closest pair of points(O(nlogn)) solution

#include <algorithm>#include <vector>#include <iostream>#include <math.h>#include <Windows.h>using namespace std;// point class representing point in 2D planestruct Point{int x;int y;Point() = default;Point(int xx, int yy) : x(xx), y(yy) {}};// compare function using x coordinatebool comparX(const Point &p1, const Point &p2){return p1.x < p2.x;}// compare function using y coordinatebool comparY(const Point &p1, const Point &p2){return p1.y < p2.y;}// euclidean distance functionfloat dist(const Point &p1, const Point &p2){return sqrt((p1.x - p2.x) * (p1.x - p2.x) + (p1.y - p2.y) * (p1.y - p2.y));}double bruteForce(vector<Point> &points, int start, int end){double minDist = DBL_MAX;for (int i = start; i < end; ++i){for (int j = i + 1; j < end; ++j){if (dist(points[i], points[j]) < minDist)minDist = dist(points[i], points[j]);}}return minDist;}// each recursive function takes two arrays, one sorted by x coordinate, one by y coordinate// so the in each recursive function, no more need to sort the strip.// time complexity is reduced to O(nlog(n))double closestUtil(vector<Point> &points_x, vector<Point> &points_y, int start, int end){int n = end - start;// base case: if there are 2 or 3 points, use brute forceif (n <= 3)return bruteForce(points_x, start, end);int mid = n / 2;// the mid pointvector<Point> points_y_left;vector<Point> points_y_right;for (int i = start; i < end; ++i){if (points_y[i].x < points_x[mid].x)points_y_left.push_back(points_y[i]);if (points_y[i].x > points_x[mid].x)points_y_right.push_back(points_y[i]);}double delta1 = closestUtil(points_x, points_y_left, start, mid);// left partdouble delta2 = closestUtil(points_x, points_y_right, mid + 1, end);// right partdouble delta = min(delta1, delta2);// strip is the points set that all points are closer than delta to the // line passing through the mid linevector<Point> strip;for (int i = start; i < end; ++i)if (abs(points_y[i].x - points_x[mid].x) < delta)strip.push_back(points_y[i]);// O(n) complexity to find the minimum in the stripfor (int i = 0; i < strip.size(); ++i)for (int j = i + 1; j < strip.size() && strip[j].y - strip[i].y < delta; ++j)if (dist(strip[i], strip[j]) < delta)delta = dist(strip[i], strip[j]);return delta;}double closestPair(vector<Point> &points){vector<Point> points_y(points);sort(points.begin(), points.end(), comparX);// points sorted by x coordinatesort(points_y.begin(), points_y.end(), comparY);// points sorted by y coordinatereturn closestUtil(points, points_y, 0, points.size());}int main(){vector<Point> p{ { 2, 3 },{ 12, 30 },{ 40, 50 },{ 5, 1 },{ 12, 10 },{ 3, 4 },{ 100, 3 },{ 10, 10 } };cout << closestPair(p) << endl;system("PAUSE");return 0;}

reference:

http://www.cs.princeton.edu/~wayne/kleinberg-tardos/pdf/05DivideAndConquerI.pdf

http://www.geeksforgeeks.org/closest-pair-of-points-onlogn-implementation/