使用分而治之算法的最接近点对

我们在平面中获得了n个点的数组，问题是找出数组中最接近的点对。在许多应用中都会出现此问题。例如，在空中交通管制中，您可能想要监视过于靠近的平面，因为这可能表示可能发生碰撞。回忆以下关于两个点p和q之间距离的公式。

$\left \|pq \right \| = \sqrt{(p_{x}-q_{x})^{2}+ (p_{y}-q_{y})^{2}}$

蛮力解为O(n ^ 2)，计算每对之间的距离并返回最小值。我们可以使用分而治之策略计算O(nLogn)时间中的最小距离。在这篇文章中，讨论了O(nx(Logn)^ 2)方法。我们将在另一篇文章中讨论O(nLogn)方法。

算法
以下是O(n(Logn)^ 2)算法的详细步骤。
输入： n个点组成的数组P []
输出：给定数组中两点之间的最小距离。

作为预处理步骤，根据x坐标对输入数组进行排序。

1)在排序数组中找到中间点，我们可以将P [n / 2]作为中间点。

2)将给定数组分成两半。第一个子数组包含从P [0]到P [n / 2]的点。第二个子数组包含从P [n / 2 + 1]到P [n-1]的点。

3)递归地找到两个子数组中的最小距离。设距离为dl和dr。找出dl和dr的最小值。设最小值为d。

4)从以上3个步骤中，我们得到最小距离的上限d。现在我们需要考虑成对，使得成对的一个点从左半边开始，另一个点从右半边开始。考虑穿过P [n / 2]的垂直线，并找到x坐标比d靠近中间垂直线的所有点。建立所有这些点的数组strip []。

5)根据y坐标对数组strip []进行排序。此步骤为O(nLogn)。可以通过递归排序和合并将其优化为O(n)。

6)在strip []中找到最小的距离。这很棘手。从第一眼看，这似乎是一个O(n ^ 2)步骤，但实际上是O(n)。可以从几何上证明，对于条带中的每个点，我们只需要检查其后最多7个点(请注意，条带是根据Y坐标排序的)。请参阅此以进行更多分析。

7)最后返回上一步中计算出的d和距离的最小值(第6步)

执行
以下是上述算法的实现。

C++

// A divide and conquer program in C++ 
// to find the smallest distance from a 
// given set of points. 
  
#include 
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 beween the closest points of 
// strip of given size. All points in 
// strip[] are sorted accordint 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 
#include 
#include 
#include 
  
// 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;
}

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 beween the closest points of 
# strip of given size. All points in 
# strip[] are sorted accordint 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]
  
    # 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(P[:mid], Q, mid)
    dr = closestUtil(P[mid:], 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 
    strip = [] 
    for i in range(n): 
        if abs(Q[i].x - midPoint.x) < d: 
            strip.append(Q[i])
  
    # Find the closest points in strip. 
    # Return the minimum of d and closest 
    # distance is strip[] 
    return min(d, stripClosest(strip, len(strip), d))
  
# 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)

输出：

The smallest distance is 1.414214

时间复杂度让上述算法的时间复杂度为T(n)。让我们假设我们使用O(nLogn)排序算法。上面的算法将所有点分为两组，然后递归调用两组。划分后，它以O(n)时间找到条带，以O(nLogn)时间排序条带，最后找到O(n)时间中条带的最接近点。所以T(n)可以表示如下
T(n)= 2T(n / 2)+ O(n)+ O(nLogn)+ O(n)
T(n)= 2T(n / 2)+ O(nLogn)
T(n)= T(nx Logn x Logn)

笔记
1)通过优化上述算法的步骤5，可以将时间复杂度提高到O(nLogn)。我们很快将在另一篇文章中讨论优化的解决方案。
2)代码找到最小的距离。可以轻松修改它以找到距离最小的点。
3)代码使用快速排序，在最坏的情况下可以为O(n ^ 2)。要使上限为O(n(Logn)^ 2)，可以使用O(nLogn)排序算法，例如合并排序或堆排序

参考：
http://www.cs.umd.edu/class/fall2013/cmsc451/Lects/lect10.pdf
http://www.youtube.com/watch?v=vS4Zn1a9KUc
http://www.youtube.com/watch?v=T3T7T8Ym20M
http://en.wikipedia.org/wiki/Closest_pair_of_points_problem