最近點對問題[CPP]C# N*LogN複雜度解法

using System;
using System.Collections.Generic;

public class UTest
{
    static void Main()
    {
        Vec2[] points = new Vec2[ 20000 ];
        Random random = new Random(123456);
        for (int i = 0; i < points.Length; i++)
        {
            points[i] = new Vec2(random.Next(200000) / 100.0f, random.Next(200000) / 100.0f);

        }
        int startTick, endTick;

        startTick = System.Environment.TickCount;
        float divideconquer = (float)Math.Sqrt((double)MinDistanceSquared(points));         // 140ms
        endTick = System.Environment.TickCount;
        Console.WriteLine("Divide and conquer finishes in {0,6}ms, result={1}", endTick - startTick, divideconquer);

        startTick = System.Environment.TickCount;
        float bruteforce = (float)Math.Sqrt((double)BruteForceMinDistanceSquared(points));  // 13200ms
        endTick = System.Environment.TickCount;
        Console.WriteLine("Brute force method finishes in {0,6}ms, result={1}", endTick - startTick, bruteforce);
    }

    static float MinDistanceSquared(Vec2[] points)
    {
        // 如果點的數目少於一定數量，直接窮舉最短距離
        if (points.Length < 6) return BruteForceMinDistanceSquared(points);

        // 將點按照x軸排序。並分成左邊部分，和右邊部分。
        Array.Sort<Vec2>(points, Vec2.CompareByX);
        int middleIdx = points.Length / 2;
        Vec2[] leftPoints = new Vec2[middleIdx];
        Vec2[] rightPoints = new Vec2[points.Length - middleIdx];
        for (int i = 0; i < leftPoints.Length; i++)
        {
            leftPoints[i] = new Vec2(points[i].y, points[i].x);
        }
        for (int i = 0; i < rightPoints.Length; i++)
        {
            rightPoints[i] = new Vec2(points[i + middleIdx].y, points[i + middleIdx].x);
        }

        // 分而治之: 算出左邊部分的最短距離和右邊部分的最短距離。
        float l = MinDistanceSquared(leftPoints);
        float r = MinDistanceSquared(rightPoints);
        float minDistance = Math.Min(l, r);

        // 靠近中間線的點，有可能跨越中間線出現更短的距離。
        // 但該中間區域有很大侷限。假定最短距離是s，中線的x爲X，則該區域限定在X-S ~ X+S之間
        float middle = points[middleIdx].x;
        List<Vec2> middleStrip = new List<Vec2>();
        foreach (Vec2 p in points)
        {
            if ((p.x - middle) * (p.x - middle) < minDistance) middleStrip.Add(new Vec2(p.y, p.x));
        }

        // 不同於wiki的算法描述，這裏再次採用分而治之的原則（代碼簡潔很多），算出中間地帶的最短距離
        float m = MinDistanceSquared(middleStrip.ToArray());

        return Math.Min(minDistance, m);
    }

    static float BruteForceMinDistanceSquared(Vec2[] points)
    {
        float minDistance = float.MaxValue;
        // 窮舉法，O(n*n)
        for (int i = 0; i < points.Length; i++)
        {
            for (int j = i + 1; j < points.Length; j++)
            {
                float distance = (points[i] - points[j]).LengthSquare();
                if (distance < minDistance) minDistance = distance;
            }
        }
        return minDistance;
    }
}

struct Vec2
{
    public float x, y;
    public Vec2(float x, float y)
    {
        this.x = x; this.y = y;
    }
    public float LengthSquare()
    {
        return x * x + y * y;
    }
    public static int CompareByX(Vec2 v1, Vec2 v2)
    {
        return v1.x - v2.x > 0 ? 1 : (v1.x - v2.x == 0 ? 0 : -1);
    }
    public static Vec2 operator -(Vec2 v1, Vec2 v2)
    {
        return new Vec2(v1.x - v2.x, v1.y - v2.y);
    }
}