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133 changes: 133 additions & 0 deletions C++/Closest pair of points - Geometrical Algorithms
Original file line number Diff line number Diff line change
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#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]; // y sorted points on left of vertical line
Point Pyr[n-mid]; // y sorted points on right of vertical line
int li = 0, ri = 0; // indexes of left and right subarrays
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);
}


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;
}