/**
* 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;
}
/**
* 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);
}