107 lines
5.0 KiB
C++
107 lines
5.0 KiB
C++
/*
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* This program source code file is part of KiCad, a free EDA CAD application.
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*
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* Copyright (C) 2007-2014 Jean-Pierre Charras, jp.charras at wanadoo.fr
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* Copyright (C) 2007-2014 KiCad Developers, see CHANGELOG.TXT for contributors.
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*
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* This program is free software; you can redistribute it and/or
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* modify it under the terms of the GNU General Public License
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* as published by the Free Software Foundation; either version 2
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* of the License, or (at your option) any later version.
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*
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* This program is distributed in the hope that it will be useful,
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* but WITHOUT ANY WARRANTY; without even the implied warranty of
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* MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
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* GNU General Public License for more details.
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*
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* You should have received a copy of the GNU General Public License
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* along with this program; if not, you may find one here:
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* http://www.gnu.org/licenses/old-licenses/gpl-2.0.html
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* or you may search the http://www.gnu.org website for the version 2 license,
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* or you may write to the Free Software Foundation, Inc.,
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* 51 Franklin Street, Fifth Floor, Boston, MA 02110-1301, USA
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*/
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/**
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* @file polygon_test_point_inside.cpp
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*/
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#include <cmath>
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#include <vector>
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#include "polygon_test_point_inside.h"
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/* this algo uses the the Jordan curve theorem to find if a point is inside or outside a polygon:
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* It run a semi-infinite line horizontally (increasing x, fixed y)
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* out from the test point, and count how many edges it crosses.
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* At each crossing, the ray switches between inside and outside.
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* If odd count, the test point is inside the polygon
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* This is called the Jordan curve theorem, or sometimes referred to as the "even-odd" test.
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* Take care to starting and ending points of segments outlines, when the horizontal line crosses a segment outline
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* exactly on an ending point:
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* Because the starting point of a segment is also the ending point of the previous, only one must be used.
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* And we do no use twice the same segment, so we do NOT use both starting and ending points of these 2 segments.
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* So we must use only one ending point of each segment when calculating intersections
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* but it cannot be always the starting or the ending point. This depends on relative position of 2 consectutive segments
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* Here, the ending point above the Y reference position is used
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* and the ending point below or equal the Y reference position is NOT used
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* Obviously, others cases are irrelevant because there is not intersection.
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*/
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#define OUTSIDE false
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#define INSIDE true
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/* Function TestPointInsidePolygon (overlaid)
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* same as previous, but use wxPoint and aCount corners
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*/
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bool TestPointInsidePolygon( const wxPoint *aPolysList, int aCount, const wxPoint &aRefPoint )
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{
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// count intersection points to right of (refx,refy). If odd number, point (refx,refy) is inside polyline
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int ics, ice;
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int count = 0;
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// find all intersection points of line with polyline sides
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for( ics = 0, ice = aCount-1; ics < aCount; ice = ics++ )
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{
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int seg_startX = aPolysList[ics].x;
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int seg_startY = aPolysList[ics].y;
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int seg_endX = aPolysList[ice].x;
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int seg_endY = aPolysList[ice].y;
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/* Trivial cases: skip if ref above or below the segment to test */
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if( ( seg_startY > aRefPoint.y ) && (seg_endY > aRefPoint.y ) )
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continue;
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// segment below ref point, or one of its ends has the same Y pos as the ref point: skip
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// So we eliminate one end point of 2 consecutive segments.
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// Note: also we skip horizontal segments if ref point is on this horizontal line
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// So reference points on horizontal segments outlines always are seen as outside the polygon
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if( ( seg_startY <= aRefPoint.y ) && (seg_endY <= aRefPoint.y ) )
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continue;
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/* refy is between seg_startY and seg_endY.
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* note: here: horizontal segments (seg_startY == seg_endY) are skipped,
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* either by the first test or by the second test
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* see if an horizontal semi infinite line from refx is intersecting the segment
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*/
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// calculate the x position of the intersection of this segment and the semi infinite line
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// this is more easier if we move the X,Y axis origin to the segment start point:
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seg_endX -= seg_startX;
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seg_endY -= seg_startY;
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double newrefx = (double) (aRefPoint.x - seg_startX);
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double newrefy = (double) (aRefPoint.y - seg_startY);
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// Now calculate the x intersection coordinate of the line from (0,0) to (seg_endX,seg_endY)
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// with the horizontal line at the new refy position
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// the line slope = seg_endY/seg_endX;
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// and the x pos relative to the new origin is intersec_x = refy/slope
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// Note: because horizontal segments are skipped, 1/slope exists (seg_endY never == O)
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double intersec_x = (newrefy * seg_endX) / seg_endY;
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if( newrefx < intersec_x ) // Intersection found with the semi-infinite line from refx to infinite
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count++;
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}
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return count & 1 ? INSIDE : OUTSIDE;
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}
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