扫雷 点一块消一大片无雷区域 方法在升级 递归可用

加入限制条件    只在未被打开的区域 进行递归  避免了来回死递归

void saolei(int leipan[X11][Y11], char youxipan[X11][Y11],int x,int y,int x1, int y1)
{
	if ((x > 0) && (x <= x1) && (y > 0) && (y <= y1))
	{
		int m = leipan[x - 1][y - 1] + leipan[x - 1][y] + leipan[x - 1][y + 1] + leipan[x][y - 1] + leipan[x][y + 1] + leipan[x + 1][y - 1] + leipan[x + 1][y] + leipan[x + 1][y + 1];
		if (m == 0)
		{
		
			if ((x > 0) && (x <= x1) && (y > 0) && (y <= y1)&&(youxipan[x][y]=='*'))      //递归不动  栈溢出
			{
				youxipan[x][y] = ' ';
				saolei(leipan, youxipan, x-1, y-1, x1, y1);
				saolei(leipan, youxipan, x-1, y, x1, y1);
				saolei(leipan, youxipan, x-1, y+1, x1, y1);
				saolei(leipan, youxipan, x, y-1, x1, y1);
				saolei(leipan, youxipan, x, y+1, x1, y1);
				saolei(leipan, youxipan, x+1, y-1, x1, y1);
				saolei(leipan, youxipan, x+1, y, x1, y1);
				saolei(leipan, youxipan, x+1, y+1, x1, y1);

           }
			
			
		}
		else
			youxipan[x][y] = m + '0';//字符1 2 3  用的ASC||码
	}
}

演示效果

效果完全复现 扫雷小游戏