/// <summary>
/// After a flip we have two triangles and know that only one will still be
/// intersecting the edge. So decide which to contiune with and legalize the
/// other
/// </summary>
///
/// <param name="tcx"></param>
/// <param name="o">should be the result of an orient2d( eq, op, ep )</param>
/// <param name="t">triangle 1</param>
/// <param name="ot">triangle 2</param>
/// <param name="p">a point shared by both triangles</param>
/// <param name="op">another point shared by both triangles</param>
/// <returns>the triangle still intersecting the edge</returns>
private static DelaunayTriangle nextFlipTriangle(DTSweepContext tcx,
Orientation o,
DelaunayTriangle t,
DelaunayTriangle ot,
TriangulationPoint p,
TriangulationPoint op)
{
int edgeIndex;
if (o == Orientation.CCW)
{
// ot is not crossing edge after flip
edgeIndex = ot.edgeIndex(p, op);
ot.dEdge[edgeIndex] = true;
legalize(tcx, ot);
ot.clearDelunayEdges();
return(t);
}
// t is not crossing edge after flip
edgeIndex = t.edgeIndex(p, op);
t.dEdge[edgeIndex] = true;
legalize(tcx, t);
t.clearDelunayEdges();
return(ot);
}
private static bool isEdgeSideOfTriangle(DelaunayTriangle triangle,
TriangulationPoint ep,
TriangulationPoint eq)
{
int index;
index = triangle.edgeIndex(ep, eq);
if (index != -1)
{
triangle.markConstrainedEdge(index);
triangle = triangle.neighbors[index];
if (triangle != null)
{
triangle.markConstrainedEdge(ep, eq);
}
return(true);
}
return(false);
}