We can imagine "danger circles" around all coworkers increase at a constant size until there is no path to the finish. This idea is very similar to the construction of a Voronoi diagram, which can be done in \(O(N \log(N))\) time using Fortune's algorithm. If we remove the restriction that you are not allowed to leave the building's borders, the optimal path maximizing the distance to the nearest person will be a sequence of edges on the Voronoi diagram, as well as a segment from the start position to a Voronoi node and a segment from the end position to a Voronoi node. For the first and last segment, we simply want to walk to a Voronoi node that isn't *closer* to our nearest neighbor than we are, of the Voronoi nodes tangential to this neighbor's corresponding node in the Delaunay triangulation. Valid start and end Voronoi nodes can be found by iterating through all neighboring Voronoi nodes to the nearest person, and simply taking the one which is most in the direction away from the nearest coworker. Then, we can handle the path of Voronoi edges by sorting the edges and using a disjoint set (similarly to Kruskal's minimum spanning tree algorithm, though in this case as a maximum spanning tree), stopping once the start and end Voronoi nodes are unioned.

Note that there are also 2 cases in which the path might not remain on the Voronoi diagram's edges:

1. At the start (and similarly at the end), the path must move from the the starting point \(a = (X_A, Y_A)\) to some graph node. If it's in the Voronoi cell of some closest point \(x\), then it should move to some node on the boundary of that cell without getting closer to \(x\) (any such node will do). A valid choice of node is the one which is closest angle-wise (relative to \(x\)) to \(a\).

2. The path can also travel along the borders of the building's rectangle — there should effectively be a graph node at each rectangle corner and at each intersection of an (infinite) Voronoi edge with the rectangle's border, with these nodes connected around the border by edges (weighted like the other edges). A good way of constructing this situation is to begin by mirroring all \(N\) points across the rectangle's \(4\) sides, and then computing the Voronoi diagram for all \(5N\) resulting points. Provided that we then include Voronoi edges lying inside or on the border of the rectangle (excluding ones lying outside it), the set of edges will then be exactly what we need.
This is because the introduction of the mirrored points doesn't affect the portion of the Voronoi diagram strictly inside the rectangle (as no mirrored point can be the closest point to a point inside the rectangle; the corresponding original point must be closer), while at each point along the rectangle's boundary, whatever original point was closest to it, the corresponding mirrored point must now be equidistant (resulting in a Voronoi node/edge at the border).
For a visualization of this concept with a 3-point example, check out [this tool](http://alexbeutel.com/webgl/voronoi.html) with the following input data: `{"sites":[220,240,270,250,290,220,180,240,220,160,380,240,220,360,270,150,270,350,130,250,330,250,290,180,290,380,110,220,310,220],"queries":[]}`.

The time complexity of the above algorithm is \(O(N \log(N))\).

It's worth noting that it's also possible to solve this problem for many (e.g. \(10^5\)) queries of start/end points, though that extension was not included in the contest as its implementation would be rather long.

Computation of which Voronoi cell and graph node each starting/ending point corresponds to must now be done more efficiently, in less than \(O(N)\) time per such point. It's possible to augment Fortune's algorithm to determine which Voronoi cell each such point lies within. Alternatively, you can do a line sweep from left to right, finding for each query point which Voronoi edge is above it (form which you can determine which Voronoi cell it is in). For each cell, if the angles of its surrounding Voronoi vertices (relative to the cell's central point) are precomputed and sorted, it's possible to binary search over them for each starting/ending point to find its enclosing triangle. These changes take a total of \(O((N+K) \log(N))\) time.

To solve the second part of the problem, one approach involves still generating a maximum spanning tree using Kruskal's algorithm. The answer for a query is then based on the minimum edge weight on the path connecting its starting/ending points' corresponding nodes in this tree. This can be computed for each query using binary lifting (or other methods). Thus, the time complexity of this portion also comes out to \(O((N+K) \log(N))\) time.

Another approach for the second part of the problem is to use a parallel binary search. For each query, we can store its minimum and maximum possible answer, and iterate through all edges until these ranges are sufficiently small. At each iteration, the range for each query's possible answer will be halved. An iteration consists of maintaining a disjoint set of all components, processing queries and edges in decreasing order of weight. When we get to a query, we simply check if its two nodes are in the same component, and adjust either the high or low value of that queries binary search accordingly (this will change its mid for where it is sorted in the next iteration). At a Voronoi edge, we simply union the two components.