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Auto comment: topic has been updated by Errichto (previous revision, new revision, compare).Can we use CenTroid Decompositions for problem D div 2 ??

Yes. We got AC http://codeforces.com/contest/771/submission/25624241

Can you explain how you did it?

For example, the logic of the solution can be like this:

In the centroid decomposition for each centroid we calculate the distances from all its vertices and store it in map. Now we use the theorem that on the path between two vertices there is one and only one centroid that contains these vertices. Then with the help of dfs we go around the centroid decomposition, in the centroid we calculate all . All the vertices of the centroid are marked by those visited, climbing up the centroid tree to recalculate distances from the vertices that have not yet been visited.

Thank you :)

Well organized. take your time.

And why my solution for F works? I tried to prove it for an hour and only after that decided to submit anything at the end. Why is it sufficient to suggest halfplanes between nearest points (sorted by angle) and the furthest from some till 180 degrees?

The editorial contains a proof but it will be refreshed only after the system testing, it's a technical "feature".

Can someone explain how to get the points "almost opposite p_1" for a point p_j?

Sort points by angle and use two pointers.

But you have to sort points by angle with respect to p_i, not with respect to p_1. So naively, you have to sort the points one time for each point, which TLEs.

Everything should be done with respect to

p_{1}. So we sort pointsp_{2},p_{3}, ...,p_{n}with respect top_{1}.Thank you for the response. It seems I didn't read the proof carefully enough. The question is beautiful :)

771C "Tutorial is not available"? edit: nvm

Where is your code for problem C?

All tutorials are already imported from Polygon. At the end of each tutorial you can find my code in C++.

When I click on submission link, it says that I am not allowed to view it.

Should be fixed now.

in problem "Different Names"

"If the i-th given string is "YES", make names i and i + k - 1"

i think you meant if it is "NO"

Thank you, fixed!

Can "Different Names" be explained a little better?

Here's my approach.

Iterate over given array s. If value is YES, make current sequence filled with unique names, otherwise repeat one value. To keep it simple, make first and last value of current sequence same. So, in next iteration, first k-1 names will be unique. You just need to decide whether to repeat first one or make it unique.

Coding approach: First select k-1 unique names. Now, according to values of s[0] to s[n-k], if s[i] is YES, then generate new name and store it as answer of (i+k)th name. Otherwise, store answer of (i+k)th name as (i+1)th name.

25618238

Not allowed to see the code

Can we solve problem B using BFS?

Yes you can

Can u pls tell me how? TIA

Here's an idea: Use your BFS on a starting node. Keep track of every node you could reach on this running of the BFS (use a vector, for example). Now, for each of these nodes, you just need to check if their degree is the size of this vector -1. Do this for every non-visited node. While this condition is true, answer is YES. Otherwise, NO.

Yes its your choice bfs or dfs. Although dfs seems easier.

For the Different Names editorial, it should be

First generate n different names. If the i-th given string isNO, make names i and i + k - 1 equal. Note that it doesn't affect other groups of k consecutive names._/_-^-

I am not allowed to see the code !!!!

My solution to the tree jump problem is O(nk). Precompute total path sums and count for each modulo for each subtree starting from bottom, then solve it again going down by subtracting your subtree from your parent node. Then sum up all the nodes, for runtime of 2nk+n

Code:25623074

mine too but due to smaller value of k our codes are slower than n*k*k due to recursion and extra constants maybe

My code is probably slower because I copy k length vectors around because I'm lazy, probably at least k times faster if I just use an array to store k values.

If not inconvenient to you, can you explain more detailed your solution? "total path sums" — to root? "count for each modulo for each subtree starting from bottom" — ?

There is something I don't understand about 771C - Bear and Tree Jumps solution.

what am I missing here? Isn't that the ceiling function? @.@

And, suppose

k= 5 and a simple tree with edges 1 - 2 and 1 - 3. There are two paths with remainder 1, they havef(L,k) = 4 so I must add 2·4 to the answer?Sorry if this is an extremely stupid question

Yeah, I'd like to understand that too...

Pretty sure that its a typo. What he probably wants to say is that, if we find

Sand if we find , then . Here,countPaths(x,y) denotes number of paths of lengthxwhich give a remainder ofymodulok.You are right. Should be fixed now.

