Partitioning a set of points in 2D plane

问题

The problem statement is - "You are given a set of N points, where N is even and N <= 1000. You have to find the number of pairs of points, such that if you draw a line through that pair, each side of the line will contains equal number of points(N/2-1)." I can't figure out, how to solve this problem in O(n^2) or less time? Here is my brute-force solution-

class Point{
public:
   int x, y;
   Point(){x = y = 0;}
   void make_point(int X, int Y){ x = X; y = Y; }
   int Point:: orientation (Point &p0, Point &p1){
     Point p2 = *this;
     Point a = p1 - p0;
     Point b = p2 - p0;
     int area = (a.x * b.y) - (b.x * a.y);
     if (area > 0)return 1;
     if (area < 0)return -1;
     return 0;
   }
};
int main() {
  Point p[4];
  p[0].make_point(0, 0);
  p[1].make_point(0, 1);
  p[2].make_point(1, 1);
  p[3].make_point(1, 0);
  int sz = sizeof(p) / sizeof(p[0]);
  int ans = 0;
  for (int i = 0; i < sz; i++){
    for (int j = i+1; j < sz; j++){
        int leftCnt = 0, rightCnt = 0;
        for (int k = 0; k < sz; k++){
            if (k == i || k == j)continue;
            if (p[k].orientation(p[i], p[j]) == 1)leftCnt++;
            if (p[k].orientation(p[i], p[j]) == -1)rightCnt++;
        }
        if (leftCnt == rightCnt && leftCnt == (sz/2-1))ans++;
    }
  }
  cout << ans << '\n';

  return 0;
}

Is there any way to optimize the solution?

回答1:

There's a simple way do do this in O(n^2 log n) time.

for each point O in the set
   for each point A /= O
      calculate the slope of the ray OA 
   sort the points by the slope (this is the n log n limiting step) 
   for each point A /= O
      determine how many points are at either side of the line OA (this is an O(1) job)

Perhaps the sorting time could be cut down because a sorted array of slopes would become nearly sorted (only requiring O(n) time to fully sort) when transforming polar co-ordinates to another origin, but I can't prove this at the moment.

来源：https://stackoverflow.com/questions/52908115/partitioning-a-set-of-points-in-2d-plane