Wednesday, July 16, 2014

IOI 2014 day 1 analysis

IOI 2014 day 1

Since there is no online judge, I haven't tried actually coding any of these. So these ideas are not validated yet. You can find the problems here.


I found this the most difficult of the three to figure out, although coding it will not be particularly challenging.

Firstly, we can note that distances are symmetric: a route from A to B can be reflected in the two tracks to give a route from B to A. So having only \(\frac{n(n-1)}{2}\) queries is not a limitation, as we can query all distances. This might be useful in tackling the first three subtasks, but I'll go directly to the hardest subtask.

If we know the position and type of a station, there is one other that we can immediately locate: the closest one. It must have the opposite type and be reached by a direct route. Let station X be the closest to station 0. The other stations can be split into three groups:
  1. d(X, Y) < d(X, 0): these are reached directly from station X and of type C, so we can locate them exactly.
  2. d(0, X) + d(X, Y) = d(0, Y), but not of type 1: these are reached from station 0 via station X, so they lie to the left of station 0.
  3. All other stations lie to the right of station X.
Let's now consider just the stations to the right of X, and see how to place them. Let's take them in increasing distance from 0. This ensures that we encounter all the type D stations in order, and any type C station will be encountered at some point after the type D station used to reach it. Suppose Y is the right-most type D station already encountered, and consider the distances for a new station Z. Let \(z = d(0, Z) - d(0, Y) - d(Y, Z)\). If Z is type C, then there must be a type D at distance \(\frac{z}{2}\) to the left of Y. On the other hand, if Z is of type D (and lies to the right of Y), then there must be a type C station at distance \(\frac{z}{2}\) to the left of Y. In the first case, we will already have encountered the station, so we can always distinguish the two cases, and hence determine the position and type of Z.
The stations to the left of station zero can be handled similarly, using station X as the reference point instead of station 0.
How many queries is this? Every station Z except 0 and X accounts for at most three queries: d(0, Z), d(X, Z) and d(Y, Z), where Y can be different for each Z. This gives \(3(n-2) + 1\), which I think can be improved to \(3(n-2)\) just by counting more carefully. Either way, it is sufficient to solve all the subtasks.


This is a fairly standard interval tree structure problem, similar to Mountain from IOI 2005 (but a little easier). Each node of the tree contains a range to which its children are clamped. To determine the value of any element of the wall, start at the leaf with a value of 0 and read up the tree, clamping the value to the range in each node in turn. Initially, each node has the range [0, inf). When applying a new instruction, it is done top-down, and clamps are pushed down the tree whenever recursion is necessary.

An interesting aspect of the problem is that it is offline, in that only the final configuration is requested and all the information is provided up-front. This makes me think that there may be an alternative solution that processes the data in a different order, but I can't immediately see a nicer solution than the one above.


I liked this problem, partly because I could reverse-engineer a solution from the assumption that it is always possible to win, and partly because it requires neither much algorithm/data-structure training (like Wall) nor tricky consideration of cases (like Rails). Suppose Mei-Yu knows that certain cities are connected. If there are any flights between the cities that she has not asked about, then she can win simply by saving one of these flights for last, since it will not affect whether the country is connected. It follows that for Jian-Jia to win, he must always answer no when asked about a flight between two components that Mei-Yu does not know to be connected, unless this is the last flight between these components?

What if he always answers yes to the last flight between two components? In this case he will win. As long as there are at least two components left, there are uncertain edges between every pair of them, so Mei-Yu can't know whether any of them is connected any other. All edges within a component are known, so the number of components can only become one after the last question.

What about complexity? We need to keep track of the number of edges between each pair of components, which takes \(O(N^2)\) space. Most operations will just decrement one of these counts. There will be \(N - 1\) component-merging operations, each of which requires a linear-time merging of these edge counts and updating a vertex-to-component table. Thus, the whole algorithm requires \(O(N^2)\) time. This is optimal given that Mei-Yu will ask \(O(N^2)\) questions.


Haidar Abboud said...
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