如何赋予有限边缘无限的优势?

时间:2013-01-11 15:50:46

标签: c++ boost voronoi

我发现在boost库中,对于voronoi图,一些边缘数据是无限的。根据指示,它必须被剪裁。但我无法找到如何做到这一点。 请问有人可以给我一个示例代码吗?

THX

3 个答案:

答案 0 :(得分:6)

这是一个相当陈旧的问题,但我在尝试解决完全相同的问题时找到了它。分享我遇到的解决方案是公平的,所以下一个可怜的傻瓜并不需要自己解决。

我不知道这是否是一种特别好的做事方式,我怀疑如果你开始使用弯曲的细胞会很困难,但它对我的目的来说是可行的。

基本问题是无限边只有一个顶点,所以你必须自己计算方向向量。使用的方向垂直于由边缘分隔的两个点之间的矢量。

#include <vector>
#include <boost/polygon/voronoi.hpp>
using boost::polygon::voronoi_builder;
using boost::polygon::voronoi_diagram;

typedef boost::polygon::point_data<int> point;
typedef voronoi_diagram<double>::cell_type cell_type;
typedef voronoi_diagram<double>::edge_type edge_type;
typedef voronoi_diagram<double>::vertex_type vertex_type;

int main(int argc, char *argv[])
{
   std::vector<point> points;

   // Populate with random points
   for (int i = 0; i < 50; i++)
   {
      points.push_back(point(60 + rand() % 500, 60 + rand() % 500));
   }

   voronoi_diagram<double> vd;
   construct_voronoi(points.begin(), points.end(), &vd);

   // vd now contains the voronoi diagram. Let's visualise it
   // pseudocode 'draw_line(x1, y1, x2, y2)'

   for (voronoi_diagram<double>::const_cell_iterator it = vd.cells().begin();
        it != vd.cells().end(); ++it)
   {
      const cell_type& cell = *it;
      const edge_type* edge = cell.incident_edge();

      do
      {
         if (edge->is_primary())
         {
            // Easy finite edge case
            if (edge->is_finite())
            {
               // Without this check each edge would be drawn twice
               // as they are really half-edges
               if (edge->cell()->source_index() < 
                   edge->twin()->cell()->source_index())
               {
                  draw_line(edge->vertex0()->x(), edge->vertex0()->y(),
                            edge->vertex1()->x(), edge->vertex1()->y());
               }
            }
            else
            {
               // This covers the infinite edge case in question.
               const vertex_type* v0 = edge->vertex0();
               // Again, only consider half the half-edges, ignore edge->vertex1()
               // to avoid overdrawing the lines
               if (v0)
               {
                  // Direction of infinite edge if perpendicular to direction
                  // between the points owning the two half edges. 
                  // Take the rotated right vector and multiply by a large 
                  // enough number to reach your bounding box
                  point p1 = points[edge->cell()->source_index()];
                  point p2 = points[edge->twin()->cell()->source_index()];
                  int end_x = (p1.y() - p2.y()) * 640;
                  int end_y = (p1.x() - p2.x()) * -640;

                  draw_line(v0->x(), v0->y(),
                            end_x, end_y);
               }
            }
         }
         edge = edge->next();
      } while (edge != cell.incident_edge());
   }
}

答案 1 :(得分:2)

我在此处找到了此代码段:http://www.boost.org/doc/libs/1_55_0/libs/polygon/example/voronoi_visualizer.cpp

void clip_infinite_edge(
      const edge_type& edge, std::vector<point_type>* clipped_edge) {
    const cell_type& cell1 = *edge.cell();
    const cell_type& cell2 = *edge.twin()->cell();
    point_type origin, direction;
    // Infinite edges could not be created by two segment sites.
    if (cell1.contains_point() && cell2.contains_point()) {
      point_type p1 = retrieve_point(cell1);
      point_type p2 = retrieve_point(cell2);
      origin.x((p1.x() + p2.x()) * 0.5);
      origin.y((p1.y() + p2.y()) * 0.5);
      direction.x(p1.y() - p2.y());
      direction.y(p2.x() - p1.x());
    } else {
      origin = cell1.contains_segment() ?
          retrieve_point(cell2) :
          retrieve_point(cell1);
      segment_type segment = cell1.contains_segment() ?
          retrieve_segment(cell1) :
          retrieve_segment(cell2);
      coordinate_type dx = high(segment).x() - low(segment).x();
      coordinate_type dy = high(segment).y() - low(segment).y();
      if ((low(segment) == origin) ^ cell1.contains_point()) {
        direction.x(dy);
        direction.y(-dx);
      } else {
        direction.x(-dy);
        direction.y(dx);
      }
    }
    coordinate_type side = xh(brect_) - xl(brect_);
    coordinate_type koef =
        side / (std::max)(fabs(direction.x()), fabs(direction.y()));
    if (edge.vertex0() == NULL) {
      clipped_edge->push_back(point_type(
          origin.x() - direction.x() * koef,
          origin.y() - direction.y() * koef));
    } else {
      clipped_edge->push_back(
          point_type(edge.vertex0()->x(), edge.vertex0()->y()));
    }
    if (edge.vertex1() == NULL) {
      clipped_edge->push_back(point_type(
          origin.x() + direction.x() * koef,
          origin.y() + direction.y() * koef));
    } else {
      clipped_edge->push_back(
          point_type(edge.vertex1()->x(), edge.vertex1()->y()));
    }
  }

它可能缺少一些类变量或方法,但逻辑在这里很重要。

答案 2 :(得分:0)

我编写了一个程序来裁剪Voronoi单元中的无限边缘,以便将边缘限制在矩形边界框内。代码在 https://github.com/datajaguar/jaguardb/blob/master/src/JagCGAL.cc

getIntersectionPointWithBox()方法基本上检查边界框的每一侧(总共4侧),如果有则获取交点。 fillInCorners()方法检查边界框周围的缺失点,并将角坐标添加到边缘。要为每个单元格制作一个多边形,它会在末尾添加第一个点以使多边形闭合。