Board covering problems

    

     Incomplete board is a 2 k x 2 k (k: 1) a square board, which just have a box missing. Figure 4-7 give k = 1 all possible incomplete board, including incomplete squares with shadow said.

     (残缺棋盘是一个有2k×2k （k≥1）个方格的棋盘，其中恰有一个方格残缺。图4-7给出k=1时各种可能的残缺棋盘，其中残缺的方格用阴影表示。)

      Song ti board so we called "three case board", incomplete problem is to use this board four three case board over a large incomplete board. On the cover of requirements:

1) two three case board can't overlap

2) three case board can't cover incomplete box, but must cover all other squares

In this restricted condition, need three case board for a total of (2 k x 2 k-1) / 3.

     (像这样的棋盘我们称作“三格板”，残缺棋盘问题就是要用这四种三格板覆盖更大的残缺棋盘。在此覆盖中要求：

1）两个三格板不能重叠

2）三格板不能覆盖残缺方格，但必须覆盖其他所有的方格

在这种限制条件下，所需要的三格板总数为(2k×2k -1 )/3。  

)

program：

//

//棋盘覆盖

#include <stdio.h>

int amount=0;

int Board[100][100];

int Cover(int tr,int tc,int dr,int dc,int size)

{

int s,t;

if(size<2)//结束条件

return 0;

amount=amount+1;

t=amount;

s=size/2;//s右偏，下偏

if((dr<tr+s)&&(dc<tc+s))//左上角

{

Cover(tr,tc,dr,dc,s);//

Board[tr+s-1][tc+s]=t;//

Board[tr+s][tc+s-1]=t;

Board[tr+s][tc+s]=t;

Cover(tr,tc+s,tr+s-1,tc+s,s);//

Cover(tr+s,tc,tr+s,tc+s-1,s);//

Cover(tr+s,tc+s,tr+s,tc+s,s);//

}

else if((dr<tr+s)&&(dc>=tc+s))//右上角

{

Cover(tr,tc+s,dr,dc,s);

Board[tr+s-1][tc+s-1]=t;

Board[tr+s][tc+s-1]=t;

Board[tr+s][tc+s]=t;

Cover(tr,tc,tr+s-1,tc+s-1,s);

Cover(tr+s,tc,tr+s,tc+s-1,s);

Cover(tr+s,tc+s,tr+s,tc+s,s);

}

else if((dr>=tr+s)&&(dc<tc+s))//左下角

{

Cover(tr+s,tc,dr,dc,s);

Board[tr+s-1][tc+s-1]=t;

Board[tr+s-1][tc+s]=t;

Board[tr+s][tc+s]=t;

Cover(tr,tc,tr+s-1,tc+s-1,s);//上边

Cover(tr,tc+s,tr+s-1,tc+s,s);//对角

Cover(tr+s,tc+s,tr+s,tc+s,s);//右边

}

else if((dr>=tr+s)&&(dc>=tc+s))//右下角

{

Cover(tr+s,tc+s,dr,dc,s);//

Board[tr+s-1][tc+s-1]=t;

Board[tr+s-1][tc+s]=t;

Board[tr+s][tc+s-1]=t;

Cover(tr,tc,tr+s-1,tc+s-1,s);//

Cover(tr,tc+s,tr+s-1,tc+s,s);//

Cover(tr+s,tc,tr+s,tc+s-1,s);//

}

return 0;

}

OutputBoard(int size)

{

int i,j;

for(i=0;i<size;++i)

{

for(j=0;j<size;++j)

printf("%d--",Board[i][j]);

printf("\n==================\n");

}

}

int main()

{

int size=1;

int x,y,i,k;

printf("k:");

scanf("%d",&k);

for(i=1;i<=k;++i)

size=size*2;

printf("input incomplete pane:");

scanf("%d %d",&x,&y);

Cover(0,0,x,y,size);

OutputBoard(size);

return 0;

}