SIFT、SURF特征匹配点的坐标获取\及匹配应用

/ / 特征点匹配.cpp : 定义控制台应用程序的入口点。

//

#include "stdafx.h"

#include

#include

#include "opencv2/core/core.hpp"

#include "opencv2/features2d/features2d.hpp"

#include "opencv2/highgui/highgui.hpp"

#include

using namespace cv;

int main( int argc, char** argv )

{

//Mat img_1=imread("image0.jpg");

//Mat img_2=imread("image1.jpg");

//Mat img_1=imread("box.png");

//Mat img_2=imread("box_in_scene.png");

Mat img_1=imread("Lena.jpg");

Mat img_2=imread("Lena_wrap_rotate.jpg");

if( !img_1.data || !img_2.data )

{ std::cout<< " --(!) Errorreading images " << std::endl; return-1; }

//-- Step 1: Detect the keypoints using SURF Detector

int minHessian = 200;

SiftFeatureDetector detector( minHessian );

//SurfFeatureDetector detector( minHessian );

std::vector keypoints_1, keypoints_2;

detector.detect( img_1, keypoints_1 );

detector.detect( img_2, keypoints_2 );

//-- Step 2: Calculate descriptors (feature vectors)

SiftDescriptorExtractor extractor;

//SurfDescriptorExtractor extractor;

Mat descriptors_1, descriptors_2;

extractor.compute( img_1, keypoints_1, descriptors_1 );

extractor.compute( img_2, keypoints_2, descriptors_2 );

//-- Step 3: Matching descriptor vectors using FLANNmatcher

BFMatcher matcher(NORM_L2);

std::vector< DMatch >matches;

matcher.match( descriptors_1, descriptors_2, matches );

//提取匹配点的前20个距离最近（最相似的点），其他的点去除
//std::nth_element(matches[i].begin(),matches[i].begin()+19,matches[i].end());
  // matches[i].erase(matches[i].begin()+6,matches[i].end());

double max_dist = 0; double min_dist = 100;

//-- Quick calculation of max and min distances betweenkeypoints

for( int i = 0; i < descriptors_1.rows; i++)

{ double dist = matches[i].distance;

if( dist < min_dist ) min_dist = dist;

if( dist > max_dist ) max_dist = dist;

}

printf("-- Max dist : %f \n", max_dist );

printf("-- Min dist : %f \n", min_dist );

//-- Draw only "good" matches (i.e. whose distance is lessthan 3*min_dist )

//-- PS.- radiusMatch can also be used here.

std::vector< DMatch >good_matches;

for( int i = 0; i < descriptors_1.rows; i++)

{ if( matches[i].distance < 2*min_dist )

{ good_matches.push_back( matches[i]); }

}

//-- Draw only "good" matches

Mat img_matches;

drawMatches( img_1, keypoints_1, img_2, keypoints_2,

good_matches, img_matches, Scalar::all(-1),Scalar::all(-1),

vector(), DrawMatchesFlags::NOT_DRAW_SINGLE_POINTS );

//-- Show detected matches

imshow( "Good Matches", img_matches );

//imwrite("Lena_match_surf.jpg",img_matches);

//imwrite("Lena_match_sift.jpg",img_matches);

for( int i = 0; i < good_matches.size(); i++)

//{ 

// printf( "-- Good Match [%d] Keypoint 1: %d  -- Keypoint 2: %d   \n", i,  good_matches[i].queryIdx,good_matches[i].trainIdx); 

//}

{

//good_matches[i].queryIdx保存着第一张图片匹配点的序号，keypoints_1[good_matches[i].queryIdx].pt.x为该序号对应的点的x坐标。y坐标同理

//good_matches[i].trainIdx保存着第二张图片匹配点的序号，keypoints_2[good_matches[i].trainIdx].pt.x为为该序号对应的点的x坐标。y坐标同理

printf( "-- Good Match [%d] Keypoint 1(%f,%f): %d  -- Keypoint 2(%f,%f): %d   \n",i, 

keypoints_1[good_matches[i].queryIdx].pt.x,keypoints_1[good_matches[i].queryIdx].pt.y,good_matches[i].queryIdx, 

keypoints_2[good_matches[i].trainIdx].pt.x,keypoints_2[good_matches[i].trainIdx].pt.y,good_matches[i].trainIdx);

int  surfsumget=0；
for( int i = 0; i < good_matches.size(); i++ ){
            
            printf("--good_matches[%d].distance: %f \n",i,good_matches[i].distance);
        
            surfsumget=surfsumget+good_matches[i].distance ;
        }


    printf("-- good_matches.size() : %d  \n", good_matches.size() );
    printf("-- Max dist : %f  \n", max_dist );
    printf("-- Min dist : %f  \n", min_dist );
    printf("--surfsumget: %f  \n",surfsumget);

 // if((surfsumget <= threshold)&(matches.size()>20)){   //特征点数目必须足够多
 //           printf(" yes,match \n");return 1;
 //       }else{
   //         printf("not match \n");return 0;
  //      }

}

waitKey(0);

return 0;            //PS：如果视频中获取的特征点不稳定，可以利用特征点之间的空间位置信息进行辅助匹配分析
                                   //如取特征点中心、去掉最远处特征点等。。。也可利用特征点间距离方差等信息，

}

运行结果：