Saturday, January 20, 2018

COCI 2017/2018 r5 analysis

For the problem texts, see here. I found this contest easier than usual for a COCI.


Given two wands A < B and two boxes X > Y, there is never any point in putting A in X and B in Y; if they fit, then they also fit the other way around. So sort the boxes and wands, match them up in this order and check if it all fits.


This is really just an implementation challenge. For each spiral, walk along it, filling in the grid where the spiral overlaps it. With more than one spiral, set each grid point to the minimum of the values induced by each spiral. Of course, it is not necessary to go to \(10^{100}\); about 10000 is enough to ensure that the grid is completely covered by the partial spiral.


The rule for determining which employee does the next task is a red herring. Each employee does exactly one task, and the order doesn't matter. The money earned by an employee is the sum of the distance from that employee to all underlings. The can be computed by a tree DP (computing both this sum and the size of each subtree).


Suppose there is a solution. It can be modified by a series of transformations into a canonical form.Firstly, moving a false claim below an adjacent true claim will not alter either of them; thus, we can push all the false claims to the bottom and true claims to the top. At this point, the true claims can be freely reordered amongst themselves. Similarly, if we have a false claim with small a above one with a larger a, we can swap them without making either true. So we can assume that the false claims are increasing from bottom to top in the deck. Finally, given a true claim with large a and a false claim with small a, we can swap them and they will both flip (so the positions of false claims stay the same).

After these transformations, the deck will, from bottom to top, contain the largest K cards in increasing order, followed by the rest in arbitrary order (let's say increasnig). To solve the problem, we simply construct this canonical form, then check if it indeed satisfies the conditions.


When building roads on the day with factor F, we don't actually need to build roads between every pair of multiples of F: it is equivalent to connect every multiple of F to F. This gives O(N log M) roads. We can represent the connected components after each day with a standard union-find structure. For reasons we'll see later, we won't use path compression, but always putting the smaller component under the larger one in the tree is sufficient to ensure a shallow tree (in theory O(log N), but I found the maximum depth was 6). A slow solution would be to check every remaining query after each day to see whether the two mathematicians are in the same component yet.

To speed this up, we can record extra information in the tree: for each edge, we record the day on which it was added. If the largest label on a path from A to B is D, then D was the first day on which they were connected (this property would be broken by path compression, which is why we cannot use it). Thus, to answer a query, we need only walk up the tree from each side to the least common ancestor; and given the shallowness of the tree, is this cheap.


I really liked this problem. Take a single starting peak P. Let A be the size of the maximum matching the graph, and let B be the size of the maximum matching excluding P. Suppose A = B. Then Mirko can win as follows: take the latter matching, and whenever Slavko moves to a valley, Mirko moves to the matched peak. Slavko can never move to a valley without a match, because otherwise the journey would form an augmenting path that would give a matching for the graph of size B + 1.

Conversely, suppose A > B. Then take a whole-graph matching, which by the assumption must include a match for P. Slavko can win by always moving to the matched valley. By a similar argument, Mirko can never reach an unmatched peak, because otherwise toggling all the edges on their journey would give a maximum matching that excludes P.

To implement it, it may not be efficient enough to construct a new subgraph matching from every peak. Instead, one can start with a full-graph matching, remove P and its match from the graph (if any), then re-augment starting from that match. This should give an O(NM) algorithm (O(NM) for the initial matching, then O(M) per query).


张晴川 said...

The last problem can be done in O(max matching + bfs).

Bruce Merry said...

That's true. It doesn't affect the big-O complexity, since it's still going to be dominated by the O(NM) for the max matching.

张晴川 said...

isn't max matching for bipartite graph O(msqrt(n))?

Bruce Merry said...

Maybe, I don't follow the more advanced algorithms. My solution uses Ford-Fulkerson with DFS, which is O(NM).

Anonymous said...

If a O(M*sqrt(N)) solution was required by the task the Hopcroft Karp algorithm could be used instead of Ford Fulkerson. That's the theory at least, I still haven't implemented the faster one.