使用一組點畫出平滑的曲線
阿新 • • 發佈:2018-11-09
今天在專案中需要人臉上的點來勾勒出人臉的輪廓,我的想法是將要畫的點存入一個數組,使用了UIBezierPath來連線每一個點。但是這樣畫出來的圖是折線,顯得過於生硬。查了若干資料,後來終於在stackoverflow上找到了一個很好的解決辦法。
這個做法的原理其實是在每兩個點之間加入一些點,來使得兩個點可以平滑的過度。這其實就是Centripetal Catmull–Rom spline的思想。
看效果:
於是便照葫蘆畫瓢,寫出了我的程式碼
#define POINT(_INDEX_) [(NSValue *)[points objectAtIndex:_INDEX_] CGPointValue]
- (void)smoothedPathWithPoints:(NSArray *) pointsArray andGranularity:(NSInteger)granularity {
NSMutableArray *points = [pointsArray mutableCopy];
CGContextRef context = UIGraphicsGetCurrentContext();
CGContextSetAllowsAntialiasing(context, YES);
CGContextSetStrokeColorWithColor(context, [UIColor greenColor].CGColor);
CGContextSetLineWidth(context, 0.6);
UIBezierPath *smoothedPath = [UIBezierPath bezierPath];
// Add control points to make the math make sense
[points insertObject:[points objectAtIndex:0] atIndex:0];
[points addObject:[points lastObject]];
[smoothedPath moveToPoint:POINT(0)];
for (NSUInteger index = 1; index < points.count - 2; index++) {
CGPoint p0 = POINT(index - 1 );
CGPoint p1 = POINT(index);
CGPoint p2 = POINT(index + 1);
CGPoint p3 = POINT(index + 2);
// now add n points starting at p1 + dx/dy up until p2 using Catmull-Rom splines
for (int i = 1; i < granularity; i++) {
float t = (float) i * (1.0f / (float) granularity);
float tt = t * t;
float ttt = tt * t;
CGPoint pi; // intermediate point
pi.x = 0.5 * (2*p1.x+(p2.x-p0.x)*t + (2*p0.x-5*p1.x+4*p2.x-p3.x)*tt + (3*p1.x-p0.x-3*p2.x+p3.x)*ttt);
pi.y = 0.5 * (2*p1.y+(p2.y-p0.y)*t + (2*p0.y-5*p1.y+4*p2.y-p3.y)*tt + (3*p1.y-p0.y-3*p2.y+p3.y)*ttt);
[smoothedPath addLineToPoint:pi];
}
// Now add p2
[smoothedPath addLineToPoint:p2];
}
// finish by adding the last point
[smoothedPath addLineToPoint:POINT(points.count - 1)];
CGContextAddPath(context, smoothedPath.CGPath);
CGContextDrawPath(context, kCGPathStroke);
}