我编辑了CircularNotchedRectangle.

I edited CircularNotchedRectangle.

使用CircularOuterNotchedRectangle代替CircularNotchedRectangle.

PS.我添加了extraOffset参数以获得额外的厚度.但是它不能完全正确地工作.但我只是想向您展示如何处理.

PS. I added extraOffset param for extra thick. but it's not working exactly correct. But I just wanted to show you how to approach.

class CircularOuterNotchedRectangle extends NotchedShape {
  /// Creates a [CircularOuterNotchedRectangle].
  ///
  /// The same object can be used to create multiple shapes.
  const CircularOuterNotchedRectangle({this.extraOffset = 10.0});

  final double extraOffset;

  /// Creates a [Path] that describes a rectangle with a smooth circular notch.
  ///
  /// `host` is the bounding box for the returned shape. Conceptually this is
  /// the rectangle to which the notch will be applied.
  ///
  /// `guest` is the bounding box of a circle that the notch accommodates. All
  /// points in the circle bounded by `guest` will be outside of the returned
  /// path.
  ///
  /// The notch is curve that smoothly connects the host's top edge and
  /// the guest circle.
  // TODO(amirh): add an example diagram here.
  @override
  Path getOuterPath(Rect host, Rect guest) {
    if (guest == null || !host.overlaps(guest)) return Path()..addRect(host);

    // The guest's shape is a circle bounded by the guest rectangle.
    // So the guest's radius is half the guest width.
    final double notchRadius = guest.width / 2.0;

    // We build a path for the notch from 3 segments:
    // Segment A - a Bezier curve from the host's top edge to segment B.
    // Segment B - an arc with radius notchRadius.
    // Segment C - a Bezier curve from segment B back to the host's top edge.
    //
    // A detailed explanation and the derivation of the formulas below is

    const double s1 = 15.0;
    const double s2 = 1.0;

    final double r = notchRadius + extraOffset/2;
    final double a = -1.0 * r - s2;
    final double b = host.top + guest.center.dy;

    final double n2 = math.sqrt(b * b * r * r * (a * a + b * b - r * r));
    final double p2xA = ((a * r * r) - n2) / (a * a + b * b);
    final double p2xB = ((a * r * r) + n2) / (a * a + b * b);
    final double p2yA = math.sqrt(r * r - p2xA * p2xA) - extraOffset/2;
    final double p2yB = math.sqrt(r * r - p2xB * p2xB) - extraOffset/2;

    final List<Offset> p = List<Offset>(6);

    // p0, p1, and p2 are the control points for segment A.
    p[0] = Offset(a - s1, b);
    p[1] = Offset(a, b);
    p[2] = p2yA > p2yB ? Offset(p2xA, -p2yA) : Offset(p2xB, p2yB);

    // p3, p4, and p5 are the control points for segment B, which is a mirror
    // of segment A around the y axis.
    p[3] = Offset(-1.0 * p[2].dx, -p[2].dy);
    p[4] = Offset(-1.0 * p[1].dx, p[1].dy);
    p[5] = Offset(-1.0 * p[0].dx, p[0].dy);

    // translate all points back to the absolute coordinate system.
    for (int i = 0; i < p.length; i += 1) p[i] += guest.center;

    return Path()
      ..moveTo(host.left, -host.top)
      ..lineTo(p[0].dx, p[0].dy)
      ..quadraticBezierTo(p[1].dx, p[1].dy, p[2].dx, -p[2].dy)
      ..arcToPoint(
        p[3],
        radius: Radius.circular(notchRadius),
        clockwise: true,
      )
      ..quadraticBezierTo(p[4].dx, p[4].dy, p[5].dx, p[5].dy)
      ..lineTo(host.right, host.top)
      ..lineTo(host.right, host.bottom)
      ..lineTo(host.left, host.bottom)
      ..close();
  }
}