标签：处理 方法 ++ bsp 中间 idt 技术 size block
图像拼接是图像处理中最为常见的一个功能，在不考虑旋转的情况下，也就是需要求取拼接亮就行了，然而当光源是点光源时，往往成像都是高斯性质，
这样会导致中间亮，两边暗或者四周暗，在这种情况下拼接，图像会存在很明显的拼接痕迹。
因此，本文针对上述拼接痕迹，设计了一种线性平滑的方法。
上代码：
1 CString leftimagepath;
2 CString rightimagepath;
3 leftimagepath.Format("0-%d-%d.jpg",i,j);
4 leftimagepath=pathIn+leftimagepath;
5 rightimagepath.Format("0-%d-%d.jpg",i,j+1);
6 rightimagepath=pathIn+rightimagepath;
7 IplImage* leftimg=cvLoadImage(leftimagepath,-1);
8 IplImage* rightimg=cvLoadImage(rightimagepath,-1);
9 sourewidth=rightimg->width;
10 soureheight=rightimg->height;
11 IplImage* rightresult=cvCreateImage(cvSize(sourewidth*2,soureheight),8,rightimg->nChannels);
12 cvZero(rightresult);
13 cvSetImageROI(rightresult,cvRect(sourewidth-xsum,0,sourewidth,soureheight));
14 cvCopy(rightimg,rightresult);
15 cvResetImageROI(rightresult);
16 IplImage* result=cvCreateImage(cvSize(sourewidth*2,soureheight),8,rightimg->nChannels);
17 cv::Mat imleft(leftimg,0); //左边的图像
18 cv::Mat imright(rightresult,0); //右边的图像，但是右边的有效区域开始于（左边的图像大小减去重合量）
19 cv::Mat dst(result,0); //结果图像
20 int channle = imright.channels();
21 for(int jx = 0;jx < imright.rows;jx++)
22 {
23 // 行定位
24 uchar *data =dst.ptr<uchar>(jx); // 最终要得到的拼接图
25 uchar *data2 = imleft.ptr<uchar>(jx); // 左图
26 uchar *datar = imright.ptr<uchar>(jx); // 单应变换后（右图）
27 for (int iy=0,q =0;iy < imright.cols;iy++)
28 {
29
30 if (iy<imleft.cols && datar[iy*channle+0]==0 && datar[iy*channle+1]==0 && datar[iy*channle+2]==0)
31 { //只在左图部分
32 data[iy*channle+0] = data2[iy*channle+0];
33 data[iy*channle+1] = data2[iy*channle+1];
34 data[iy*channle+2] = data2[iy*channle+2];
35 q++; // 进入右图的标记
36 }
37 else if(iy<imleft.cols )
38 { //在左图也在右图部分
39 float overlapwidth = float(imleft.cols - q); // 重叠区域的宽度
40 float overlapratio = float(iy-q); // 当前位置的重叠比例
41 float thr = overlapratio/overlapwidth; // 阈值
42 data[iy*channle+0] = (1-thr) * data2[iy*channle+0] + thr * datar[iy*channle+0];
43 data[iy*channle+1] = (1-thr) * data2[iy*channle+1] + thr * datar[iy*channle+1];
44 data[iy*channle+2] = (1-thr) * data2[iy*channle+2] + thr * datar[iy*channle+2];
45 }
46 else if(iy>=imleft.cols)
47 { //只在右图部分
48 data[iy*channle+0] = datar[iy*channle+0];
49 data[iy*channle+1] = datar[iy*channle+1];
50 data[iy*channle+2] = datar[iy*channle+2];
51 }
52 }//iy
53 }//jx
54 //保存图像
55 //保存当前拼接后的
56 cvSetImageROI(result,cvRect(0,0,rightimg->width-xsum,rightimg->height));
57 CString saveleftimagepath;
58 saveleftimagepath.Format("0-%d-%d.jpg",i,j);
59 saveleftimagepath=pathIn+saveleftimagepath;
60 cvSaveImage(saveleftimagepath,result);
61 cvResetImageROI(result);
62 //覆盖原始图像
63 CString saverigimagepath;
64 saverigimagepath.Format("0-%d-%d.jpg",i,j+1);
65 saverigimagepath=pathIn+saverigimagepath;
66 cvSetImageROI(result,cvRect(rightimg->width-xsum,0,rightimg->width,rightimg->height));
67 cvSaveImage(saverigimagepath,result);
68 cvResetImageROI(result);
69 /*cvNamedWindow("rs",0);
70 cvShowImage("rs",result);
71 cvWaitKey(0);*/
没有采用上述拼接结果：
采用上述过渡算法的结果：
