Esempio n. 1
0
  /**
   * Find the innermost enclosing shell EdgeRing containing the argument EdgeRing, if any. The
   * innermost enclosing ring is the <i>smallest</i> enclosing ring. The algorithm used depends on
   * the fact that: <br>
   * ring A contains ring B iff envelope(ring A) contains envelope(ring B) <br>
   * This routine is only safe to use if the chosen point of the hole is known to be properly
   * contained in a shell (which is guaranteed to be the case if the hole does not touch its shell)
   *
   * @return containing EdgeRing, if there is one or null if no containing EdgeRing is found
   */
  public static EdgeRing findEdgeRingContaining(EdgeRing testEr, List shellList) {
    LinearRing testRing = testEr.getRing();
    Envelope testEnv = testRing.getEnvelopeInternal();
    Coordinate testPt = testRing.getCoordinateN(0);

    EdgeRing minShell = null;
    Envelope minShellEnv = null;
    for (Iterator it = shellList.iterator(); it.hasNext(); ) {
      EdgeRing tryShell = (EdgeRing) it.next();
      LinearRing tryShellRing = tryShell.getRing();
      Envelope tryShellEnv = tryShellRing.getEnvelopeInternal();
      // the hole envelope cannot equal the shell envelope
      // (also guards against testing rings against themselves)
      if (tryShellEnv.equals(testEnv)) continue;
      // hole must be contained in shell
      if (!tryShellEnv.contains(testEnv)) continue;

      testPt =
          CoordinateArrays.ptNotInList(testRing.getCoordinates(), tryShellRing.getCoordinates());
      boolean isContained = false;
      if (CGAlgorithms.isPointInRing(testPt, tryShellRing.getCoordinates())) isContained = true;

      // check if this new containing ring is smaller than the current minimum ring
      if (isContained) {
        if (minShell == null || minShellEnv.contains(tryShellEnv)) {
          minShell = tryShell;
          minShellEnv = minShell.getRing().getEnvelopeInternal();
        }
      }
    }
    return minShell;
  }
Esempio n. 2
0
 /**
  * Tests whether a point lies in the envelopes of both input segments. A correctly computed
  * intersection point should return <code>true</code> for this test. Since this test is for
  * debugging purposes only, no attempt is made to optimize the envelope test.
  *
  * @return <code>true</code> if the input point lies within both input segment envelopes
  */
 private boolean isInSegmentEnvelopes(Coordinate intPt) {
   Envelope env0 = new Envelope(inputLines[0][0], inputLines[0][1]);
   Envelope env1 = new Envelope(inputLines[1][0], inputLines[1][1]);
   return env0.contains(intPt) && env1.contains(intPt);
 }