/** * @file polygon_test_point_inside.cpp */ #include #include #include "PolyLine.h" /* this algo uses the the Jordan curve theorem to find if a point is inside or outside a polygon: * It run a semi-infinite line horizontally (increasing x, fixed y) * out from the test point, and count how many edges it crosses. * At each crossing, the ray switches between inside and outside. * If odd count, the test point is inside the polygon * This is called the Jordan curve theorem, or sometimes referred to as the "even-odd" test. * Take care to starting and ending points of segments outlines, when the horizontal line crosses a segment outline * exactly on an ending point: * Because the starting point of a segment is also the ending point of the previous, only one must be used. * And we do no use twice the same segment, so we do NOT use both starting and ending points of these 2 segments. * So we must use only one ending point of each segment when calculating intersections * but it cannot be always the starting or the ending point. This depend on relative position of 2 consectutive segments * Here, the ending point above the Y reference position is used * and the ending point below or equal the Y reference position is NOT used * Obviously, others cases are irrelevant because there is not intersection. */ #define OUTSIDE false #define INSIDE true bool TestPointInsidePolygon( std::vector aPolysList, int aIdxstart, int aIdxend, int aRefx, int aRefy) /** Function TestPointInsidePolygon * test if a point is inside or outside a polygon. * the polygon must have only lines (not arcs) for outlines. * @param aPolysList: the list of polygons * @param aIdxstart: the starting point of a given polygon in m_FilledPolysList. * @param aIdxend: the ending point of this polygon in m_FilledPolysList. * @param aRefx, aRefy: the point coordinate to test * @return true if the point is inside, false for outside */ { // count intersection points to right of (refx,refy). If odd number, point (refx,refy) is inside polyline int ics, ice; int count = 0; // find all intersection points of line with polyline sides for( ics = aIdxstart, ice = aIdxend; ics <= aIdxend; ice = ics++ ) { int seg_startX = aPolysList[ics].x; int seg_startY = aPolysList[ics].y; int seg_endX = aPolysList[ice].x; int seg_endY = aPolysList[ice].y; /* Trivial cases: skip if ref above or below the segment to test */ if( ( seg_startY > aRefy ) && (seg_endY > aRefy ) ) continue; // segment below ref point, or one of its ends has the same Y pos as the ref point: skip // So we eliminate one end point of 2 consecutive segments. // Note: also we skip horizontal segments if ref point is on this horizontal line // So reference points on horizontal segments outlines always are seen as outside the polygon if( ( seg_startY <= aRefy ) && (seg_endY <= aRefy ) ) continue; /* refy is between seg_startY and seg_endY. * note: here: horizontal segments (seg_startY == seg_endY) are skipped, * either by the first test or by the second test * see if an horizontal semi infinite line from refx is intersecting the segment */ // calculate the x position of the intersection of this segment and the semi infinite line // this is more easier if we move the X,Y axis origin to the segment start point: seg_endX -= seg_startX; seg_endY -= seg_startY; double newrefx = (double) (aRefx - seg_startX); double newrefy = (double) (aRefy - seg_startY); // Now calculate the x intersection coordinate of the line from (0,0) to (seg_endX,seg_endY) // with the horizontal line at the new refy position // the line slope = seg_endY/seg_endX; // and the x pos relative to the new origin is intersec_x = refy/slope // Note: because horizontal segments are skipped, 1/slope exists (seg_end_y never == O) double intersec_x = newrefy * seg_endX / seg_endY; if( newrefx < intersec_x ) // Intersection found with the semi-infinite line from refx to infinite count++; } return count & 1 ? INSIDE : OUTSIDE; } /* Function TestPointInsidePolygon (overlaid) * same as previous, but use wxPoint and aCount corners */ bool TestPointInsidePolygon( wxPoint *aPolysList, int aCount,wxPoint aRefPoint ) { // count intersection points to right of (refx,refy). If odd number, point (refx,refy) is inside polyline int ics, ice; int count = 0; // find all intersection points of line with polyline sides for( ics = 0, ice = aCount-1; ics < aCount; ice = ics++ ) { int seg_startX = aPolysList[ics].x; int seg_startY = aPolysList[ics].y; int seg_endX = aPolysList[ice].x; int seg_endY = aPolysList[ice].y; /* Trivial cases: skip if ref above or below the segment to test */ if( ( seg_startY > aRefPoint.y ) && (seg_endY > aRefPoint.y ) ) continue; // segment below ref point, or one of its ends has the same Y pos as the ref point: skip // So we eliminate one end point of 2 consecutive segments. // Note: also we skip horizontal segments if ref point is on this horizontal line // So reference points on horizontal segments outlines always are seen as outside the polygon if( ( seg_startY <= aRefPoint.y ) && (seg_endY <= aRefPoint.y ) ) continue; /* refy is between seg_startY and seg_endY. * note: here: horizontal segments (seg_startY == seg_endY) are skipped, * either by the first test or by the second test * see if an horizontal semi infinite line from refx is intersecting the segment */ // calculate the x position of the intersection of this segment and the semi infinite line // this is more easier if we move the X,Y axis origin to the segment start point: seg_endX -= seg_startX; seg_endY -= seg_startY; double newrefx = (double) (aRefPoint.x - seg_startX); double newrefy = (double) (aRefPoint.y - seg_startY); // Now calculate the x intersection coordinate of the line from (0,0) to (seg_endX,seg_endY) // with the horizontal line at the new refy position // the line slope = seg_endY/seg_endX; // and the x pos relative to the new origin is intersec_x = refy/slope // Note: because horizontal segments are skipped, 1/slope exists (seg_end_y never == O) double intersec_x = newrefy * seg_endX / seg_endY; if( newrefx < intersec_x ) // Intersection found with the semi-infinite line from refx to infinite count++; } return count & 1 ? INSIDE : OUTSIDE; }