问题描述:已知两幅图像Image1和Image2,计算出两幅图像的重叠区域,并在Image1和Image2标识出重叠区域。
算法思想:
若两幅图像存在重叠区域,则进行图像匹配后,会得到一张完整的全景图,因而可以转换成图像匹配问题。
图像匹配问题,可以融合两幅图像,得到全景图,但无法标识出在原图像的重叠区域。
将两幅图像都理解为多边形,则其重叠区域的计算,相当于求多边形的交集。
通过多边形求交,获取重叠区域的点集,然后利用单应矩阵还原在原始图像的点集信息,从而标识出重叠区域。
算法步骤:
1.图像匹配计算,获取单应矩阵。
2.根据单应矩阵,计算图像2的顶点转换后的点集。
3.由图像1的顶点集合和图像2的转换点集,计算多边形交集。
4.根据单应矩阵的逆,计算多边形的交集在图像2中的原始点集。
代码实现如下所示:
1 bool ImageOverlap(cv::Mat &img1,cv::Mat &img2,std::vector<cv::Point> &vPtsImg1,std::vector<cv::Point> &vPtsImg2) 2 { 3 cv::Mat g1(img1,Rect(0,0,img1.cols,img1.rows)); 4 cv::Mat g2(img2,Rect(0,0,img2.cols,img2.rows)); 5 6 cv::cvtColor(g1,g1,CV_BGR2GRAY); 7 cv::cvtColor(g2,g2,CV_BGR2GRAY); 8 9 std::vector<cv::KeyPoint> keypoints_roi, keypoints_img; /* keypoints found using SIFT */ 10 cv::Mat descriptor_roi, descriptor_img; /* Descriptors for SIFT */ 11 cv::FlannBasedMatcher matcher; /* FLANN based matcher to match keypoints */ 12 std::vector<cv::DMatch> matches, good_matches; 13 cv::SIFT sift; 14 int i, dist=80; 15 16 sift(g1, Mat(), keypoints_roi, descriptor_roi); /* get keypoints of ROI image */ 17 sift(g2, Mat(), keypoints_img, descriptor_img); /* get keypoints of the image */ 18 matcher.match(descriptor_roi, descriptor_img, matches); 19 20 double max_dist = 0; double min_dist = 1000; 21 22 //-- Quick calculation of max and min distances between keypoints 23 for( int i = 0; i < descriptor_roi.rows; i++ ) 24 { 25 double dist = matches[i].distance; 26 if( dist < min_dist ) min_dist = dist; 27 if( dist > max_dist ) max_dist = dist; 28 } 29 30 for (i=0; i < descriptor_roi.rows; i++) 31 { 32 if (matches[i].distance < 3*min_dist) 33 { 34 good_matches.push_back(matches[i]); 35 } 36 } 37 38 //printf("%ld no. of matched keypoints in right image\n", good_matches.size()); 39 /* Draw matched keypoints */ 40 41 //Mat img_matches; 42 //drawMatches(img1, keypoints_roi, img2, keypoints_img, 43 // good_matches, img_matches, Scalar::all(-1), 44 // Scalar::all(-1), vector<char>(), 45 // DrawMatchesFlags::NOT_DRAW_SINGLE_POINTS); 46 //imshow("matches",img_matches); 47 48 vector<Point2f> keypoints1, keypoints2; 49 for (i=0; i<good_matches.size(); i++) 50 { 51 keypoints1.push_back(keypoints_img[good_matches[i].trainIdx].pt); 52 keypoints2.push_back(keypoints_roi[good_matches[i].queryIdx].pt); 53 } 54 //计算单应矩阵 55 Mat H = findHomography( keypoints1, keypoints2, CV_RANSAC ); 56 57 //show stitchImage 58 // cv::Mat stitchedImage; 59 // int mRows = img2.rows; 60 // if (img1.rows> img2.rows) 61 // { 62 // mRows = img1.rows; 63 // } 64 // stitchedImage = Mat::zeros(img2.cols+img1.cols, mRows, CV_8UC3); 65 // warpPerspective(img2,stitchedImage,H,Size(img2.cols+img1.cols,mRows)); 66 // Mat half(stitchedImage,Rect(0,0,img1.cols,img1.rows)); 67 // img1.copyTo(half); 68 // imshow("stitchedImage",stitchedImage); 69 70 std::vector<cv::Point> vSrcPtsImg1; 71 std::vector<cv::Point> vSrcPtsImg2; 72 73 vSrcPtsImg1.push_back(cv::Point(0,0)); 74 vSrcPtsImg1.push_back(cv::Point(0,img1.rows)); 75 vSrcPtsImg1.push_back(cv::Point(img1.cols,img1.rows)); 76 vSrcPtsImg1.push_back(cv::Point(img1.cols,0)); 77 78 vSrcPtsImg2.push_back(cv::Point(0,0)); 79 vSrcPtsImg2.push_back(cv::Point(0,img2.rows)); 80 vSrcPtsImg2.push_back(cv::Point(img2.cols,img2.rows)); 81 vSrcPtsImg2.push_back(cv::Point(img2.cols,0)); 82 83 //计算图像2在图像1中对应坐标信息 84 std::vector<cv::Point> vWarpPtsImg2; 85 for(int i = 0;i < vSrcPtsImg2.size();i++ ) 86 { 87 cv::Mat srcMat = Mat::zeros(3,1,CV_64FC1); 88 srcMat.at<double>(0,0) = vSrcPtsImg2[i].x; 89 srcMat.at<double>(1,0) = vSrcPtsImg2[i].y; 90 srcMat.at<double>(2,0) = 1.0; 91 92 cv::Mat warpMat = H * srcMat; 93 cv::Point warpPt; 94 warpPt.x = cvRound(warpMat.at<double>(0,0)/warpMat.at<double>(2,0)); 95 warpPt.y = cvRound(warpMat.at<double>(1,0)/warpMat.at<double>(2,0)); 96 97 vWarpPtsImg2.push_back(warpPt); 98 } 99 //计算图像1和转换后的图像2的交点 100 if(!PolygonClip(vSrcPtsImg1,vWarpPtsImg2,vPtsImg1)) 101 return false; 102 103 for (int i = 0;i < vPtsImg1.size();i++) 104 { 105 cv::Mat srcMat = Mat::zeros(3,1,CV_64FC1); 106 srcMat.at<double>(0,0) = vPtsImg1[i].x; 107 srcMat.at<double>(1,0) = vPtsImg1[i].y; 108 srcMat.at<double>(2,0) = 1.0; 109 110 cv::Mat warpMat = H.inv() * srcMat; 111 cv::Point warpPt; 112 warpPt.x = cvRound(warpMat.at<double>(0,0)/warpMat.at<double>(2,0)); 113 warpPt.y = cvRound(warpMat.at<double>(1,0)/warpMat.at<double>(2,0)); 114 vPtsImg2.push_back(warpPt); 115 } 116 return true; 117 }
其中,多边形求交集可参考:http://www.cnblogs.com/dwdxdy/p/3232110.html
最终,程序运行的示意图如下:
时间: 2024-10-25 11:44:12