Here are the solution ideas for the problems from the last week's summary. The first problem described there was about picking the right order to apply skill improvements. The solution is explained very well in the editorial (heading "521D - Shop"), but the basic steps one needed to make were:

- All multiplications should be applied in the end, in arbitrary order, and higher multiplier is better than lower multiplier.
- All assignments should be applied before all other operations, and we should use at most one assignment per skill. Since we only apply one, we can imagine that we have an addition instead of an assignment: highest value that can be assigned minus initial value.
- For each particular skill, higher addition is always better than lower addition, so we should sort the additions in decreasing order and imagine applying them in that order. In this case, for each addition we know for sure the value of the corresponding skill before and after the addition, and thus we can imagine that we have a multiplication instead (new value divided by old value)!
- Now all our operations are multiplications, and we should simply sort them in decreasing order.

And the third problem was about reconstructing the smallest parallelepiped drawn on the grid that contains the two given cells. Here's the solution that avoids a bulk of case studies that I was referring to: when the two given points are close to each other, we can just try all small parallelpipeds until we find one that covers both; when they are far from each other (let's say at least 10 apart), it's not hard to see that there's a simple lower boundary on the answer that is achievable. More specifically, let the parallelepiped horizontal side be a, vertical side be b, and the diagonal side be c. y-coordinate can be increased at most b+c-2 times (and similarly x-coordinate can be increased at most a+c-2 times) inside the parallelepiped, so b+c-2 must be at least y2-y1, and thus a+b+c (which is almot the answer) must be at least 5+y2-y1. Also, the difference between the coordinates doesn't change when we move diagonally, so it can only increase at most a+b-2 times, so the answer must be at least 5+(y2-x2)-(y1-x1). The only remaining step is to understand that the maximum of those lower bounds is always achievable if the two given points are sufficiently far apart.

Now, let's finish digging into the Open Cup archives! Open Cup 2014-15 Grand Prix of China happened 2 weeks ago (results, top 5 on the left). Let me describe the problem that I couldn't solve. You're given a simple undirected connected graph with at most 8 vertices. Let's say each edge has a random uniform real weight between 0 and 1. What's the expected value of the minimum spanning tree of this graph? It would seem that with 8 vertices any solution should work, but I've managed to implement one that times out :) Of course, the key to creating a fast solution lies in linearity of expectation. Can you see how to use it?

And finally, Open Cup 2014-15 Grand Prix of Tatarstan happened this Sunday (results, top 5 on the left). I've learned a new beautiful data structure at this contest - a rare event in my old age. Here's the problem that required it: you are given a tree with at most 10

^{5}vertices, where each edge has an integer length, and a sequence of 10

^{5}updates and queries. Each update tells to color all vertices in the tree that are at most the given distance from the given vertex with the given color. Each query requires you to output the current color of a given vertex.

I couldn't invent the data structure during the contest time, nor did I know it in advance, so I've only managed to come up with a O(N*sqrtN) solution for the case where all update distances are equal. But it turns out that the problem is solvable in O(N*logN*logN) for arbitrary updates. Do you know which data structure helps here?

Thanks for reading, and see you next week!

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