放棋子

电玩女神 2022-08-08 14:00 159阅读 0赞

今有 6 x 6 的棋盘格。其中某些格子已经预先放好了棋子。现在要再放上去一些，使得：每行每列都正好有3颗棋子。我们希望推算出所有可能的放法。下面的代码就实现了这个功能。  
![Image 1][]

初始数组中，“1”表示放有棋子，“0”表示空白。

请分析代码逻辑，并推测划线处的代码。  
  
  
答案写在 “解答.txt” 文件中  
  
  
注意：只写划线处应该填的内容，划线前后的内容不要抄写。

#include<stdio.h>
    
    int N = 0;
    
    bool CheckStoneNum(int x[][6])
    {
    	for(int k=0; k<6; k++)
    	{
    		int NumRow = 0;
    		int NumCol = 0;
    		for(int i=0; i<6; i++)
    		{
    			if(x[k][i]) 
    				NumRow++;
    			if(x[i][k]) 
    				NumCol++;
    		}
    		if(NumRow!=3||NumCol!=3)  // 填空
    			return false;
    	}
    	return true;
    }
    
    int GetRowStoneNum(int x[][6], int r)
    {
    	int sum = 0;
    	for(int i=0; i<6; i++) 	
    		if(x[r][i]) sum++;
    	return sum;
    }
    
    int GetColStoneNum(int x[][6], int c)
    {
    	int sum = 0;
    	for(int i=0; i<6; i++) 	
    		if(x[i][c]) sum++;
    	return sum;
    }
    
    void show(int x[][6])
    {
    	for(int i=0; i<6; i++)
    	{
    		for(int j=0; j<6; j++) printf("%2d", x[i][j]);
    		printf("\n");
    	}
    	printf("\n");
    }
    
    void f(int x[][6], int r, int c);
    
    void GoNext(int x[][6],  int r,  int c)
    {
    	if(c<6)
    		f(x, r, c+1);   // 填空
    	else
    		f(x, r+1, 0);
    }
    
    void f(int x[][6], int r, int c)
    {
    	if(r==6)
    	{
    		if(CheckStoneNum(x))
    		{
    			N++;
    			show(x);
    		}
    		return;
    	}
    
    	if(x[r][c])  // 已经放有了棋子
    	{
    		GoNext(x,r,c);
    		return;
    	}
    	
    	int rr = GetRowStoneNum(x,r);
    	int cc = GetColStoneNum(x,c);
    
    	if(cc>=3)  // 本列已满
    		GoNext(x,r,c);  
    	else 
    		if(rr>=3)  // 本行已满
    			f(x, r+1, 0);   
    	else
    	{
    		x[r][c] = 1;
    		GoNext(x,r,c);
    		x[r][c] = 0;
    		
    		if(!(3-rr >= 6-c || 3-cc >= 6-r))  // 本行或本列严重缺子，则本格不能空着！
    			GoNext(x,r,c);  
    	}
    }
    
    int main(int argc, char* argv[])
    {
    	int x[6][6] = {
    		{1,0,0,0,0,0},
    		{0,0,1,0,1,0},
    		{0,0,1,1,0,1},
    		{0,1,0,0,1,0},
    		{0,0,0,1,0,0},
    		{1,0,1,0,0,1}
    	};
    
    	f(x, 0, 0);
    	
    	printf("%d\n", N);
    
    	return 0;
    }

![SouthEast][] ![SouthEast 1][] ![SouthEast 2][] ![SouthEast 3][] ![SouthEast 4][] ![SouthEast 5][]

[Image 1]: 
[SouthEast]: https://img-blog.csdn.net/20150523164054328?watermark/2/text/aHR0cDovL2Jsb2cuY3Nkbi5uZXQveHVrdW41MTM3/font/5a6L5L2T/fontsize/400/fill/I0JBQkFCMA==/dissolve/70/gravity/SouthEast
[SouthEast 1]: https://img-blog.csdn.net/20150523163950733?watermark/2/text/aHR0cDovL2Jsb2cuY3Nkbi5uZXQveHVrdW41MTM3/font/5a6L5L2T/fontsize/400/fill/I0JBQkFCMA==/dissolve/70/gravity/SouthEast
[SouthEast 2]: https://img-blog.csdn.net/20150523164203686?watermark/2/text/aHR0cDovL2Jsb2cuY3Nkbi5uZXQveHVrdW41MTM3/font/5a6L5L2T/fontsize/400/fill/I0JBQkFCMA==/dissolve/70/gravity/SouthEast
[SouthEast 3]: https://img-blog.csdn.net/20150523164323308?watermark/2/text/aHR0cDovL2Jsb2cuY3Nkbi5uZXQveHVrdW41MTM3/font/5a6L5L2T/fontsize/400/fill/I0JBQkFCMA==/dissolve/70/gravity/SouthEast
[SouthEast 4]: https://img-blog.csdn.net/20150523164224549?watermark/2/text/aHR0cDovL2Jsb2cuY3Nkbi5uZXQveHVrdW41MTM3/font/5a6L5L2T/fontsize/400/fill/I0JBQkFCMA==/dissolve/70/gravity/SouthEast
[SouthEast 5]: /images/20220805/8fe90e50efd64e61a6251981b954be75.png