我正在使用OpenCV 3.1并想知道如何在Subdiv2D Delaunay三角剖分中获得边的顶点索引,而不仅仅是顶点坐标?我想使用此索引来保持顶点值以进行进一步插值。
例如:
cv::Subdiv2D subdiv(rect);
// Insert some points to subdiv....
cv::Point2f org, dst;
subdiv.edgeOrg(edge, &org); // starting point of edge
这只给出了org
中顶点的坐标。
我已经在OpenCV 2.4中看到过,我们可以使用CvSubdiv2DPoint
类型,除了标准的Point32f
之外还有id。在OpenCV 3.1中,我找不到这个结构,看起来它因为某些原因被删除了。
如果有其他人看这个问题。您可以创建一个继承自subdiv2D的类,并实现您自己的函数,以这种方式返回索引:
。H
#ifndef __Subdiv2DIndex__
#define __Subdiv2DIndex__
#include <vector>
#include <opencv2\opencv.hpp>
using namespace cv;
class Subdiv2DIndex : public Subdiv2D
{
public :
Subdiv2DIndex(Rect rectangle);
//Source code of Subdiv2D: https://github.com/opencv/opencv/blob/master/modules/imgproc/src/subdivision2d.cpp#L762
//The implementation tweaks getTrianglesList() so that only the indice of the triangle inside the image are returned
void getTrianglesIndices(std::vector<int> &ind) const;
};
#endif
.cpp #include“Subdiv2DIndex.h”
Subdiv2DIndex::Subdiv2DIndex(Rect rectangle) : Subdiv2D{rectangle}
{
}
void Subdiv2DIndex::getTrianglesIndices(std::vector<int> &triangleList) const
{
triangleList.clear();
int i, total = (int)(qedges.size() * 4);
std::vector<bool> edgemask(total, false);
const bool filterPoints = true;
Rect2f rect(topLeft.x, topLeft.y, bottomRight.x - topLeft.x, bottomRight.y - topLeft.y);
for (i = 4; i < total; i += 2)
{
if (edgemask[i])
continue;
Point2f a, b, c;
int edge_a = i;
int indexA = edgeOrg(edge_a, &a) -4;
if (filterPoints && !rect.contains(a))
continue;
int edge_b = getEdge(edge_a, NEXT_AROUND_LEFT);
int indexB = edgeOrg(edge_b, &b) - 4;
if (filterPoints && !rect.contains(b))
continue;
int edge_c = getEdge(edge_b, NEXT_AROUND_LEFT);
int indexC = edgeOrg(edge_c, &c) - 4;
if (filterPoints && !rect.contains(c))
continue;
edgemask[edge_a] = true;
edgemask[edge_b] = true;
edgemask[edge_c] = true;
triangleList.push_back(indexA);
triangleList.push_back(indexB);
triangleList.push_back(indexC);
}
}
事实证明,在OpenCV 3中,cv::Subdiv2D
中的一些函数现在是void,它们返回一个整数。在我的例子中,我发现subdiv.edgeOrg(...)
和subdiv.edgeDst(...)
将顶点id作为整数返回。