Just use findContours() in your image, then decide whether the contour is closed or not by examining the hierarchy passed to the findContours() function. From the second figure it is clearer that no contour has child contour as compared to the first image, you will get this data from hierarchy parameter which is optional output vector, containing information about the image topology. It has as many elements as the number of contours.
Here we will use hierarchy as
vector< Vec4i > hierarchy
where for an i-th contour
hierarchy[i][0] = next contour at the same hierarchical level
hierarchy[i][1] = previous contour at the same hierarchical level
hierarchy[i][2] = denotes its first child contour
hierarchy[i][3] = denotes index of its parent contour
If for the contour i there are no next, previous, parent, or nested contours, the corresponding elements of hierarchy[i]
will be negative. See findContours() function for more details.
So by checking the value hierarchy[i][2]
you can decide the contour belongs to closed or not, that is for a contour if the hierarchy[i][2] = -1
then no child and it belongs to opened.
And one more thing is that in findContours() function you should use CV_RETR_CCOMP which retrieves all of the contours and organizes them into a two-level hierarchy.
Here is the C++ code how to implement this.
Mat tmp,thr;
Mat src=imread("1.png",1);
cvtColor(src,tmp,CV_BGR2GRAY);
threshold(tmp,thr,200,255,THRESH_BINARY_INV);
vector< vector <Point> > contours; // Vector for storing contour
vector< Vec4i > hierarchy;
findContours( thr, contours, hierarchy,CV_RETR_CCOMP, CV_CHAIN_APPROX_SIMPLE );
for( int i = 0; i< contours.size(); i=hierarchy[i][0] ) // iterate through each contour.
{
Rect r= boundingRect(contours[i]);
if(hierarchy[i][2]<0) //Check if there is a child contour
rectangle(src,Point(r.x-10,r.y-10), Point(r.x+r.width+10,r.y+r.height+10), Scalar(0,0,255),2,8,0); //Opened contour
else
rectangle(src,Point(r.x-10,r.y-10), Point(r.x+r.width+10,r.y+r.height+10), Scalar(0,255,0),2,8,0); //closed contour
}
Result: