Mostly revert ef01f37
That commit message was a lie, it did not improve performance but actually reduce it significantly... I had accidentally added a lot of complexity due to the std::list data structure. this approach might seem more redundant, but it's much faster with rectangles (like 100x faster) and doesn't seem to make a difference in the other cases.
This commit is contained in:
parent
23fef77aaa
commit
e832407ea0
36
Quickhull.h
36
Quickhull.h
@ -149,38 +149,38 @@ private:
|
|||||||
// If the input vector is empty, we're done
|
// If the input vector is empty, we're done
|
||||||
if (input.empty()) return;
|
if (input.empty()) return;
|
||||||
|
|
||||||
// Find the points which are furthest away from the line
|
// Find the point which is furthest away from the line, add it to the output
|
||||||
|
Point furthest_point;
|
||||||
float furthest_distance = -1.0;
|
float furthest_distance = -1.0;
|
||||||
|
|
||||||
std::list<Point> furthest_points;
|
|
||||||
|
|
||||||
for (const Point &point : input)
|
for (const Point &point : input)
|
||||||
{
|
{
|
||||||
float this_distance = line.distance_squared_to(point);
|
float this_distance = line.distance_squared_to(point);
|
||||||
// It's possible for there to be multiple closests points (e.g. in the case)
|
|
||||||
// of a rectangle). We need to handle all these properly, so we make a list
|
|
||||||
// of all points at the furthest distance.
|
|
||||||
if (this_distance > furthest_distance)
|
if (this_distance > furthest_distance)
|
||||||
{
|
{
|
||||||
furthest_distance = this_distance;
|
furthest_distance = this_distance;
|
||||||
furthest_points.clear();
|
furthest_point = point;
|
||||||
furthest_points.emplace_back(point);
|
|
||||||
}
|
|
||||||
else if (this_distance == furthest_distance)
|
|
||||||
{
|
|
||||||
furthest_points.emplace_back(point);
|
|
||||||
}
|
}
|
||||||
}
|
}
|
||||||
|
|
||||||
for (const Point &point : furthest_points)
|
// This seems unnecessarily complicated, but it actually yielded the best results,
|
||||||
|
// especially when the rectangle edge case is taken into account.
|
||||||
|
for (const Point &point : input)
|
||||||
{
|
{
|
||||||
// All furthest points we found are part of the hull, so add them to the output
|
float this_distance = line.distance_squared_to(point);
|
||||||
// and remove them from the input.
|
if (this_distance == furthest_distance)
|
||||||
output.emplace_back(point);
|
{
|
||||||
input.remove(point);
|
output.emplace_back(point);
|
||||||
|
}
|
||||||
}
|
}
|
||||||
|
// Remove the previously checked points from the input -- we can't do that in
|
||||||
|
// the previous loop because we can't delete from the list we're iterating over
|
||||||
|
input.remove_if([furthest_distance, line](Point point)
|
||||||
|
{
|
||||||
|
return furthest_distance == line.distance_squared_to(point);
|
||||||
|
});
|
||||||
|
|
||||||
Point furthest_point = furthest_points.front();
|
output.emplace_back(furthest_point);
|
||||||
|
|
||||||
// Build a triangle with these 3 points
|
// Build a triangle with these 3 points
|
||||||
|
|
||||||
|
Loading…
x
Reference in New Issue
Block a user