直角三角形 纪中 1385 数学_斜率 英文题解

题解

We need to find an algorithm of complexity better than O(N3). Here we will describe three such
algorithms.
The basic idea of the first one is: choose the point in which the angle will be right, fix one other point
in the triangle and quickly calculate how many of the remaining points form a right triangle with the
first two points. The algorithm is based on the so-called canonical representation of a line. Each line
can be, regardless of which two points it is generated from, transformed into a unique form. For a pair
of points we use the usual formulas to calculate three integers A, B and C such that Ax + By + C = 0.
The equation can be multiplied by a constant, because this doesn’t change the line it represents. Now
divide A, B and C by their greatest common divisor so that they become relatively prime, and also
negate the entire equation if the first non-zero number is negative. With this we can build a
function/map f(line), which tells us how many points are on a line. Now it is easy to implement the
idea from the start of this paragraph, and using appropriate date structures, the complexity will be
good.
The second algorithm, after choosing the first point (where the right angle will be), sorts the remaining
points around it by angle. Now we can use two variables (the so-called “sweep” method) to count all
right triangles. We move the first variable point by point, and have the second one 90 degrees ahead of
it, moving forward, trying to form right triangles with the first variable.
The third algorithm is different from the first two, but also easier to implement. Choose a point P and
translate the coordinate plane so that the point P is the origin (more precisely, subtract the coordinates
of point P from every point). Now, for each point, first determine which quadrant it is in, and then
rotate it by 90 degrees until it is in the first quadrant. After that, sort all points by angle (y divided by x).
Two points form a right triangle with point P if they have the same angle and if they were in
neighbouring quadrants before rotating. After sorting, for each set of points with the same angle, count
how many of them were in each of the four quadrants and multiply the numbers for neighbouring
quadrants.
The complexity of all three algorithms is O(N2·logN). The official source code features the lastalgorithm.