Mercurial > hg > orthanc-stone
view OrthancStone/Resources/Graveyard/RTStructTentativeReimplementation-BGO/DicomStructureSet-BGO.cpp @ 2031:a56f7ed0cdf9 deep-learning
integration mainline->deep-learning
author | Sebastien Jodogne <s.jodogne@gmail.com> |
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date | Fri, 23 Dec 2022 17:51:07 +0100 |
parents | affde38b84de |
children |
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namespace OrthancStone { static RtStructRectangleInSlab CreateRectangle(float x1, float y1, float x2, float y2) { RtStructRectangleInSlab rect; rect.xmin = std::min(x1, x2); rect.xmax = std::max(x1, x2); rect.ymin = std::min(y1, y2); rect.ymax = std::max(y1, y2); return rect; } bool CompareRectanglesForProjection(const std::pair<RtStructRectangleInSlab,double>& r1, const std::pair<RtStructRectangleInSlab, double>& r2) { return r1.second < r2.second; } bool CompareSlabsY(const RtStructRectanglesInSlab& r1, const RtStructRectanglesInSlab& r2) { if ((r1.size() == 0) || (r2.size() == 0)) return false; return r1[0].ymax < r2[0].ymax; } } bool DicomStructureSet::ProjectStructure(std::vector< std::vector<ScenePoint2D> >& chains, const Structure& structure, const CoordinateSystem3D& sourceSlice) const { // FOR SAGITTAL AND CORONAL // this will contain the intersection of the polygon slab with // the cutting plane, projected on the cutting plane coord system // (that yields a rectangle) + the Z coordinate of the polygon // (this is required to group polygons with the same Z later) std::vector<std::pair<RtStructRectangleInSlab, double> > projected; for (Polygons::const_iterator polygon = structure.polygons_.begin(); polygon != structure.polygons_.end(); ++polygon) { double x1, y1, x2, y2; if (polygon->Project(x1, y1, x2, y2, slice, GetEstimatedNormal(), GetEstimatedSliceThickness())) { double curZ = polygon->GetGeometryOrigin()[2]; // x1,y1 and x2,y2 are in "slice" coordinates (the cutting plane // geometry) projected.push_back(std::make_pair(CreateRectangle( static_cast<float>(x1), static_cast<float>(y1), static_cast<float>(x2), static_cast<float>(y2)),curZ)); } } // projected contains a set of rectangles specified by two opposite // corners (x1,y1,x2,y2) // we need to merge them // each slab yields ONE polygon! // we need to sorted all the rectangles that originate from the same Z // into lanes. To make sure they are grouped together in the array, we // sort it. std::sort(projected.begin(), projected.end(), CompareRectanglesForProjection); std::vector<RtStructRectanglesInSlab> rectanglesForEachSlab; rectanglesForEachSlab.reserve(projected.size()); double curZ = 0; for (size_t i = 0; i < projected.size(); ++i) { #if 0 rectanglesForEachSlab.push_back(RtStructRectanglesInSlab()); #else if (i == 0) { curZ = projected[i].second; rectanglesForEachSlab.push_back(RtStructRectanglesInSlab()); } else { // this check is needed to prevent creating a new slab if // the new polygon is at the same Z coord than last one if (!LinearAlgebra::IsNear(curZ, projected[i].second)) { rectanglesForEachSlab.push_back(RtStructRectanglesInSlab()); curZ = projected[i].second; } } #endif rectanglesForEachSlab.back().push_back(projected[i].first); // as long as they have the same y, we should put them into the same lane // BUT in Sebastien's code, there is only one polygon per lane. //std::cout << "rect: xmin = " << rect.xmin << " xmax = " << rect.xmax << " ymin = " << rect.ymin << " ymax = " << rect.ymax << std::endl; } // now we need to sort the slabs in increasing Y order (see ConvertListOfSlabsToSegments) std::sort(rectanglesForEachSlab.begin(), rectanglesForEachSlab.end(), CompareSlabsY); std::vector< std::pair<ScenePoint2D, ScenePoint2D> > segments; ConvertListOfSlabsToSegments(segments, rectanglesForEachSlab, projected.size()); chains.resize(segments.size()); for (size_t i = 0; i < segments.size(); i++) { chains[i].resize(2); chains[i][0] = segments[i].first; chains[i][1] = segments[i].second; } #endif return true; }