使用OpenCV检测图像中的矩形

//检测矩形
//第一个参数是传入的原始图像，第二是输出的图像。
void findSquares(const Mat& image,Mat &out)
{
 int thresh = 50, N = 5;
 vector<vector<Point> > squares;
 squares.clear();

 Mat src,dst, gray_one, gray;

 src = image.clone();
 out = image.clone();
 gray_one = Mat(src.size(), CV_8U);
 //滤波增强边缘检测
 medianBlur(src, dst, 9);
 //bilateralFilter(src, dst, 25, 25 * 2, 35);

 vector<vector<Point> > contours;
 vector<Vec4i> hierarchy;

 //在图像的每个颜色通道中查找矩形
 for (int c = 0; c < image.channels(); c++)
 {
 int ch[] = { c, 0 };

 //通道分离
 mixChannels(&dst, 1, &gray_one, 1, ch, 1);

 // 尝试几个阈值
 for (int l = 0; l < N; l++)
 {
  // 用canny()提取边缘
  if (l == 0)
  {
  //检测边缘
  Canny(gray_one, gray, 5, thresh, 5);
  //膨脹
  dilate(gray, gray, Mat(), Point(-1, -1));
  imshow("dilate", gray);
  }
  else
  {
  gray = gray_one >= (l + 1) * 255 / N;
  }

  // 轮廓查找
  //findContours(gray, contours, RETR_CCOMP, CHAIN_APPROX_SIMPLE);
  findContours(gray, contours, hierarchy, RETR_CCOMP, CHAIN_APPROX_SIMPLE);

  vector<Point> approx;
  
  // 检测所找到的轮廓
  for (size_t i = 0; i < contours.size(); i++)
  {
  //使用图像轮廓点进行多边形拟合
  approxPolyDP(Mat(contours[i]), approx, arcLength(Mat(contours[i]), true)*0.02, true);

  //计算轮廓面积后，得到矩形4个顶点
  if (approx.size() == 4 &&fabs(contourArea(Mat(approx))) > 1000 &&isContourConvex(Mat(approx)))
  {
   double maxCosine = 0;

   for (int j = 2; j < 5; j++)
   {
   // 求轮廓边缘之间角度的最大余弦
   double cosine = fabs(angle(approx[j % 4], approx[j - 2], approx[j - 1]));
   maxCosine = MAX(maxCosine, cosine);
   }

   if (maxCosine < 0.3)
   {
   squares.push_back(approx);
   }
  }
  }
 }
 }

 
 for (size_t i = 0; i < squares.size(); i++)
 {
 const Point* p = &squares[i][0];

 int n = (int)squares[i].size();
 if (p->x > 3 && p->y > 3)
 {
  polylines(out, &p, &n, 1, true, Scalar(0, 255, 0), 3, LINE_AA);
 }
 }
 imshow("dst",out);
}

static double angle(Point pt1, Point pt2, Point pt0)
{
 double dx1 = pt1.x - pt0.x;
 double dy1 = pt1.y - pt0.y;
 double dx2 = pt2.x - pt0.x;
 double dy2 = pt2.y - pt0.y;
 return (dx1*dx2 + dy1*dy2) / sqrt((dx1*dx1 + dy1*dy1)*(dx2*dx2 + dy2*dy2) + 1e-10);
}