我正在尝试实现此算法的更简单版本,但它比二次算法效果更好。我的想法基本上是只按 x 坐标对点进行排序,然后尝试从那里解决。一旦我按 x 坐标对我的点数组进行排序,我想遍历数组并基本上跳过距离大于我拍摄的前两个点的点。
比如我的currentminDist = x;
如果我正在查看的两对点的距离 > x(仅通过其 x 坐标 dist),我将忽略该点并在数组中移过它。
我已经记下了这个想法,但我对如何实际实现它(尤其是条件部分)有点困惑。我有一个函数可以根据 x 坐标返回两点之间的距离。
我对如何实际编写我的循环条件感到困惑,因为如果距离碰巧太远我想忽略一个点并且仍然填写我的数组,该数组将包含每个 i (i是我正在查看的当前点)。
如有任何提示或指示,我们将不胜感激。我对编码算法不是很了解,所以这很令人沮丧。
这是我的部分代码:
for (i = 0; i < numofmypoints; i++)
{
for (int j = i + 1; (j < numpofmypoints) && ((inputpoints[j].x - inputpoints[i].x) < currbest); j++ )
{
currdist = Auxilary.distbyX(inputpoints[i],inputpoints[j]);
if (currdist < bestdist)
{
closest[i] = j;
bestdist = currdist;
}
}
}
distbyX 是我的函数,它只返回两点之间的距离。
谢谢!
最佳答案
使用 KD 树的快速算法
该算法创建一个 kd 树,然后为每个点找到最接近的对。创建 kd-tree 是 O(n log2n),找到一个点的最近邻居是 O(logn)。信用必须转到Wikipedia ,其中一篇文章解释了如何创建 kd 树以及如何使用它们找到最近的邻居。
import java.util.*;
public class Program
{
public static void main(String[] args)
{
List<Point> points = generatePoints();
Point[] closest = new Point[points.size()];
KDTree tree = new KDTree(points, 0); // WILL MODIFY 'points'
for (int i = 0; i < points.size(); i++)
{
closest[i] = tree.findClosest(points.get(i));
}
for (int i = 0; i < points.size(); i++)
{
System.out.println(points.get(i) + " is closest to " + closest[i]);
}
}
private static List<Point> generatePoints()
{
ArrayList<Point> points = new ArrayList<Point>();
Random r = new Random();
for (int i = 0; i < 1000; i++)
{
points.add(new Point(r.nextInt() % 1000, r.nextInt() % 1000));
}
return points;
}
}
class Point
{
public static final Point INFINITY
= new Point(Double.POSITIVE_INFINITY,
Double.POSITIVE_INFINITY);
public double[] coord; // coord[0] = x, coord[1] = y
public Point(double x, double y)
{
coord = new double[] { x, y };
}
public double getX() { return coord[0]; }
public double getY() { return coord[1]; }
public double distance(Point p)
{
double dX = getX() - p.getX();
double dY = getY() - p.getY();
return Math.sqrt(dX * dX + dY * dY);
}
public boolean equals(Point p)
{
return (getX() == p.getX()) && (getY() == p.getY());
}
public String toString()
{
return "(" + getX() + ", " + getY() + ")";
}
public static class PointComp implements Comparator<Point>
{
int d; // the dimension to compare in (0 => x, 1 => y)
public PointComp(int dimension)
{
d = dimension;
}
public int compare(Point a, Point b)
{
return (int) (a.coord[d] - b.coord[d]);
}
}
}
class KDTree
{
// 2D k-d tree
private KDTree childA, childB;
private Point point; // defines the boundary
private int d; // dimension: 0 => left/right split, 1 => up/down split
public KDTree(List<Point> points, int depth)
{
childA = null;
childB = null;
d = depth % 2;
// find median by sorting in dimension 'd' (either x or y)
Comparator<Point> comp = new Point.PointComp(d);
Collections.sort(points, comp);
int median = (points.size() - 1) / 2;
point = points.get(median);
// Create childA and childB recursively.
// WARNING: subList() does not create a true copy,
// so the original will get modified.
if (median > 0)
{
childA = new KDTree(
points.subList(0, median),
depth + 1);
}
if (median + 1 < points.size())
{
childB = new KDTree(
points.subList(median + 1, points.size()),
depth + 1);
}
}
public Point findClosest(Point target)
{
Point closest = point.equals(target) ? Point.INFINITY : point;
double bestDist = closest.distance(target);
double spacing = target.coord[d] - point.coord[d];
KDTree rightSide = (spacing < 0) ? childA : childB;
KDTree otherSide = (spacing < 0) ? childB : childA;
/*
* The 'rightSide' is the side on which 'target' lies
* and the 'otherSide' is the other one. It is possible
* that 'otherSide' will not have to be searched.
*/
if (rightSide != null)
{
Point candidate = rightSide.findClosest(target);
if (candidate.distance(target) < bestDist)
{
closest = candidate;
bestDist = closest.distance(target);
}
}
if (otherSide != null && (Math.abs(spacing) < bestDist))
{
Point candidate = otherSide.findClosest(target);
if (candidate.distance(target) < bestDist)
{
closest = candidate;
bestDist = closest.distance(target);
}
}
return closest;
}
}
修复问题中的代码
如果您真的不担心复杂性,那么您的代码的唯一问题就是您向前看而不是向后看。只需复制内部循环并使 j
从 (i - 1)
到 0
:
Point[] points = sort(input());
int[] closest = new int[points.length];
for (int i = 0; i < points.length; i++)
{
double bestdist = Double.POSITIVE_INFINITY;
for (int j = i + 1; (j < points.length) && ((points[j].x - points[i].x) < bestdist); j++ )
{
double currdist = dist(points[i], points[j]);
if (currdist < bestdist)
{
closest[i] = j;
bestdist = currdist;
}
}
for (int j = i - 1; (j >= 0) && ((points[i].x - points[j].x) < bestdist); j-- )
{
double currdist = dist(points[i], points[j]);
if (currdist < bestdist)
{
closest[i] = j;
bestdist = currdist;
}
}
}
关于java - 最近点对算法,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/9405660/