Пример #1
0
 /**
  * @param grid
  * @return an array of vertex coordinates, one vertex per edge end, 3 coords per vertex
  */
 public double[] getCoords(float[][] grid) {
   double[] ret = new double[edges.size() * 3];
   Iterator<Integer> it = edges.iterator();
   switch (direction) {
     case 0:
       for (int i = 0; i < ret.length; i += 3) {
         ret[i + 0] = grid[0][it.next().intValue()];
         ret[i + 1] = grid[1][c1];
         ret[i + 2] = grid[2][c2];
       }
       break;
     case 1:
       for (int i = 0; i < ret.length; i += 3) {
         ret[i + 0] = grid[0][c1];
         ret[i + 1] = grid[1][it.next().intValue()];
         ret[i + 2] = grid[2][c2];
       }
       break;
     case 2:
       for (int i = 0; i < ret.length; i += 3) {
         ret[i + 0] = grid[0][c1];
         ret[i + 1] = grid[1][c2];
         ret[i + 2] = grid[2][it.next().intValue()];
       }
       break;
   }
   return ret;
 }
Пример #2
0
 /**
  * Add an edge along this line. Overlapping edges are merged.
  *
  * @param e1 The first edge end
  * @param e2 The second edge end
  */
 public void add(int e1, int e2) {
   // sort the ends
   int min = (e1 < e2) ? e1 : e2;
   e2 = (e1 > e2) ? e1 : e2;
   e1 = min;
   // state of the current segment end : first end or last end of the
   // current segment
   boolean first = true;
   boolean inMerge = false;
   for (int i = 0; i < edges.size(); ++i) {
     int end = edges.get(i).intValue();
     // wait till newsegment match this interval in edges list order
     if (e1 > end) {
       first = !first;
       continue;
     }
     // now e1<=end. Handle case where end is the first segment end
     if (first) {
       // If not already merging, see if merging is necessary
       if (!inMerge) {
         // segments overlap => extend existing segment to match
         // the new one
         if (end <= e2) {
           edges.set(i, new Integer(e1));
           inMerge = true; // now merging the new segment
         } else {
           // Segments do not overlap => add a new segment,
           // finished
           edges.add(i, new Integer(e2)); // insert e2 before
           // the current end
           edges.add(i, new Integer(e1)); // insert e1 before
           // e2
           return;
         }
         // In merge already => it normal that e1<=end. Let's
         // check e2
       } else {
         // new segment terminates before current segment
         // => add the new point, finished
         if (e2 < end) {
           edges.add(i, new Integer(e2));
           return;
           // new segment terminates after or inside current
           // segment => merge this end
         } else if (e2 >= end) {
           // remove this end, as the segments merge to make
           // one bigger segment
           edges.remove(i--); // remove corresponding index too
         }
       }
     }
     // Handle second case : end is the last end of the segment.
     // Note: e1<=end, see above
     else {
       // new segment terminates inside current one
       // => end of the merge, keep this segment's end
       if (e2 <= end) return;
       // New segment terminates strictly after current segment
       // remove this end, as the segments merge to make one bigger
       // segment
       edges.remove(i--); // remove corresponding index too
       // If not already merging, this is necessary
       inMerge = true; // now merging the new segment
     }
     first = !first;
   }
   // Still here? => new segment terminates after any existing segment
   // If not merging, must add the first end of the new segment since
   // it was larger than all existing segment ends
   if (!inMerge) edges.add(new Integer(e1));
   // Close the new segment
   edges.add(new Integer(e2));
 }
Пример #3
0
 /** @return */
 public int getNumberOfEdges() {
   if ((edges.size() & 1) != 0) {
     System.out.println("Error: Odd number of vertex in edge line : " + edges.size());
   }
   return edges.size() / 2;
 }