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| 1702 | - | 1 | /// -*- tab-width: 4; Mode: C++; c-basic-offset: 4; indent-tabs-mode: nil -*- |
| 2 | /* |
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| 3 | * polygon.cpp |
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| 4 | * Copyright (C) Andrew Tridgell 2011 |
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| 5 | * |
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| 6 | * This file is free software: you can redistribute it and/or modify it |
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| 7 | * under the terms of the GNU General Public License as published by the |
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| 8 | * Free Software Foundation, either version 3 of the License, or |
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| 9 | * (at your option) any later version. |
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| 10 | * |
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| 11 | * This file is distributed in the hope that it will be useful, but |
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| 12 | * WITHOUT ANY WARRANTY; without even the implied warranty of |
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| 13 | * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. |
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| 14 | * See the GNU General Public License for more details. |
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| 15 | * |
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| 16 | * You should have received a copy of the GNU General Public License along |
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| 17 | * with this program. If not, see <http://www.gnu.org/licenses/>. |
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| 18 | */ |
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| 19 | |||
| 20 | #include "AP_Math.h" |
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| 21 | |||
| 22 | /* |
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| 23 | The point in polygon algorithm is based on: |
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| 24 | http://www.ecse.rpi.edu/Homepages/wrf/Research/Short_Notes/pnpoly.html |
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| 25 | */ |
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| 26 | |||
| 27 | |||
| 28 | /* |
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| 29 | Polygon_outside(): test for a point in a polygon |
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| 30 | Input: P = a point, |
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| 31 | V[] = vertex points of a polygon V[n+1] with V[n]=V[0] |
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| 32 | Return: true if P is outside the polygon |
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| 33 | |||
| 34 | This does not take account of the curvature of the earth, but we |
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| 35 | expect that to be very small over the distances involved in the |
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| 36 | fence boundary |
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| 37 | */ |
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| 38 | bool Polygon_outside(const Vector2l &P, const Vector2l *V, unsigned n) |
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| 39 | { |
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| 40 | unsigned i, j; |
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| 41 | bool outside = true; |
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| 42 | for (i = 0, j = n-1; i < n; j = i++) { |
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| 43 | if ((V[i].y > P.y) == (V[j].y > P.y)) { |
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| 44 | continue; |
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| 45 | } |
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| 46 | int32_t dx1, dx2, dy1, dy2; |
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| 47 | dx1 = P.x - V[i].x; |
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| 48 | dx2 = V[j].x - V[i].x; |
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| 49 | dy1 = P.y - V[i].y; |
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| 50 | dy2 = V[j].y - V[i].y; |
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| 51 | int8_t dx1s, dx2s, dy1s, dy2s, m1, m2; |
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| 52 | #define sign(x) ((x)<0?-1:1) |
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| 53 | dx1s = sign(dx1); |
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| 54 | dx2s = sign(dx2); |
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| 55 | dy1s = sign(dy1); |
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| 56 | dy2s = sign(dy2); |
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| 57 | m1 = dx1s * dy2s; |
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| 58 | m2 = dx2s * dy1s; |
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| 59 | // we avoid the 64 bit multiplies if we can based on sign checks. |
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| 60 | if (dy2 < 0) { |
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| 61 | if (m1 > m2) { |
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| 62 | outside = !outside; |
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| 63 | } else if (m1 < m2) { |
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| 64 | continue; |
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| 65 | } else if ( dx1 * (int64_t)dy2 > dx2 * (int64_t)dy1 ) { |
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| 66 | outside = !outside; |
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| 67 | } |
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| 68 | } else { |
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| 69 | if (m1 < m2) { |
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| 70 | outside = !outside; |
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| 71 | } else if (m1 > m2) { |
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| 72 | continue; |
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| 73 | } else if ( dx1 * (int64_t)dy2 < dx2 * (int64_t)dy1 ) { |
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| 74 | outside = !outside; |
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| 75 | } |
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| 76 | } |
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| 77 | } |
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| 78 | return outside; |
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| 79 | } |
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| 80 | |||
| 81 | /* |
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| 82 | check if a polygon is complete. |
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| 83 | |||
| 84 | We consider a polygon to be complete if we have at least 4 points, |
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| 85 | and the first point is the same as the last point. That is the |
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| 86 | minimum requirement for the Polygon_outside function to work |
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| 87 | */ |
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| 88 | bool Polygon_complete(const Vector2l *V, unsigned n) |
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| 89 | { |
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| 90 | return (n >= 4 && V[n-1].x == V[0].x && V[n-1].y == V[0].y); |
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| 91 | } |