凸壳船库

时间:2013-02-03 09:45:23

标签: c# convex-hull

我是C#的新手,并且很难计算凸包。 C#是否有某种数学库?如果没有,那么我想我只需要实现自己的。

4 个答案:

答案 0 :(得分:10)

MIConvexHull - https://designengrlab.github.io/MIConvexHull/ - 是C#中的高性能凸包实现,也支持更高维的凸包。 LGPL许可证。

答案 1 :(得分:3)

这是我使用Monotone Chain算法编写的二维凸包算法,a.k.a。安德鲁算法。

public static IListSource<Point> ComputeConvexHull(List<Point> points, bool sortInPlace = false)
{
    if (!sortInPlace)
        points = new List<Point>(points);
    points.Sort((a, b) => 
        a.X == b.X ? a.Y.CompareTo(b.Y) : a.X.CompareTo(b.X));

    // Importantly, DList provides O(1) insertion at beginning and end
    DList<Point> hull = new DList<Point>();
    int L = 0, U = 0; // size of lower and upper hulls

    // Builds a hull such that the output polygon starts at the leftmost point.
    for (int i = points.Count - 1; i >= 0 ; i--)
    {
        Point p = points[i], p1;

        // build lower hull (at end of output list)
        while (L >= 2 && (p1 = hull.Last).Sub(hull[hull.Count-2]).Cross(p.Sub(p1)) >= 0) {
            hull.RemoveAt(hull.Count-1);
            L--;
        }
        hull.PushLast(p);
        L++;

        // build upper hull (at beginning of output list)
        while (U >= 2 && (p1 = hull.First).Sub(hull[1]).Cross(p.Sub(p1)) <= 0)
        {
            hull.RemoveAt(0);
            U--;
        }
        if (U != 0) // when U=0, share the point added above
            hull.PushFirst(p);
        U++;
        Debug.Assert(U + L == hull.Count + 1);
    }
    hull.RemoveAt(hull.Count - 1);
    return hull;
}

它依赖于假设存在的一些内容,有关详细信息,请参阅我的blog post

答案 2 :(得分:2)

下面是对Qwertie的答案中使用的相同Java源代码的C#的音译,但是没有依赖于具有双字段的Point类之外的非标准类。

class ConvexHull
{
    public static double cross(Point O, Point A, Point B)
    {
        return (A.X - O.X) * (B.Y - O.Y) - (A.Y - O.Y) * (B.X - O.X);
    }

    public static List<Point> GetConvexHull(List<Point> points)
    {
        if (points == null)
            return null;

        if (points.Count() <= 1)
            return points;

        int n = points.Count(), k = 0;
        List<Point> H = new List<Point>(new Point[2 * n]);

        points.Sort((a, b) =>
             a.X == b.X ? a.Y.CompareTo(b.Y) : a.X.CompareTo(b.X));

        // Build lower hull
        for (int i = 0; i < n; ++i)
        {
            while (k >= 2 && cross(H[k - 2], H[k - 1], points[i]) <= 0)
                k--;
            H[k++] = points[i];
        }

        // Build upper hull
        for (int i = n - 2, t = k + 1; i >= 0; i--)
        {
            while (k >= t && cross(H[k - 2], H[k - 1], points[i]) <= 0)
                k--;
            H[k++] = points[i];
        }

        return H.Take(k - 1).ToList();
    }
}

答案 3 :(得分:2)

我将许多Convex Hull算法/实现与所提供的所有代码进行了比较。一切都包含在CodeProject文章中。

算法比较:

  • 单调链
  • MiConvexHull(Delaunay三角测量和Voronoi网格)
  • 格雷厄姆扫描
  • Ouellet(我的)

文章: