这个问题自己一开始写的实现代码虽然和书上的在思想上是一样的,但是罗嗦了很多,因为对过程情况的划分不合理,导致冗余代码过多。书上:把问题分为四个子问题,分别覆盖四个子棋盘,然后在覆盖每一个子棋盘的时候再分为两种情况,即特殊方格是否在子棋盘中。而我的划分是:上来就以特殊方格的位置进行划分,那么代码就。。。

#include<iostream>
using namespace std;
const int SIZE = 100;
int cardNum=1;
int board[SIZE][SIZE];
//step既可以标志此时问题的规模,同时也可以将相对坐标转化为绝对坐标
void chessBoard(int step,int board_x,int board_y,int black_x,int black_y){
if(step==2){
int temp = board[black_x][black_y];
board[board_x][board_y]
=board[board_x+1][board_y]
=board[board_x][board_y+1]=board[board_x+1][board_y+1]=cardNum;
cardNum
++;
board[black_x][black_y]
=temp;//还原回来
return;
}
//计算中线的绝对坐标
int mid_x=board_x+step/2;
int mid_y=board_y+step/2;

//黑快在左上角的情形
if(black_x<mid_x&&black_y<mid_y){
board[mid_x][mid_y]
=board[mid_x-1][mid_y]=board[mid_x][mid_y-1]=cardNum;
cardNum
++;
chessBoard(step
/2,board_x,board_y,black_x,black_y);
chessBoard(step
/2,mid_x,board_y,mid_x,mid_y-1);
chessBoard(step
/2,board_x,mid_y,mid_x-1,mid_y);
chessBoard(step
/2,mid_x,mid_y,mid_x,mid_y);
return ;
}
//黑快在右上的情形
if(black_x>=mid_x&&black_y<mid_y){
board[mid_x][mid_y]
=board[mid_x-1][mid_y-1]=board[mid_x-1][mid_y]=cardNum;
cardNum
++;
chessBoard(step
/2,board_x,board_y,mid_x-1,mid_y-1);
chessBoard(step
/2,mid_x,board_y,black_x,black_y);
chessBoard(step
/2,board_x,mid_y,mid_x-1,mid_y);
chessBoard(step
/2,mid_x,mid_y,mid_x,mid_y);
return;
}
//左下的情形
if(black_x<mid_x&&black_y>=mid_y){
board[mid_x][mid_y]
=board[mid_x-1][mid_y-1]=board[mid_x][mid_y-1]=cardNum;
cardNum
++;
chessBoard(step
/2,board_x,board_y,mid_x-1,mid_y-1);
chessBoard(step
/2,mid_x,board_y,mid_x,mid_y-1);
chessBoard(step
/2,board_x,mid_y,black_x,black_y);
chessBoard(step
/2,mid_x,mid_y,mid_x,mid_y);
return;
}
//右下的情形
if(black_x>=mid_x&&black_y>=mid_y){
board[mid_x
-1][mid_y-1]=board[mid_x-1][mid_y]=board[mid_x][mid_y-1]=cardNum;
cardNum
++;
chessBoard(step
/2,board_x,board_y,mid_x-1,mid_y-1);
chessBoard(step
/2,mid_x,board_y,mid_x,mid_y-1);
chessBoard(step
/2,board_x,mid_y,mid_x-1,mid_y);
chessBoard(step
/2,mid_x,mid_y,black_x,black_y);
return;
}
}
int main(){
int step;
cin
>>step;
int x,y;
cin
>>x>>y;
chessBoard(step,
0,0,x,y);
for(int i=0;i<step;i++){
for(int j=0;j<step;j++)
cout
<<board[i][j]<<" ";
cout
<<endl;
}
return 0;
}
posted on 2011-09-19 15:53  geeker  阅读(293)  评论(0编辑  收藏  举报