private void ComputeOverlaps(int start0, int end0, MonotoneChain mc, int start1, int end1, MonotoneChainOverlapAction mco)
{
// terminating condition for the recursion
if (end0 - start0 == 1 && end1 - start1 == 1)
{
mco.Overlap(this, start0, mc, start1);
return;
}
// nothing to do if the envelopes of these sub-chains don't overlap
if (!Overlaps(start0, end0, mc, start1, end1))
{
return;
}
// the chains overlap, so split each in half and iterate (binary search)
int mid0 = (start0 + end0) / 2;
int mid1 = (start1 + end1) / 2;
// Assert: mid != start or end (since we checked above for end - start <= 1)
// check terminating conditions before recursing
if (start0 < mid0)
{
if (start1 < mid1)
{
ComputeOverlaps(start0, mid0, mc, start1, mid1, mco);
}
if (mid1 < end1)
{
ComputeOverlaps(start0, mid0, mc, mid1, end1, mco);
}
}
if (mid0 < end0)
{
if (start1 < mid1)
{
ComputeOverlaps(mid0, end0, mc, start1, mid1, mco);
}
if (mid1 < end1)
{
ComputeOverlaps(mid0, end0, mc, mid1, end1, mco);
}
}
}
/// <summary>
/// Determine all the line segments in two chains which may overlap, and process them.
/// </summary>
/// <remarks>
/// The monotone chain search algorithm attempts to optimize
/// performance by not calling the overlap action on chain segments
/// which it can determine do not overlap.
/// However, it *may* call the overlap action on segments
/// which do not actually interact.
/// This saves on the overhead of checking intersection
/// each time, since clients may be able to do this more efficiently.
/// </remarks>
/// <param name="mc">The monotone chain</param>
/// <param name="mco">The overlap action to execute on selected segments</param>
public void ComputeOverlaps(MonotoneChain mc, MonotoneChainOverlapAction mco)
{
ComputeOverlaps(_start, _end, mc, mc._start, mc._end, mco);
}