Fix #8437: Planes landing at non-rectangular airports could be drawn at the wrong height

Only the rotated intercontinental airport, don't get excited

(cherry picked from commit 190528c2a9aaf4e90d0e768e6ad6afe6870af73d)

src/aircraft_cmd.cpp

@@ -1115,6 +1115,15 @@ static bool AircraftController(Aircraft *v)
 			z = GetAircraftFlightLevel(v);
 		}
 
+		/* Some airports (like a rotated intercontinental) are non-rectangular and their primary (north-most) tile might not be at the same height as the runway.
+		 * Therefore, as the hangar must be on flat ground (and we must have a hangar, as we're landing/braking and so not a helicopter), use that as the z position instead. */
+		int airport_z = v->z_pos;
+		if ((amd.flag & (AMED_LAND | AMED_BRAKE)) && st != nullptr) {
+			assert(st->airport.HasHangar());
+			TileIndex hangar_tile = st->airport.GetHangarTile(0);
+			airport_z = TilePixelHeight(hangar_tile) + 1; // To avoid clashing with the shadow
+		}
+
 		if (amd.flag & AMED_LAND) {
 			if (st->airport.tile == INVALID_TILE) {
 				/* Airport has been removed, abort the landing procedure */
@@ -1126,27 +1135,24 @@ static bool AircraftController(Aircraft *v)
 				continue;
 			}
 
-			int curz = GetSlopePixelZ(x + amd.x, y + amd.y) + 1;
-
 			/* We're not flying below our destination, right? */
-			assert(curz <= z);
+			assert(airport_z <= z);
 			int t = max(1U, dist - 4);
-			int delta = z - curz;
+			int delta = z - airport_z;
 
 			/* Only start lowering when we're sufficiently close for a 1:1 glide */
 			if (delta >= t) {
-				z -= CeilDiv(z - curz, t);
+				z -= CeilDiv(z - airport_z, t);
 			}
-			if (z < curz) z = curz;
+			if (z < airport_z) z = airport_z;
 		}
 
 		/* We've landed. Decrease speed when we're reaching end of runway. */
 		if (amd.flag & AMED_BRAKE) {
-			int curz = GetSlopePixelZ(x, y) + 1;
 
-			if (z > curz) {
+			if (z > airport_z) {
 				z--;
-			} else if (z < curz) {
+			} else if (z < airport_z) {
 				z++;
 			}