Graphics.DrawCurve的算法
2013-11-13 17:50
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Graphics.DrawCurve的算法为Cardinal Spline，中文可能叫做'基数样条'。
它计算并不复杂，如下代码：
C# code?
| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 | public static class Spline {[System.Diagnostics.DebuggerDisplay( "({X},{Y})")] public partial struct Vec2 {public float X, Y; public Vec2( float x, float y) {this .X = x; this .Y = y;} public static implicit operator PointF(Vec2v) {return new PointF(v.X, v.Y); } public static implicit operator Vec2(PointF p) {return new Vec2(p.X, p.Y); } public static Vec2 operator +(Vec2v1, Vec2v2) { return new Vec2(v1.X + v2.X, v1.Y + v2.Y); } public static Vec2 operator -(Vec2v1, Vec2v2) { return new Vec2(v1.X - v2.X, v1.Y - v2.Y); } public static Vec2 operator *(Vec2v, float f) {return new Vec2(v.X * f, v.Y * f); } public static Vec2 operator /(Vec2v, float f) {return new Vec2(v.X / f, v.Y / f); } } /// <summary> /// '贝塞尔'内插。结果不包括头尾点 /// </summary> public static PointF[] InterpolateBezier(PointF p0, PointF p1, PointF p2, PointF p3, int samples) {PointF[] result = new PointF[samples]; for ( int i = 0; i < samples; i++) {float t = (i + 1) / (samples + 1.0f); result[i] = (Vec2)p0 * (1 - t) * (1 - t) * (1 - t) + (Vec2)p1 * (3 * (1 - t) * (1 - t) * t) + (Vec2)p2 * (3 * (1 - t) * t * t) + (Vec2)p3 * (t * t * t); } return result; } public static PointF[] InterpolateCardinalSpline(PointF p0, PointF p1, PointF p2, PointF p3, int samples) {const float tension = 0.5f; Vec2u = ((Vec2)p2 - (Vec2)p0) * (tension / 3) + p1; Vec2v = ((Vec2)p1 - (Vec2)p3) * (tension / 3) + p2; return InterpolateBezier(p1, u, v, p2, samples); } /// <summary> /// '基数样条'内插法。 points为通过点，samplesInSegment为两个样本点之间的内插数量。 /// </summary> public static PointF[] CardinalSpline(PointF[] points, int samplesInSegment) {List<PointF> result = new List<PointF>(); for ( int i = 0; i < points.Length - 1; i++) {result.Add(points[i]); result.AddRange( InterpolateCardinalSpline( points[Math.Max(i-1, 0)], points[i], points[i+1], points[Math.Min(i+2, points.Length-1)], samplesInSegment )); } result.Add(points[points.Length - 1]); return result.ToArray(); } } |
C# code?
| 1 2 3 4 5 6 7 8 9 10 11 12 13 | public partial class Form1 : Form {protected override void OnPaint(PaintEventArgs e) {PointF[] ps = {new PointF(50,50), new PointF(100, 80), new PointF(120, 10), new PointF(200,100)}; // 系统的Graphics.DrawCurve，桃色 e.Graphics.DrawCurve( new Pen(Brushes.PeachPuff, 5), ps); // 自己取样，蓝色 e.Graphics.DrawLines(Pens.Blue, Spline.CardinalSpline(ps, 10)); } } |
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