Tenemos una array de n puntos en el plano, y el problema es encontrar el par de puntos más cercano en la array. Este problema surge en varias aplicaciones. Por ejemplo, en el control del tráfico aéreo, es posible que desee controlar los aviones que se acercan demasiado, ya que esto puede indicar una posible colisión. Recuerda la siguiente fórmula para la distancia entre dos puntos p y q. 
La solución de fuerza bruta es O (n ^ 2), calcule la distancia entre cada par y devuelva la más pequeña. Podemos calcular la distancia más pequeña en tiempo O(nLogn) usando la estrategia Divide and Conquer. En esta publicación, se analiza un enfoque O(nx (Logn)^2). Discutiremos un enfoque O (nLogn) en una publicación separada.
Algoritmo
Los siguientes son los pasos detallados de un algoritmo O(n (Log)^2).
Entrada: Una array de n puntos P[]
Salida: La distancia más pequeña entre dos puntos en la array dada.
Como paso de preprocesamiento, la array de entrada se ordena según las coordenadas x.
1) Encuentre el punto medio en la array ordenada, podemos tomar P[n/2] como punto medio.
2) Divide la array dada en dos mitades. El primer subarreglo contiene puntos de P[0] a P[n/2]. El segundo subarreglo contiene puntos de P[n/2+1] a P[n-1].
3) Encuentre recursivamente las distancias más pequeñas en ambos subarreglos. Sean las distancias dl y dr. Encuentre el mínimo de dl y dr. Sea el mínimo d.
C++
// A divide and conquer program in C++
// to find the smallest distance from a
// given set of points.
#include <bits/stdc++.h>
using namespace std;
// A structure to represent a Point in 2D plane
class Point
{
public:
int x, y;
};
/* Following two functions are needed for library function qsort().
Refer: http://www.cplusplus.com/reference/clibrary/cstdlib/qsort/ */
// Needed to sort array of points
// according to X coordinate
int compareX(const void* a, const void* b)
{
Point *p1 = (Point *)a, *p2 = (Point *)b;
return (p1->x - p2->x);
}
// Needed to sort array of points according to Y coordinate
int compareY(const void* a, const void* b)
{
Point *p1 = (Point *)a, *p2 = (Point *)b;
return (p1->y - p2->y);
}
// A utility function to find the
// distance between two points
float dist(Point p1, Point p2)
{
return sqrt( (p1.x - p2.x)*(p1.x - p2.x) +
(p1.y - p2.y)*(p1.y - p2.y)
);
}
// A Brute Force method to return the
// smallest distance between two points
// in P[] of size n
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;
}
// A utility function to find
// minimum of two float values
float min(float x, float y)
{
return (x < y)? x : y;
}
// A utility function to find the
// distance between the closest points of
// strip of given size. All points in
// strip[] are sorted according to
// y coordinate. They all have an upper
// bound on minimum distance as d.
// Note that this method seems to be
// a O(n^2) method, but it's a O(n)
// method as the inner loop runs at most 6 times
float stripClosest(Point strip[], int size, float d)
{
float min = d; // Initialize the minimum distance as d
qsort(strip, size, sizeof(Point), compareY);
// Pick all points one by one and try the next points till the difference
// between y coordinates is smaller than d.
// This is a proven fact that this loop runs at most 6 times
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;
}
// A recursive function to find the
// smallest distance. The array P contains
// all points sorted according to x coordinate
float closestUtil(Point P[], int n)
{
// If there are 2 or 3 points, then use brute force
if (n <= 3)
return bruteForce(P, n);
// Find the middle point
int mid = n/2;
Point midPoint = P[mid];
// Consider the vertical line passing
// through the middle point calculate
// the smallest distance dl on left
// of middle point and dr on right side
float dl = closestUtil(P, mid);
float dr = closestUtil(P + mid, n - mid);
// Find the smaller of two distances
float d = min(dl, dr);
// Build an array strip[] that contains
// points close (closer than d)
// to the line passing through the middle point
Point strip[n];
int j = 0;
for (int i = 0; i < n; i++)
if (abs(P[i].x - midPoint.x) < d)
strip[j] = P[i], j++;
// Find the closest points in strip.
// Return the minimum of d and closest
// distance is strip[]
return min(d, stripClosest(strip, j, d) );
}
// The main function that finds the smallest distance
// This method mainly uses closestUtil()
float closest(Point P[], int n)
{
qsort(P, n, sizeof(Point), compareX);
// Use recursive function closestUtil()
// to find the smallest distance
return closestUtil(P, n);
}
// Driver code
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;
}
// This is code is contributed by rathbhupendra
C
// A divide and conquer program in C/C++ to find the smallest distance from a
// given set of points.
#include <stdio.h>
#include <float.h>
#include <stdlib.h>
#include <math.h>
// A structure to represent a Point in 2D plane
struct Point
{
int x, y;
};
/* Following two functions are needed for library function qsort().
Refer: http://www.cplusplus.com/reference/clibrary/cstdlib/qsort/ */
// Needed to sort array of points according to X coordinate
int compareX(const void* a, const void* b)
{
Point *p1 = (Point *)a, *p2 = (Point *)b;
return (p1->x - p2->x);
}
// Needed to sort array of points according to Y coordinate
int compareY(const void* a, const void* b)
{
Point *p1 = (Point *)a, *p2 = (Point *)b;
return (p1->y - p2->y);
}
// A utility function to find the distance between two points
float dist(Point p1, Point p2)
{
return sqrt( (p1.x - p2.x)*(p1.x - p2.x) +
(p1.y - p2.y)*(p1.y - p2.y)
);
}
// A Brute Force method to return the smallest distance between two points
// in P[] of size n
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;
}
// A utility function to find a minimum of two float values
float min(float x, float y)
{
return (x < y)? x : y;
}
// A utility function to find the distance between the closest points of
// strip of a given size. All points in strip[] are sorted according to
// y coordinate. They all have an upper bound on minimum distance as d.
// Note that this method seems to be a O(n^2) method, but it's a O(n)
// method as the inner loop runs at most 6 times
float stripClosest(Point strip[], int size, float d)
{
float min = d; // Initialize the minimum distance as d
qsort(strip, size, sizeof(Point), compareY);
// Pick all points one by one and try the next points till the difference
// between y coordinates is smaller than d.
// This is a proven fact that this loop runs at most 6 times
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;
}
// A recursive function to find the smallest distance. The array P contains
// all points sorted according to x coordinate
float closestUtil(Point P[], int n)
{
// If there are 2 or 3 points, then use brute force
if (n <= 3)
return bruteForce(P, n);
// Find the middle point
int mid = n/2;
Point midPoint = P[mid];
// Consider the vertical line passing through the middle point
// calculate the smallest distance dl on left of middle point and
// dr on right side
float dl = closestUtil(P, mid);
float dr = closestUtil(P + mid, n-mid);
// Find the smaller of two distances
float d = min(dl, dr);
// Build an array strip[] that contains points close (closer than d)
// to the line passing through the middle point
Point strip[n];
int j = 0;
for (int i = 0; i < n; i++)
if (abs(P[i].x - midPoint.x) < d)
strip[j] = P[i], j++;
// Find the closest points in strip. Return the minimum of d and closest
// distance is strip[]
return min(d, stripClosest(strip, j, d) );
}
// The main function that finds the smallest distance
// This method mainly uses closestUtil()
float closest(Point P[], int n)
{
qsort(P, n, sizeof(Point), compareX);
// Use recursive function closestUtil() to find the smallest distance
return closestUtil(P, 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]);
printf("The smallest distance is %f ", closest(P, n));
return 0;
}
Java
import java.text.DecimalFormat;
import java.util.Arrays;
import java.util.Comparator;
// A divide and conquer program in Java
// to find the smallest distance from a
// given set of points.
// A structure to represent a Point in 2D plane
class Point {
public int x;
public int y;
Point(int x, int y) {
this.x = x;
this.y = y;
}
// A utility function to find the
// distance between two points
public static float dist(Point p1, Point p2) {
return (float) Math.sqrt((p1.x - p2.x) * (p1.x - p2.x) +
(p1.y - p2.y) * (p1.y - p2.y)
);
}
// A Brute Force method to return the
// smallest distance between two points
// in P[] of size n
public static float bruteForce(Point[] P, int n) {
float min = Float.MAX_VALUE;
float currMin = 0;
for (int i = 0; i < n; ++i) {
for (int j = i + 1; j < n; ++j) {
currMin = dist(P[i], P[j]);
if (currMin < min) {
min = currMin;
}
}
}
return min;
}
// A utility function to find the
// distance between the closest points of
// strip of given size. All points in
// strip[] are sorted according to
// y coordinate. They all have an upper
// bound on minimum distance as d.
// Note that this method seems to be
// a O(n^2) method, but it's a O(n)
// method as the inner loop runs at most 6 times
public static float stripClosest(Point[] strip, int size, float d) {
float min = d; // Initialize the minimum distance as d
Arrays.sort(strip, 0, size, new PointYComparator());
// Pick all points one by one and try the next points till the difference
// between y coordinates is smaller than d.
// This is a proven fact that this loop runs at most 6 times
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;
}
// A recursive function to find the
// smallest distance. The array P contains
// all points sorted according to x coordinate
public static float closestUtil(Point[] P,
int startIndex,
int endIndex)
{
// If there are 2 or 3 points, then use brute force
if ((endIndex - startIndex) <= 3) {
return bruteForce(P, endIndex);
}
// Find the middle point
int mid = startIndex + (endIndex - startIndex) / 2;
Point midPoint = P[mid];
// Consider the vertical line passing
// through the middle point calculate
// the smallest distance dl on left
// of middle point and dr on right side
float dl = closestUtil(P, startIndex, mid);
float dr = closestUtil(P, mid, endIndex);
// Find the smaller of two distances
float d = Math.min(dl, dr);
// Build an array strip[] that contains
// points close (closer than d)
// to the line passing through the middle point
Point[] strip = new Point[endIndex];
int j = 0;
for (int i = 0; i < endIndex; i++) {
if (Math.abs(P[i].x - midPoint.x) < d) {
strip[j] = P[i];
j++;
}
}
// Find the closest points in strip.
