圖像拼接（七）：OpenCV單應變換模型拼接多幅圖像

上篇博客圖像拼接（六）：OpenCV單應變換模型拼接兩幅圖像 實現了兩幅圖像的拼接，主要是使用了單應矩陣和warpPerspective()這個庫函數。

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#求取每相鄰兩幅圖像的單應矩陣
拼接多幅圖像，需要計算每相鄰兩幅圖像的單應矩陣，上篇已經封裝了求取單應矩陣的類，可以拿來用。
現有4幅圖像：$img1$，$img2$，$img3$，$img4$。依次從右向左排列，拼接圖像以最左側的$img4$爲參考圖像。

Homography homo12(img1,img2);
Homography homo23(img2, img3);
Homography homo34(img3, img4);

Mat h12 = homo12.getHomography();
Mat h23 = homo23.getHomography();
Mat h34 = homo34.getHomography();

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#透視變換

warpPerspective(InputArray src, OutputArray dst, InputArray M, Size dsize)

這個函數卻顯得不夠靈活。一方面，目標投影的圖像不能夠選取興趣區域，使得每拼接兩幅都會產生一個結果圖像；另一方面，投影參考原點爲圖像左上角，如果投影后的圖像在左方，就不能顯示出，所以需要左側的圖像爲參考圖像。

爲了實現方便，拼接4幅圖像以最左側圖像作爲參考（其實這種方案會產生累加誤差和圖像變形，最佳的方案應該是選擇中間幅作爲參考基準），計算其餘3幅圖像到它的變換。

	Mat h24 = h34*h23;
	Mat h14 = h24*h12;

	float scale_h24 = h24.at<double>(2, 2);
	float scale_h14 = h14.at<double>(2, 2);
	h24 = h24 / scale_h24;
	h14 = h14 / scale_h14;

	Mat warp1;
	warpPerspective(img1_color, warp1, h14, Size(img1.cols * 4, img1.rows));
	Mat warp2;
	warpPerspective(img2_color, warp2, h24, Size(img1.cols * 4, img1.rows));
	Mat warp3;
	warpPerspective(img3_color, warp3, h34, Size(img1.cols * 4, img1.rows));

投影結果:

img3->img4

img2->img4

img1->img4

再把上述中間結果放在一個最終的結果圖像中。怎麼自動地確定水平方向剪切的邊界呢？我的一個解決思路是：根據單應矩陣的8個參數，提取出水平位移量。將單應變換公式展開

$\begin{bmatrix} x^\prime \ y^\prime \1 \end{bmatrix} =\begin{bmatrix} 1+h_{00} &h_{01} & h_{02} \ h_{10} & 1+h_{11}&h_{12} \h_{20}&h_{21}&1 \end{bmatrix} \begin{bmatrix} x \ y \1 \end{bmatrix} $

$x^\prime=\dfrac{(1+h_{00})x+h_{01}y+h_{02}}{h_{20}x+h_{21}y+1}$

對不同的$x$和$y$，位移量不同，不可能提取出一個統一的值。這裏採取粗略的方式，取$x=0$和$y=0$時得出的$x^\prime$值作爲水平位移量。因爲在邊緣剪容易出現黑色像素，所以再加上$\frac{2}{5}$原始圖像寬度，選取大概靠圖像中間的位置作爲剪切位置。當然這個中間位置的計算是及其粗略的，但至少避開了邊緣。

剪切代碼

	int d = img1.cols*2/5;

	int x3 = h34.at<double>(0,2)+d;
	int x2 = x3 + h23.at<double>(0, 2);
	int x1 = x2 + h12.at<double>(0, 2);

	Mat canvas(img1.rows,img1.cols*4,CV_8UC3);
	img4_color.copyTo(canvas(Range::all(), Range(0, img1.cols)));
	warp3(Range::all(), Range(x3, x2)).copyTo(canvas(Range::all(), Range(x3, x2)));
	warp2(Range::all(), Range(x2, x1)).copyTo(canvas(Range::all(), Range(x2, x1)));
	warp1(Range::all(), Range(x1, x1 + img1.cols)).copyTo(canvas(Range::all(), Range(x1, x1 + img1.cols)));

