Saturday, June 28, 2014

ICPC Problem L: Wires

While this problem wasn't too conceptually difficult, it requires a lot of code (my solution is about 400 lines), and careful implementation of a number of geometric algorithms. A good chunk of the code comes from implementing a rational number class in order to precisely represent the intersection points of the wires. It is also very easy to suffer from overflow: I spent a long time banging my head against an assertion failure on the server until I upgraded my rational number class to use 128-bit integers everywhere, instead of just for comparisons.

The wires will divide the space up into connected regions. The regions can be represented in a planar graph, with edges between regions that share an edge. The problem is then to find the shortest path between the regions containing the two new end-points.

My solution works in a number of steps:
  1. Find all the intersection points between wires, and the segments between intersection points. This just tests every wire against every other wire. The case of two parallel wires sharing an endpoint needs to be handled carefully. For each line, I sort the intersection points along the line. I used a dot produce for this, which is where my rational number class overflowed, but would probably have been safer to just sort lexicographically. More than two lines can meet at an intersection point, so I used a std::map to assign a unique ID to each intersection point (I'll call them vertices from here on).
  2. Once the intersection points along a line have been sorted, one can identify the segments connecting them. I create two copies of each segment, one in each direction. With each vertex A I store a list of all segments A->B. Each pair is stored contiguously so that it is trivial to find its partner. Each segment is considered to belong to the region to its left as one travels A->B.
  3. The segments emanating from each vertex are sorted by angle. These comparisons could easily cause overflows again, but one can use a handy trick: instead of using the vector for the segment in an angle comparison, one can use the vector for the entire wire. It has identical direction but has small integer coordinates.
  4. Using the sorted lists from the previous step, each segment is given a pointer to its following segment from the same region. In other words, if one is tracing the boundary of the region and one has just traced A->B, the pointer will point to B->C.
  5. I extract the contours of the regions. A region typically consists of an outer contour and optionally some holes. The outermost region lacks an outer contour (one could add a big square if one needed to, but I didn't). A contour is found by following the next pointers. A case that turns out to be inconvenient later is that some segments might be part of the contour but not enclose any area. This can make a contour disappear completely, in which case it is discarded. Any remaining contours have the property that two adjacent segments are not dual to each other, although it is still possible to both sides of an edge to belong to the same contour.
  6. Each contour is identified as an outer contour or a hole. With integer coordinates I could just measure the signed area of the polygon, but that gets nasty with rational coordinates. Instead, I pick the lexicographically smallest vertex in the contour and examine the angle between the two incident segments (this is why it is important that there is a non-trivial angle between them). I also sort the contours by this lexicographically smallest vertex, which causes any contour to sort before any other contours it encloses.
  7. For each segment I add an edge of weight 1 from its containing region to the containing region of its dual.
  8. For each hole, I search backwards through the other contours to find the smallest non-hole that contains it. I take one vertex of the hole and do a point-in-polygon test. Once again, some care is needed to avoid overflows, and using the vectors for the original wires proves useful. One could then associate the outer contour and the holes it contains into a single region object, but instead I just added an edge to the graph to join them with weight 0. In other words, one can travel from the boundary of a region to the outside of a hole at no cost.
  9. Finally, I identify the regions containing the endpoints of the new wire, using the same search as in the previous step.
After all this, we still need to implement a shortest path search - but by this point that seems almost trivial in comparison.

What is the complexity? There can be \(O(M^2)\) intersections and hence also \(O(M^2)\) contours, but only \(O(M)\) of them can be holes (because two holes cannot be part of the same connected component). The slowest part is the fairly straightforward point-in-polygon test which tests each hole against each non-hole segment, giving \(O(M^3)\) time. There are faster algorithms for doing point location queries, so it is probably theoretically possible to reduce this to \(O(M^2\log N)\) or even \(O(M^2)\), but certainly not necessary for this problem.


Unknown said...

hi,can you post this problem code here,thanks,and my email is

Unknown said...

could you please send me source code of this problem. Thanks my email :

Bruce Merry said...

Unfortunately I don't seem to have the code any more.