MonotoneChainEdge.java
/*
* Copyright (c) 2016 Vivid Solutions.
*
* All rights reserved. This program and the accompanying materials
* are made available under the terms of the Eclipse Public License 2.0
* and Eclipse Distribution License v. 1.0 which accompanies this distribution.
* The Eclipse Public License is available at http://www.eclipse.org/legal/epl-v20.html
* and the Eclipse Distribution License is available at
*
* http://www.eclipse.org/org/documents/edl-v10.php.
*/
package org.locationtech.jts.geomgraph.index;
import org.locationtech.jts.geom.Coordinate;
import org.locationtech.jts.geom.Envelope;
import org.locationtech.jts.geomgraph.Edge;
/**
* MonotoneChains are a way of partitioning the segments of an edge to
* allow for fast searching of intersections.
* They have the following properties:
* <ol>
* <li>the segments within a monotone chain will never intersect each other
* <li>the envelope of any contiguous subset of the segments in a monotone chain
* is simply the envelope of the endpoints of the subset.
* </ol>
* Property 1 means that there is no need to test pairs of segments from within
* the same monotone chain for intersection.
* Property 2 allows
* binary search to be used to find the intersection points of two monotone chains.
* For many types of real-world data, these properties eliminate a large number of
* segment comparisons, producing substantial speed gains.
* @version 1.7
*/
public class MonotoneChainEdge {
Edge e;
Coordinate[] pts; // cache a reference to the coord array, for efficiency
// the lists of start/end indexes of the monotone chains.
// Includes the end point of the edge as a sentinel
int[] startIndex;
public MonotoneChainEdge(Edge e) {
this.e = e;
pts = e.getCoordinates();
MonotoneChainIndexer mcb = new MonotoneChainIndexer();
startIndex = mcb.getChainStartIndices(pts);
}
public Coordinate[] getCoordinates() { return pts; }
public int[] getStartIndexes() { return startIndex; }
public double getMinX(int chainIndex)
{
double x1 = pts[startIndex[chainIndex]].x;
double x2 = pts[startIndex[chainIndex + 1]].x;
return x1 < x2 ? x1 : x2;
}
public double getMaxX(int chainIndex)
{
double x1 = pts[startIndex[chainIndex]].x;
double x2 = pts[startIndex[chainIndex + 1]].x;
return x1 > x2 ? x1 : x2;
}
public void computeIntersects(MonotoneChainEdge mce, SegmentIntersector si)
{
for (int i = 0; i < startIndex.length - 1; i++) {
for (int j = 0; j < mce.startIndex.length - 1; j++) {
computeIntersectsForChain( i,
mce, j,
si );
}
}
}
public void computeIntersectsForChain(
int chainIndex0,
MonotoneChainEdge mce,
int chainIndex1,
SegmentIntersector si)
{
computeIntersectsForChain(startIndex[chainIndex0], startIndex[chainIndex0 + 1],
mce,
mce.startIndex[chainIndex1], mce.startIndex[chainIndex1 + 1],
si );
}
private void computeIntersectsForChain(
int start0, int end0,
MonotoneChainEdge mce,
int start1, int end1,
SegmentIntersector ei)
{
//Debug.println("computeIntersectsForChain:" + p00 + p01 + p10 + p11);
// terminating condition for the recursion
if (end0 - start0 == 1 && end1 - start1 == 1) {
ei.addIntersections(e, start0, mce.e, start1);
return;
}
// nothing to do if the envelopes of these chains don't overlap
if (! overlaps(start0, end0, mce, start1, end1)) return;
// the chains overlap, so split each in half and iterate (binary search)
int mid0 = (start0 + end0) / 2;
int mid1 = (start1 + end1) / 2;
// Assert: mid != start or end (since we checked above for end - start <= 1)
// check terminating conditions before recursing
if (start0 < mid0) {
if (start1 < mid1) computeIntersectsForChain(start0, mid0, mce, start1, mid1, ei);
if (mid1 < end1) computeIntersectsForChain(start0, mid0, mce, mid1, end1, ei);
}
if (mid0 < end0) {
if (start1 < mid1) computeIntersectsForChain(mid0, end0, mce, start1, mid1, ei);
if (mid1 < end1) computeIntersectsForChain(mid0, end0, mce, mid1, end1, ei);
}
}
/**
* Tests whether the envelopes of two chain sections overlap (intersect).
*
* @param start0
* @param end0
* @param mce
* @param start1
* @param end1
* @return true if the section envelopes overlap
*/
private boolean overlaps(
int start0, int end0,
MonotoneChainEdge mce,
int start1, int end1)
{
return Envelope.intersects(pts[start0], pts[end0], mce.pts[start1], mce.pts[end1]);
}
}