最終結果

由拼接結果可以看出，圖像越往右側，圖像變形越大，出現了明顯的拼接誤差，這個可以通過後期融合來消除。另一方面，由於接縫位置選擇不當，第4張圖像只顯示了一部分和出現了不對齊的情況，可以採用最佳拼接縫算法，使不對齊的情況最小化。本篇博客未實現接縫融合和最佳拼接縫尋找。

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#完整代碼和圖像素材

#include"Homography.h"

int main()
{
	//從右向左升序
	string imgPath1 = "trees_003.jpg";
	string imgPath2 = "trees_002.jpg";
	string imgPath3 = "trees_001.jpg";
	string imgPath4 = "trees_000.jpg";

	Mat img1 = imread(imgPath1, CV_LOAD_IMAGE_GRAYSCALE);
	Mat img2 = imread(imgPath2, CV_LOAD_IMAGE_GRAYSCALE);
	Mat img3 = imread(imgPath3, CV_LOAD_IMAGE_GRAYSCALE);
	Mat img4 = imread(imgPath4, CV_LOAD_IMAGE_GRAYSCALE);

	Mat img1_color = imread(imgPath1, CV_LOAD_IMAGE_COLOR);
	Mat img2_color = imread(imgPath2, CV_LOAD_IMAGE_COLOR);
	Mat img3_color = imread(imgPath3, CV_LOAD_IMAGE_COLOR);
	Mat img4_color = imread(imgPath4, CV_LOAD_IMAGE_COLOR);

	Homography homo12(img1,img2);
	Homography homo23(img2, img3);
	Homography homo34(img3, img4);

	Mat h12 = homo12.getHomography();
	Mat h23 = homo23.getHomography();
	Mat h34 = homo34.getHomography();

	/*homo12.drawMatches();
	homo23.drawMatches();
	homo34.drawMatches();*/

	Mat h24 = h34*h23;
	Mat h14 = h24*h12;

	float scale_h24 = h24.at<double>(2, 2);
	float scale_h14 = h14.at<double>(2, 2);
	h24 = h24 / scale_h24;
	h14 = h14 / scale_h14;

	Mat warp1;
	warpPerspective(img1_color, warp1, h14, Size(img1.cols * 4, img1.rows));
	Mat warp2;
	warpPerspective(img2_color, warp2, h24, Size(img1.cols * 4, img1.rows));
	Mat warp3;
	warpPerspective(img3_color, warp3, h34, Size(img1.cols * 4, img1.rows));

	imshow("warp1", warp1);
	imshow("warp2",warp2);
	imshow("warp3", warp3);
	
	imwrite("warp1.jpg", warp1);
	imwrite("warp2.jpg", warp2);
	imwrite("warp3.jpg", warp3);

	int d = img1.cols*2/5;
	int x3 = h34.at<double>(0,2)+d;
	int x2 = x3 + h23.at<double>(0, 2);
	int x1 = x2 + h12.at<double>(0, 2);

	Mat canvas(img1.rows,img1.cols*4,CV_8UC3);
	img4_color.copyTo(canvas(Range::all(), Range(0, img1.cols)));
	warp3(Range::all(), Range(x3, x2)).copyTo(canvas(Range::all(), Range(x3, x2)));
	warp2(Range::all(), Range(x2, x1)).copyTo(canvas(Range::all(), Range(x2, x1)));
	warp1(Range::all(), Range(x1, x1 + img1.cols)).copyTo(canvas(Range::all(), Range(x1, x1 + img1.cols)));
	

	imwrite("canvas.jpg",canvas);
	imshow("canvas",canvas);

	waitKey(0);
	return 0;
}

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img4

img3

img2

img1