


/*
 * 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.geomgraph.index;

import com.vividsolutions.jts.geom.Coordinate;
import com.vividsolutions.jts.geom.Envelope;
import com.vividsolutions.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;
  // these envelopes are created once and reused
  Envelope env1 = new Envelope();
  Envelope env2 = new Envelope();

  public MonotoneChainEdge(Edge e) {
    this.e = e;
    pts = e.getCoordinates();
    MonotoneChainIndexer mcb = new MonotoneChainIndexer();
    startIndex = mcb.getChainStartIndices(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 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)
  {
    Coordinate p00 = pts[start0];
    Coordinate p01 = pts[end0];
    Coordinate p10 = mce.pts[start1];
    Coordinate p11 = mce.pts[end1];
//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
    env1.init(p00, p01);
    env2.init(p10, p11);
    if (! env1.intersects(env2)) 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);
    }
  }
}
