com.vividsolutions.jts.algorithm.RectangleLineIntersector Maven / Gradle / Ivy
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/*
* The JTS Topology Suite is a collection of Java classes that
* implement the fundamental operations required to validate a given
* geo-spatial data set to a known topological specification.
*
* Copyright (C) 2001 Vivid Solutions
*
* This library is free software; you can redistribute it and/or
* modify it under the terms of the GNU Lesser General Public
* License as published by the Free Software Foundation; either
* version 2.1 of the License, or (at your option) any later version.
*
* This library is distributed in the hope that it will be useful,
* but WITHOUT ANY WARRANTY; without even the implied warranty of
* MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
* Lesser General Public License for more details.
*
* You should have received a copy of the GNU Lesser General Public
* License along with this library; if not, write to the Free Software
* Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA 02111-1307 USA
*
* For more information, contact:
*
* Vivid Solutions
* Suite #1A
* 2328 Government Street
* Victoria BC V8T 5G5
* Canada
*
* (250)385-6040
* www.vividsolutions.com
*/
package com.vividsolutions.jts.algorithm;
import com.vividsolutions.jts.geom.*;
/**
* Computes whether a rectangle intersects line segments.
*
* Rectangles contain a large amount of inherent symmetry
* (or to put it another way, although they contain four
* coordinates they only actually contain 4 ordinates
* worth of information).
* The algorithm used takes advantage of the symmetry of
* the geometric situation
* to optimize performance by minimizing the number
* of line intersection tests.
*
* @author Martin Davis
*
*/
public class RectangleLineIntersector
{
// for intersection testing, don't need to set precision model
private LineIntersector li = new RobustLineIntersector();
private Envelope rectEnv;
private Coordinate diagUp0;
private Coordinate diagUp1;
private Coordinate diagDown0;
private Coordinate diagDown1;
/**
* Creates a new intersector for the given query rectangle,
* specified as an {@link Envelope}.
*
*
* @param rectEnv the query rectangle, specified as an Envelope
*/
public RectangleLineIntersector(Envelope rectEnv)
{
this.rectEnv = rectEnv;
/**
* Up and Down are the diagonal orientations
* relative to the Left side of the rectangle.
* Index 0 is the left side, 1 is the right side.
*/
diagUp0 = new Coordinate(rectEnv.getMinX(), rectEnv.getMinY());
diagUp1 = new Coordinate(rectEnv.getMaxX(), rectEnv.getMaxY());
diagDown0 = new Coordinate(rectEnv.getMinX(), rectEnv.getMaxY());
diagDown1 = new Coordinate(rectEnv.getMaxX(), rectEnv.getMinY());
}
/**
* Tests whether the query rectangle intersects a
* given line segment.
*
* @param p0 the first endpoint of the segment
* @param p1 the second endpoint of the segment
* @return true if the rectangle intersects the segment
*/
public boolean intersects(Coordinate p0, Coordinate p1)
{
// TODO: confirm that checking envelopes first is faster
/**
* If the segment envelope is disjoint from the
* rectangle envelope, there is no intersection
*/
Envelope segEnv = new Envelope(p0, p1);
if (! rectEnv.intersects(segEnv))
return false;
/**
* If either segment endpoint lies in the rectangle,
* there is an intersection.
*/
if (rectEnv.intersects(p0)) return true;
if (rectEnv.intersects(p1)) return true;
/**
* Normalize segment.
* This makes p0 less than p1,
* so that the segment runs to the right,
* or vertically upwards.
*/
if (p0.compareTo(p1) > 0) {
Coordinate tmp = p0;
p0 = p1;
p1 = tmp;
}
/**
* Compute angle of segment.
* Since the segment is normalized to run left to right,
* it is sufficient to simply test the Y ordinate.
* "Upwards" means relative to the left end of the segment.
*/
boolean isSegUpwards = false;
if (p1.y > p0.y)
isSegUpwards = true;
/**
* Since we now know that neither segment endpoint
* lies in the rectangle, there are two possible
* situations:
* 1) the segment is disjoint to the rectangle
* 2) the segment crosses the rectangle completely.
*
* In the case of a crossing, the segment must intersect
* a diagonal of the rectangle.
*
* To distinguish these two cases, it is sufficient
* to test intersection with
* a single diagonal of the rectangle,
* namely the one with slope "opposite" to the slope
* of the segment.
* (Note that if the segment is axis-parallel,
* it must intersect both diagonals, so this is
* still sufficient.)
*/
if (isSegUpwards) {
li.computeIntersection(p0, p1, diagDown0, diagDown1);
}
else {
li.computeIntersection(p0, p1, diagUp0, diagUp1);
}
if (li.hasIntersection())
return true;
return false;
}
}