Exemplo n.º 1
0
 public boolean envelopesOverlap(LineSegment seg1, LineSegment seg2) {
   if (seg1.maxX() <= seg2.minX()) return false;
   if (seg2.maxX() <= seg1.minX()) return false;
   if (seg1.maxY() <= seg2.minY()) return false;
   if (seg2.maxY() <= seg1.minY()) return false;
   return true;
 }
Exemplo n.º 2
0
    public int XcompareTo(Object obj) {
      DepthSegment other = (DepthSegment) obj;

      // if segments are collinear and vertical compare endpoints
      if (isVertical() && other.isVertical() && upwardSeg.p0.x == other.upwardSeg.p0.x)
        return compareX(this.upwardSeg, other.upwardSeg);
      // check if segments are trivially ordered along X
      if (upwardSeg.maxX() <= other.upwardSeg.minX()) return -1;
      if (upwardSeg.minX() >= other.upwardSeg.maxX()) return 1;
      /**
       * try and compute a determinate orientation for the segments. Test returns 1 if other is left
       * of this (i.e. this > other)
       */
      int orientIndex = upwardSeg.orientationIndex(other.upwardSeg);
      // if orientation is determinate, return it
      if (orientIndex != 0) return orientIndex;

      /**
       * If comparison between this and other is indeterminate, try the opposite call order.
       * orientationIndex value is 1 if this is left of other, so have to flip sign to get proper
       * comparison value of -1 if this is leftmost
       */
      if (orientIndex == 0) orientIndex = -1 * other.upwardSeg.orientationIndex(upwardSeg);

      // if orientation is determinate, return it
      if (orientIndex != 0) return orientIndex;

      // otherwise, segs must be collinear - sort based on minimum X value
      return compareX(this.upwardSeg, other.upwardSeg);
    }