Closest pair of points - Geometrical Algorithms

#include <iostream>
#include <float.h>
#include <stdlib.h>
#include <math.h>
using namespace std;

struct Point
{
	int x, y;
};

int compareX(const void* a, const void* b)
{
	Point *p1 = (Point *)a, *p2 = (Point *)b;
	return (p1->x != p2->x) ? (p1->x - p2->x) : (p1->y - p2->y);
}

int compareY(const void* a, const void* b)
{
	Point *p1 = (Point *)a, *p2 = (Point *)b;
	return (p1->y != p2->y) ? (p1->y - p2->y) : (p1->x - p2->x);
}


float dist(Point p1, Point p2)
{
	return sqrt( (p1.x - p2.x)*(p1.x - p2.x) +
				(p1.y - p2.y)*(p1.y - p2.y)
			);
}

float bruteForce(Point P[], int n)
{
	float min = FLT_MAX;
	for (int i = 0; i < n; ++i)
		for (int j = i+1; j < n; ++j)
			if (dist(P[i], P[j]) < min)
				min = dist(P[i], P[j]);
	return min;
}

float min(float x, float y)
{
	return (x < y)? x : y;
}


float stripClosest(Point strip[], int size, float d)
{
	float min = d; // Initialize the minimum distance as d


	for (int i = 0; i < size; ++i)
		for (int j = i+1; j < size && (strip[j].y - strip[i].y) < min; ++j)
			if (dist(strip[i],strip[j]) < min)
				min = dist(strip[i], strip[j]);

	return min;
}

float closestUtil(Point Px[], Point Py[], int n)
{
	
	if (n <= 3)
		return bruteForce(Px, n);


	int mid = n/2;
	Point midPoint = Px[mid];


	Point Pyl[mid];
	Point Pyr[n-mid]; 
	int li = 0, ri = 0; 
	for (int i = 0; i < n; i++)
	{
	if ((Py[i].x < midPoint.x || (Py[i].x == midPoint.x && Py[i].y < midPoint.y)) && li<mid)
		Pyl[li++] = Py[i];
	else
		Pyr[ri++] = Py[i];
	}

	float dl = closestUtil(Px, Pyl, mid);
	float dr = closestUtil(Px + mid, Pyr, n-mid);


	float d = min(dl, dr);

	Point strip[n];
	int j = 0;
	for (int i = 0; i < n; i++)
		if (abs(Py[i].x - midPoint.x) < d)
			strip[j] = Py[i], j++;


	return stripClosest(strip, j, d);
}

float closest(Point P[], int n)
{
	Point Px[n];
	Point Py[n];
	for (int i = 0; i < n; i++)
	{
		Px[i] = P[i];
		Py[i] = P[i];
	}

	qsort(Px, n, sizeof(Point), compareX);
	qsort(Py, n, sizeof(Point), compareY);


	return closestUtil(Px, Py, n);
}

// Driver program to test above functions
int main()
{
	Point P[] = {{2, 3}, {12, 30}, {40, 50}, {5, 1}, {12, 10}, {3, 4}};
	int n = sizeof(P) / sizeof(P[0]);
	cout << "The smallest distance is " << closest(P, n);
	return 0;
}