Opencv: pointPolygonTest函数使用

参考： 
http://blog.csdn.net/fuyou25/article/details/53467204 
https://docs.opencv.org/2.4/modules/imgproc/doc/structural_analysis_and_shape_descriptors.html?highlight=pointpolygontest#pointpolygontest 
https://docs.opencv.org/2.4/doc/tutorials/imgproc/shapedescriptors/point_polygon_test/point_polygon_test.html

根据Opencv文档，pointPolygonTest的参数及解释如下：

函数形式
C++: double pointPolygonTest(InputArray contour, Point2f pt, bool measureDist)
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函数作用
判断一个点是否在一个contour的内部还是外部。
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参数解释
contour – 输入findContour提取到的边缘.
pt – 需要检测的点.
measureDist – 为真，则计算检测点到边缘的距离，为负值在外部，0在边上，正值在内部。为假，则返回-1（在contour外部）、0（在contour上）、1（在contour内部。
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Demo(代码和示意图版权归Opencv所有，此处仅作示例，参考链接在文首）
#include "opencv2/highgui/highgui.hpp"
#include "opencv2/imgproc/imgproc.hpp"
#include
#include
#include

using namespace cv;
using namespace std;

/** @function main */
int main( int argc, char** argv )
{
  /// Create an image
  const int r = 100;
  Mat src = Mat::zeros( Size( 4*r, 4*r ), CV_8UC1 );

  /// Create a sequence of points to make a contour:
  vector vert(6);

  vert[0] = Point( 1.5*r, 1.34*r );
  vert[1] = Point( 1*r, 2*r );
  vert[2] = Point( 1.5*r, 2.866*r );
  vert[3] = Point( 2.5*r, 2.866*r );
  vert[4] = Point( 3*r, 2*r );
  vert[5] = Point( 2.5*r, 1.34*r );

  /// Draw it in src
  for( int j = 0; j < 6; j++ )
     { line( src, vert[j],  vert[(j+1)%6], Scalar( 255 ), 3, 8 ); }

  /// Get the contours
  vector > contours; vector hierarchy;
  Mat src_copy = src.clone();

  findContours( src_copy, contours, hierarchy, RETR_TREE, CHAIN_APPROX_SIMPLE);

  /// Calculate the distances to the contour
  Mat raw_dist( src.size(), CV_32FC1 );

  for( int j = 0; j < src.rows; j++ )
     { for( int i = 0; i < src.cols; i++ )
          { raw_dist.at(j,i) = pointPolygonTest( contours[0], Point2f(i,j), true ); }
     }

  double minVal; double maxVal;
  minMaxLoc( raw_dist, &minVal, &maxVal, 0, 0, Mat() );
  minVal = abs(minVal); maxVal = abs(maxVal);

  /// Depicting the  distances graphically
  Mat drawing = Mat::zeros( src.size(), CV_8UC3 );

  for( int j = 0; j < src.rows; j++ )
     { for( int i = 0; i < src.cols; i++ )
          {
            if( raw_dist.at(j,i) < 0 )
              { drawing.at(j,i)[0] = 255 - (int) abs(raw_dist.at(j,i))*255/minVal; }
            else if( raw_dist.at(j,i) > 0 )
              { drawing.at(j,i)[2] = 255 - (int) raw_dist.at(j,i)*255/maxVal; }
            else
              { drawing.at(j,i)[0] = 255; drawing.at(j,i)[1] = 255; drawing.at(j,i)[2] = 255; }
          }
     }

  /// Create Window and show your results
  char* source_window = "Source";
  namedWindow( source_window, CV_WINDOW_AUTOSIZE );
  imshow( source_window, src );
  namedWindow( "Distance", CV_WINDOW_AUTOSIZE );
  imshow( "Distance", drawing );

  waitKey(0);
  return(0);
}

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作者：Stone_Yannn 
来源：CSDN 
原文：https://blog.csdn.net/u012348774/article/details/78882084 
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