The Dec 17 - Dec 23 week contained a surprise extra Open Cup round: Open Cup 2018-19 Grand Prix of Xi'An (results, top 5 on the left, official round results, overall Open Cup standings so far). The Moscow SU team solved three problems in the last hour, including problem A which was solved by only three teams out of 150 attempts, and ended up winning by a problem. Congratulations on the great performance!

Codeforces Round 528 followed on Sunday evening (problems, results, top 5 on the left, analysis). It was mnbvmar's turn to be the only contestant to solve the hardest problem F, and thus win with a huge margin. Congratulations on the victory!

In my previous summary, I have mentioned several problems. The first one came from TopCoder: you are given a rooted tree with 100000 vertices. Each vertex has a demand

Let's look at the top-most vertices with non-zero value in the solution (such that all their parents and ancestors have zero value). Those vertices are not parents of one another, and thus they satisfy exactly one unit of demand of themselves and their descendants. If we now reduce their value by 1 and repeat, we can split into our solution into "waves" each wave satisfying one unit of demand in all remaining vertices with non-zero demand.

We can also notice that each wave can be constructed as follows: first, start with a wave that contains only the root vertex. Then repeatedly do the following: take any vertex with zero demand that is in the wave, remove it from the wave but add all its children. Each such operation changes the cost of the wave by the sum of costs of the children minus the cost of the removed vertex. Let's call this value the

Now we'll build the waves one by one while always maintaining the minimum cost wave. Initially such wave is just a root vertex. As soon as the number of waves we've built reaches the demand of a vertex, this vertex becomes available for replacement, which would change the weight of the wave by its delta. If the delta is positive, we don't need to do anything. If the delta is negative, and the vertex is in the minimum cost wave, then we need to do the replacement. If the delta is negative but the vertex is not yet in the minimum cost wave, we will need to do its replacement as soon as it gets into the wave, in other word we should add its delta to the delta of its parent. If the delta of the parent becomes negative, then we propagate it to its parent, and so on, so that at every moment we have only the optimal wave plus some non-negative deltas remaining.

In order for this to run fast, we need to directly propagate to the closest non-zero ancestor instead of going through all parents in the chain to it, which can be accomplished by keeping a disjoint-set union data structure where each set has a vertex with non-zero delta as its representative, and contains all vertices with zero deltas which have it as the closest non-zero ancestor.

This was quite tricky to come up with, but the implementation was rather simple and without any corner cases or data structures more complex than disjoint-set union.

The second problem I mentioned came from AtCoder: you are given

The first step is to think when is this actually impossible. The sample cases provided a helpful example: there was a case where there were two vertices that belonged to just one set, and moreover to the same set. Since only one edge will come out of this set, we will not be able to connect both of those vertices to the rest of the graph.

This obstacle can be generalized as follows: if there's a set of vertices of size

But wait, we have seen such obstacles before! Consider the bipartite graph where the left part is the input vertices and the right part is the input sets, with edges connecting each set with its members. If there's no obstacle as described above, then by Hall's theorem in this graph there's a matching covering any (

Let's find any such matching, it covers some (

Now we can see that if we take any vertex covered by matching, its previous vertex in the augmenting path tree (which will be in the right part, so represent a set), and the set's previous vertex in the augmenting path tree (which will be in the left part, so represent a vertex), then we get two vertices belonging to the corresponding set, and those

Finally, there was an Open Cup problem: two players are playing a game on a rooted tree with 100000 vertices. They make moves in turn. In each move of the first player, she colors one of the leaves of the tree black (initially all leaves are colored white). The root of the tree also initially contains a token. In each move of the second player, she can move the token from a vertex to any adjacent vertex. The second player wins if she moves the token to a leaf that is still white. The first player wins when all leaves are colored black and the second player has not yet won. Who will win if both players play optimally?

First, we can notice that it doesn't make sense for the second player to go up, as they will end up in a strictly worse state than before. So for each subtree, there's at most one moment in the game when the token enters this subtree, and then the game continues just in this subtree.

Moreover, suppose that the first player has made

Thanks for reading, and check back tomorrow!

Codeforces Round 528 followed on Sunday evening (problems, results, top 5 on the left, analysis). It was mnbvmar's turn to be the only contestant to solve the hardest problem F, and thus win with a huge margin. Congratulations on the victory!

In my previous summary, I have mentioned several problems. The first one came from TopCoder: you are given a rooted tree with 100000 vertices. Each vertex has a demand

*d*and a cost_{i}*c*. Your goal is to assign some non-negative value_{i}*v*to each vertex so that the demand of each vertex is less than or equal to the sum of values on the path from this vertex to the root of the tree, and the sum of_{i}*c**_{i}*v*is minimized._{i}Let's look at the top-most vertices with non-zero value in the solution (such that all their parents and ancestors have zero value). Those vertices are not parents of one another, and thus they satisfy exactly one unit of demand of themselves and their descendants. If we now reduce their value by 1 and repeat, we can split into our solution into "waves" each wave satisfying one unit of demand in all remaining vertices with non-zero demand.

We can also notice that each wave can be constructed as follows: first, start with a wave that contains only the root vertex. Then repeatedly do the following: take any vertex with zero demand that is in the wave, remove it from the wave but add all its children. Each such operation changes the cost of the wave by the sum of costs of the children minus the cost of the removed vertex. Let's call this value the

*delta*of the removed vertex.Now we'll build the waves one by one while always maintaining the minimum cost wave. Initially such wave is just a root vertex. As soon as the number of waves we've built reaches the demand of a vertex, this vertex becomes available for replacement, which would change the weight of the wave by its delta. If the delta is positive, we don't need to do anything. If the delta is negative, and the vertex is in the minimum cost wave, then we need to do the replacement. If the delta is negative but the vertex is not yet in the minimum cost wave, we will need to do its replacement as soon as it gets into the wave, in other word we should add its delta to the delta of its parent. If the delta of the parent becomes negative, then we propagate it to its parent, and so on, so that at every moment we have only the optimal wave plus some non-negative deltas remaining.

In order for this to run fast, we need to directly propagate to the closest non-zero ancestor instead of going through all parents in the chain to it, which can be accomplished by keeping a disjoint-set union data structure where each set has a vertex with non-zero delta as its representative, and contains all vertices with zero deltas which have it as the closest non-zero ancestor.

This was quite tricky to come up with, but the implementation was rather simple and without any corner cases or data structures more complex than disjoint-set union.

The second problem I mentioned came from AtCoder: you are given

*n*vertices, and*n*-1 sets of those vertices, each of size at least 2. Your goal is to pick exactly two vertices in each of the*n*-1 sets in such a way that if we draw the*n*-1 edges connecting each chosen pair to each other, we get a tree.*n*is at most 100000, and the total size of the given sets is at most 200000.The first step is to think when is this actually impossible. The sample cases provided a helpful example: there was a case where there were two vertices that belonged to just one set, and moreover to the same set. Since only one edge will come out of this set, we will not be able to connect both of those vertices to the rest of the graph.

This obstacle can be generalized as follows: if there's a set of vertices of size

*k*which is less than*n*such that the number of input sets containing at least one of those vertices is less than*k*, there's no solution.But wait, we have seen such obstacles before! Consider the bipartite graph where the left part is the input vertices and the right part is the input sets, with edges connecting each set with its members. If there's no obstacle as described above, then by Hall's theorem in this graph there's a matching covering any (

*n*-1)-element subset of the left part.Let's find any such matching, it covers some (

*n*-1)-element subset of vertices. The fact that all other (*n*-1)-element subsets can also be covered by a matching means that if we run one more step of the maximum matching algorithm, it must find augmenting paths from the only remaining uncovered vertex to all covered vertices (so that flipping this augmenting path would result in the solution for the corresponding (*n*-1)-element subset). These augmenting paths form a subtree of the residual network.Now we can see that if we take any vertex covered by matching, its previous vertex in the augmenting path tree (which will be in the right part, so represent a set), and the set's previous vertex in the augmenting path tree (which will be in the left part, so represent a vertex), then we get two vertices belonging to the corresponding set, and those

*n*-1 pairs are exactly the solution to our problem as they necessarily form a tree!Finally, there was an Open Cup problem: two players are playing a game on a rooted tree with 100000 vertices. They make moves in turn. In each move of the first player, she colors one of the leaves of the tree black (initially all leaves are colored white). The root of the tree also initially contains a token. In each move of the second player, she can move the token from a vertex to any adjacent vertex. The second player wins if she moves the token to a leaf that is still white. The first player wins when all leaves are colored black and the second player has not yet won. Who will win if both players play optimally?

First, we can notice that it doesn't make sense for the second player to go up, as they will end up in a strictly worse state than before. So for each subtree, there's at most one moment in the game when the token enters this subtree, and then the game continues just in this subtree.

Moreover, suppose that the first player has made

*x*moves into this subtree before the token enters it. Then it's clear that there's some boundary*b*such that when*x*<*b*the second player wins if the token enters this subtree, and when*x*>=*b*the first player wins if the token enters this subtree. This finally gives us a linear-time dynamic programming solution: we will compute this boundary*b*for each subtree going from the bottom to the top.Thanks for reading, and check back tomorrow!

I know that you will probably never read this, but worth a shot:

ReplyDeleteCan you make a post about, what path should a new competitive programmer should take or more like what path would you have taken if you are given a chance to go back in life and relive your years to be where you are right now, like what resources and what kinda competition would you have opted for.

I do read this, but frankly, I have no idea what to answer, sorry :(

DeleteAlright, good to know that you read this. Give it thought if sometime you feel like it.

DeleteThanks for the response.