示例#1
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        // Same as above, but avoid using Sqrt(), saves a new nanoseconds in cases where you only want
        //  to compare several distances to find the smallest or largest, but don't need the distance
        public static float DistanceToLineSegmentSquared(Vector2 A, Vector2 B, Vector2 P, out Vector2 closestPoint)
        {
            // Compute length of line segment (squared) and handle special case of coincident points
            float segmentLengthSquared = A.SqrDist(B);
            if (segmentLengthSquared < 1E-7f)  // Arbitrary "close enough for government work" value
            {
                closestPoint = A;
                return P.SqrDist(closestPoint);
            }

            // Use the magic formula to compute the "projection" of this point on the infinite line
            Vector2 lineSegment = B - A;
            float t = Vector2.Dot(P - A, lineSegment) / segmentLengthSquared;

            // Handle the two cases where the projection is not on the line segment, and the case where
            //  the projection is on the segment
            if (t <= 0)
                closestPoint = A;
            else if (t >= 1)
                closestPoint = B;
            else
                closestPoint = A + (lineSegment * t);
            return P.SqrDist(closestPoint);
        }