// Return the minimum of d and closest
// distance is strip[]
return Math.min(d, stripClosest(strip, j, d));
}
// The main function that finds the smallest distance
// This method mainly uses closestUtil()
public static float closest(Point[] P, int n) {
Arrays.sort(P, 0, n, new PointXComparator());
// Use recursive function closestUtil()
// to find the smallest distance
return closestUtil(P, 0, n);
}
}
// A structure to represent a Point in 2D plane
class PointXComparator implements Comparator<Point> {
// Needed to sort array of points
// according to X coordinate
@Override
public int compare(Point pointA, Point pointB) {
return Integer.compare(pointA.x, pointB.x);
}
}
class PointYComparator implements Comparator<Point> {
// Needed to sort array of points
// according to Y coordinate
@Override
public int compare(Point pointA, Point pointB) {
return Integer.compare(pointA.y, pointB.y);
}
}
public class ClosestPoint {
// Driver code
public static void main(String[] args) {
Point[] P = new Point[]{
new Point(2, 3),
new Point(12, 30),
new Point(40, 50),
new Point(5, 1),
new Point(12, 10),
new Point(3, 4)
};
DecimalFormat df = new DecimalFormat("#.######");
System.out.println("The smallest distance is " +
df.format(Point.closest(P, P.length)));
}
}
// This code is contributed by sanjay sharma 1
Python3
# A divide and conquer program in Python3
# to find the smallest distance from a
# given set of points.
import math
import copy
# A class to represent a Point in 2D plane
class Point():
def __init__(self, x, y):
self.x = x
self.y = y
# A utility function to find the
# distance between two points
def dist(p1, p2):
return math.sqrt((p1.x - p2.x) *
(p1.x - p2.x) +
(p1.y - p2.y) *
(p1.y - p2.y))
# A Brute Force method to return the
# smallest distance between two points
# in P[] of size n
def bruteForce(P, n):
min_val = float('inf')
for i in range(n):
for j in range(i + 1, n):
if dist(P[i], P[j]) < min_val:
min_val = dist(P[i], P[j])
return min_val
# A utility function to find the
# distance between the closest points of
# strip of given size. All points in
# strip[] are sorted according to
# y coordinate. They all have an upper
# bound on minimum distance as d.
# Note that this method seems to be
# a O(n^2) method, but it's a O(n)
# method as the inner loop runs at most 6 times
def stripClosest(strip, size, d):
# Initialize the minimum distance as d
min_val = d
# Pick all points one by one and
# try the next points till the difference
# between y coordinates is smaller than d.
# This is a proven fact that this loop
# runs at most 6 times
for i in range(size):
j = i + 1
while j < size and (strip[j].y -
strip[i].y) < min_val:
min_val = dist(strip[i], strip[j])
j += 1
return min_val
# A recursive function to find the
# smallest distance. The array P contains
# all points sorted according to x coordinate
def closestUtil(P, Q, n):
# If there are 2 or 3 points,
# then use brute force
if n <= 3:
return bruteForce(P, n)
# Find the middle point
mid = n // 2
midPoint = P[mid]
#keep a copy of left and right branch
Pl = P[:mid]
Pr = P[mid:]
# Consider the vertical line passing
# through the middle point calculate
# the smallest distance dl on left
# of middle point and dr on right side
dl = closestUtil(Pl, Q, mid)
dr = closestUtil(Pr, Q, n - mid)
# Find the smaller of two distances
d = min(dl, dr)
# Build an array strip[] that contains
# points close (closer than d)
# to the line passing through the middle point
stripP = []
stripQ = []
lr = Pl + Pr
for i in range(n):
if abs(lr[i].x - midPoint.x) < d:
stripP.append(lr[i])
if abs(Q[i].x - midPoint.x) < d:
stripQ.append(Q[i])
stripP.sort(key = lambda point: point.y) #<-- REQUIRED
min_a = min(d, stripClosest(stripP, len(stripP), d))
min_b = min(d, stripClosest(stripQ, len(stripQ), d))
# Find the self.closest points in strip.
# Return the minimum of d and self.closest
# distance is strip[]
return min(min_a,min_b)
# The main function that finds
# the smallest distance.
# This method mainly uses closestUtil()
def closest(P, n):
P.sort(key = lambda point: point.x)
Q = copy.deepcopy(P)
Q.sort(key = lambda point: point.y)
# Use recursive function closestUtil()
# to find the smallest distance
return closestUtil(P, Q, n)
# Driver code
P = [Point(2, 3), Point(12, 30),
Point(40, 50), Point(5, 1),
Point(12, 10), Point(3, 4)]
n = len(P)
print("The smallest distance is",
closest(P, n))
# This code is contributed
# by Prateek Gupta (@prateekgupta10)
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA