unity 中多边形合并问题
发表于2016-05-31
This is a very good question. I implemented the same algorithm on c# some time ago. The Algorithm constructs a common contour of two polygons (i.e. Constructs a union without holes). Here it is.
Step 1、Create graph that describes the polygons.
Input: first polygon (n points), second polygon (m points). Output: graph. Vertex - polygon point of intersection point.
We should find intersections. Iterate through all polygon sides in both polygons [O(n*m)] and find any intersections.
If an intersection is not found, simply add vertices and connect them to the edge.
If any intersections are found, sort them by length to their start point, add all vertexes (start, end and intersections) and connect them (already in sorted order) to the edge.
Step 2、Check constructed graph
If we did not find any intersection points when graph was built, we have one of the following conditions:
1、Polygon1 contains polygon2 - return polygon12、Polygon2 contains polygon1 - return polygon2
3、Polygon1 and polygon2 do not intersect. Return polygon1 AND polygon2.
Step 3、 Find left-bottom vertex.
Find the minimum x and y coordinates (minx, miny). Then find the minimum distance between (minx, miny) and the polygon's points. This point will be the left-bottom point.
Step 4、 Construct common contour.
We start to traverse the graph from the left-bottom point and continue until we get back into it. At the beginning we mark all edges as unvisited. On every iteration you should select the next point and mark it as visited.
To choose the next point, choose an edge with a maximum internal angle in counter-clockwise direction.
I calculate two vectors: vector1 for current edge and vector2 for each next unvisited edge (as presented in the picture).
For vectors I calculate:
1、Scalar product (dot product). It returns a value related to an angle between vectors.2、Vector product (cross product). It returns a new vector. If z-coordinate of this vector is positive, scalar product gives me right angle in counter-clockwise direction. Else (z-coordinate is negative), I calculate get angle between vectors as 360 - angle from scalar product.
As a result I get an edge (and a correspond next vertex) with the maximum angle.
I add to result list each passed vertex. Result list is the union polygon. 
Remarks
1、This algorithm allows us to merge multiple of polygons - to apply iteratively with polygon's pairs.2、If you have a path that consists of many bezier curves and lines, you should flatten this path first.
以上转自
http://stackoverflow.com/questions/2667748/how-do-i-combine-complex-polygons/19475433#19475433
一下是算法代码实现:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255 256 257 258 259 260 261 262 263 264 265 266 267 268 269 270 271 272 273 274 275 276 277 278 279 280 281 282 283 284 285 286 287 288 289 290 291 292 293 294 295 296 297 298 299 300 301 302 303 304 305 306 307 308 309 310 311 312 313 314 315 316 317 318 319 320 321 322 323 324 325 326 327 328 329 330 331 332 333 334 335 336 337 338 339 | public class UnitPolygonsUtils { private List polygons = new List(); private int curPgon; private Vector2 startPoint; private int curIndex; public UnitPolygonsUtils() { } private Vector2 dotV1 = Vector2.zero; private Vector2 dotV2 = Vector2.zero; private void findIntersectionVertex(Vector2 p1, Vector2 p2, Vector2 p3, Vector2 p4, ref List intersections, out float t1) { float t2 = 0; float dx12 = p2.x - p1.x; float dy12 = p2.y - p1.y; float dx34 = p4.x - p3.x; float dy34 = p4.y - p3.y; // Solve for t1 and t2 float denominator = (dy12 * dx34 - dx12 * dy34); t1 = ((p1.x - p3.x) * dy34 (p3.y - p1.y) * dx34) / denominator; if (float.IsInfinity(t1)) { return; } float dotNum = 0; float maxNum = 0; //直线重叠 if (float.IsNaN(t1)) { dotV1.Set(p2.x - p1.x, p2.y - p1.y); dotV2.Set(p2.x - p1.x, p2.y - p1.y); maxNum = Vector2.Dot(dotV1, dotV2); dotV1.Set(p3.x - p1.x, p3.y - p1.y); dotV2.Set(p2.x - p1.x, p2.y - p1.y); dotNum = Vector2.Dot(dotV1, dotV2); if (dotNum > 0 && dotNum < maxNum) addIntersectionsPoint(ref intersections, p3, p1, p2); dotV1.Set(p4.x - p1.x, p4.y - p1.y); dotV2.Set(p2.x - p1.x, p2.y - p1.y); dotNum = Vector2.Dot(dotV1, dotV2); if (dotNum > 0 && dotNum < maxNum) addIntersectionsPoint(ref intersections, p4, p1, p2); } // Find the point of intersection. t2 = ((p3.x - p1.x) * dy12 (p1.y - p3.y) * dx12) / -denominator; // The segments intersect if t1 and t2 are between 0 and 1. bool segments_intersect = ((t1 >= 0) && (t1 <= 1) && (t2 >= 0) && (t2 <= 1)); if (t1 > 0.001 && segments_intersect) { Vector2 intersectP = new Vector2(p1.x dx12 * t1, p1.y dy12 * t1); addIntersectionsPoint(ref intersections, intersectP, p1, p2); } } private void addIntersectionsPoint(ref List intersections, Vector2 intersectP, Vector2 p1, Vector2 p2) { if (indexListItem(ref intersections, intersectP) == -1) { addWithOrde(ref intersections, p1, p2, intersectP); } } private int indexListItem(ref List list, Vector2 item) { int index = -1; for(int ix = 0; ix < list.Count; ix ) { if (Vector2.Distance(list[ix],item) <= 0.01f) { index = ix; break; } } return index; } private void addWithOrde(ref List list, Vector2 p1, Vector2 p2, Vector2 intersecP) { if (intersecP.Equals(p1) || intersecP.Equals(p2)) return; bool insert = false; for (int ix = 0; ix < list.Count; ix ) { if (Vector2.Distance(list[ix], p1) > Vector2.Distance(intersecP, p1)) { list.Insert(ix, intersecP); insert = true; break; } } if (!insert) { list.Add(intersecP); } } private float dotProduct(PointF p1, PointF p2, PointF p3, PointF p4) { return (p2.X - p1.X) * (p4.X - p3.X) (p2.Y - p1.Y) * (p4.Y - p3.Y); } private void findLeftBottomVertex(List { int cur_index = 0; curPgon = 0; startPoint = polygons[curPgon][cur_index]; for (int pgon = 0; pgon < polygons.Count; pgon ) { for (int index = 0; index < polygons[pgon].Count; index ) { Vector2 test_point = polygons[pgon][index]; if ((test_point.x < startPoint.x) || ((test_point.x == startPoint.x) && (test_point.y < startPoint.y))) { curPgon = pgon; cur_index = index; startPoint = polygons[curPgon][cur_index]; } } } } public List UnitPoligons(List polygons) { if (polygons.Count == 1) { return polygons[0].ToList(); } List units = new List(); int cur_index = 0, cur_pgon = 0, cur_index2 = 0; Vector2 p1, p2, p3, p4; Listnew List List<list<< code=""> |