"Exercise 8.5" Contour length computer Cvapproxpoly approximation

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Topic Requirements

Program code

Result picture

Homilies Miao DAO

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Title Requirements:

A. Detecting contours and calculating contour Lengths

b, using 1/90,1/66,1/11,1/10 as the accuracy parameter, using Cvapproxpoly approximation, calculate the contour length and draw the result

Program code:

1 //OpenCVExerciseTesting.cpp: Defines the entry point of the console application. 2 //3 //string file_full_name = "d:\\work\\work_programming\\source\\image\\opencvexerciseimage\\ 8th chapter \\r20.jpg";4 5 6#include"stdafx.h"7#include <string>8#include <cv.h>9#include Ten#include <iostream> One#include <math.h> A  -#include <opencv2/legacy/legacy.hpp> - //#pragma comment (lib, "Opencv_legacy2411.lib") the  - using namespaceCV; - using namespacestd; -  + //function declaration---->-->--->-->--->-->--->// -  +  A //<--<--<--<--<--<--<--<--<--Function Declaration// at  - int_tmain (intARGC, _tchar*argv[]) - { -     stringFile_full_name ="d:\\work\\work_programming\\source\\image\\opencvexerciseimage\\ 8th Chapter \\r20.jpg"; -  -Iplimage * Image_source =Cvloadimage (File_full_name.c_str (), cv_load_image_grayscale); in Cv_assert (image_source); -  toIplimage * Image_binary =cvcloneimage (image_source); + Cvzero (image_binary); -  theCvthreshold (Image_source, Image_binary, the,255, cv_thresh_binary); *  $Cvmemstorage *storage =cvcreatememstorage ();Panax Notoginsengcvseq* first_contour=NULL; -     intContour_num; theContour_num = Cvfindcontours (image_binary, storage, &first_contour,sizeof(Cvcontour), cv_retr_list); +  Acout <<"Number of outlines"<< Contour_num <<Endl; the      +     Doublecontour_length; -     //For (CVSEQ * c = first_contour; c = NULL; c = c->h_next) $     //{ $     //contour_length = Cvcontourperimeter (c); -     //cout << "perimeter" << contour_length << Endl; -     //} theContour_length =Cvcontourperimeter (first_contour); -cout <<"Perimeter"<< Contour_length <<Endl;Wuyi  the     DoublePerimeter =126.7; -     Doubleparameters[4] = {126.7/ -,126.7/ the,126.7/ One,126.7/Ten }; Wucvmemstorage* Storage_approx =cvcreatememstorage (); -  AboutIplimage *image_approx =cvcloneimage (image_binary); $ Cvzero (Image_approx); -Cvseq *seq_approx=NULL; -      -     stringWindow_name ="approx window";; A      for(inti =0; I <4; ++i) +     {         theSeq_approx = Cvapproxpoly (First_contour,sizeof(Cvcontour), Storage_approx, CV_POLY_APPROX_DP, Parameters[i],0); -Contour_length =Cvcontourperimeter (Seq_approx); $cout << contour_length <<Endl; theWindow_name = Window_name +"."; theCvdrawcontours (Image_approx, Seq_approx, Cvscalar (255), Cvscalar ( the),0); the Cvshowimage (Window_name.c_str (), image_approx); the     } -      inCvwaitkey (0); the  theCvreleaseimage (&image_source); AboutCvreleaseimage (&image_binary); theCvreleaseimage (&Image_approx); the cvdestroyallwindows (); the  +     return 0; -}

Result Picture:

Homilies Miao DAO:

For reference:

"Exercise 8.5" Contour length computer Cvapproxpoly approximation