

 

 

 残缺位置所在的四种不同情况：

 

/*** 二分法不相似情况：残缺棋盘*   by* 小俱的一步步*/
public class CanquedeQP {private int size;private int[][] board;//所使用三格板的数目private int amount;/*** 初始化board* @param size 行、列的长度*/public CanquedeQP(int size){this.amount = 0;this.size = size;this.board = new int[this.size][this.size];}public static void main(String[] args) {CanquedeQP canquedeQP = new CanquedeQP(16);canquedeQP.cover(16,0,0,5,8);System.out.println(canquedeQP);}/*** 分治覆盖法* @param size 方形棋盘的长宽* @param tr 子棋盘左上角方格所在行* @param tc 子棋盘左上角方格所在列* @param dr 残缺方格所在行* @param dc 残缺方格所在列*/public void cover(int size,int tr,int tc,int dr,int dc){//当分隔为一个方块时，返回if (size==1){return;}this.amount++;int t = this.amount;size = size/2;//左上if (dr<(tr+size) && dc<(tc+size)){cover(size,tr,tc,dr,dc);board[tr+size-1][tc+size] = t;board[tr+size][tc+size-1] = t;board[tr+size][tc+size] = t;//覆盖其余部分cover(size,tr,tc+size,tr+size-1,tc+size);cover(size,tr+size,tc,tr+size,tc+size-1);cover(size,tr+size,tc+size,tr+size,tc+size);//右上}else if (dr<(tr+size)&& dc>=(tc+size)){cover(size,tr,tc+size,dr,dc);board[tr+size-1][tc+size-1] = t;board[tr+size][tc+size-1] = t;board[tr+size][tc+size] = t;//覆盖其余部分cover(size,tr,tc,tr+size-1,tc+size-1);cover(size,tr+size,tc,tr+size,tc+size-1);cover(size,tr+size,tc+size,tr+size,tc+size);//左下}else if (dr>=(tr+size) && dc<(tc+size)) {cover(size, tr + size, tc, dr, dc);board[tr + size - 1][tc + size - 1] = t;board[tr + size - 1][tc +size] = t;board[tr + size][tc + size] = t;//覆盖其余部分cover(size, tr, tc, tr + size - 1, tc +size - 1);cover(size, tr, tc + size, tr + size - 1, tc +size);cover(size, tr + size, tc + size, tr +size, tc + size);//右下}else if (dr>=(tr+size) && dc>=(tc+size)) {cover(size, tr+size, tc + size, dr, dc);board[tr + size - 1][tc + size - 1] = t;board[tr + size-1][tc + size] = t;board[tr + size][tc + size-1] = t;//覆盖其余部分cover(size, tr, tc, tr + size - 1, tc + size - 1);cover(size, tr , tc+size, tr + size-1, tc + size );cover(size, tr + size, tc , tr + size, tc + size-1);}}@Overridepublic String toString() {String res = "";for (int i = 0; i < this.size ; i++) {for (int j = 0; j <this.size ; j++) {res= res+this.board[i][j]+"\t";}res = res+"\n";}return res;}
}

测试为16*16的方格残缺位置为【5,8】，输出结果为：