"We know the sum of because its in total"This seems wrong since we're dealing with Integer Division here, right? Don't we have to add

SandP(see my definition above) before we divide?Here we use normal division, not integer. We added

f(L,k) to make normal division and integer division produce same result, and then we use only normal division.Thanks! The typo got me very confused haha

Can someone explain merging and how reminder is calculated in C, please?

http://codeforces.com/contest/771/submission/25630872 Check this one. s is the subtree size (excluding the current vertex), p[u][x] stands for the number of such vetrices v in the subtree of v that the distance between (u, v) equals k * Y + x. d[u][x] is responsible for the sum of Y-s.

Did you understood the merging part? If yes, can you please explain me?

Can someone please explain this solution of problem A(Bear and Big Brother)

submission link : 25610202

Apparently the desire number of years we are looking for is the solution for the following inequation: 3^i * a > 2^i * b, which is equivalent to i > (log(b) — log(a)) / (log(3) — log(2)). That's the O(1) solution for this problem.

I did my solution like that at first, but the double precision caused a WA on some tests. Thus it is better to be careful working with floating number.

cheater caughtanantheparty 25623078 wxh_ak 25623055just find some keywords in both codes like 0x3f enjoy the rating changes :D

Hey dude, you were out of competition in last contest......Why? and How?

Did you cheated?

I confess. This is my way of returning to the right path

Is there some book that describes it?

Where does this problem come from?

At least http://acm.timus.ru/problem.aspx?space=1&num=1371

Problem Source: IX Urals Programming Contest. Yekaterinburg, April 19-24, 2005

Are you talking about floyd warshall algorithm?

It is called the wiener index. Here are some links that might come in handy:

A diploma thesis

An article

Refer the following link.. http://codereview.stackexchange.com/questions/135915/sum-of-all-paths-between-all-pairs-of-nodes-in-a-tree/158790#158790 Hope it'll help you out. :)

I understand how it works, but it was interesting where the problem itself came from =)

Nevertheless, thank you for the link!

Your text to link here...

After I read tutorial for Div1-D (E in official round), I still don't get what exactly I have to do to improve the DP to O(n), especially how to calculate

C_{i}. Can anyone elaborate that part more specifically? Thanks in advance.Exactly :)

Imagine the slower

O(n^{2}) solution that considers only particularO(n) states, includingdp[i][i] (denoted asC_{i}). In my code in the editorial, the function`go_from`

will be run onlyO(n) times.Why will it run only O(n) times?

Read the editorial just after

"It turns out that the following values for each i are enough to solve the problem:". For each indexiwe will consider at most three states:dp[i][i], onedp[i][j], onedp[j][i]. Note that in two latter cases we havei<j.Can somebody explain this line of code in the given solution for the problem 771C Bear and Tree Jumps : int remainder = subtract(i + j, 2 * depth); why 2*depth?? Sorry if it's a silly doubt. Thanks in advance :)

The formula to calculate the distance from u to v in a rooted tree is depth(u) + depth(v) — 2 * depth(LCA(u, v)).It is the similar.

Thank You :)

Can anyone explain to me this part of code in problem C? I don't understand it even after reading the editorial and the comments.

Count_of_subtree[a][i] is no of nodes whose distance from root is i( where i=distfromroot%k ) and belongs to subtree of a including a.

Now we want distance between each pair of nodes. This is calculated using normal, dist(u, v) = dist(u) + dist(v) — LCA(u, v), here all distances are from root.

Thus we see all pairs, 0...K-1 to merge the two sub-trees. need is the f(L, K) that needs to be added( to allow total distance to be divisible by k ).

Now how many times times f(L, K) would be added, it is no of pairs i, j such that need = K- ( dist(i, j) % K ), which is nothing but count_subtree[a][i]*count_subtree[b][j].

In this way we recur up, and do it for all nodes. Here we use K(as MOD) to calculate distance between two nodes to avoid TLE by trying all nodes.

Thanks for your answer. But why do add

`total_subtree[a] += total_subtree[b]`

and`count_subtree[a][i] += count_subtree[b][i]`

after we calculate`answer`

?`total_subtree[a] += total_subtree[b]`

because b is a child of a. To find out the total number of nodes in a's subtree, you recurse into every one of its children (here, b), compute their subtrees' total number of nodes, and then add it all up to a's subtree.`count_subtree[a][i] += count_subtree[b][i]`

because`count_subtree[u][j]`

means the number of paths starting at the root of the whole tree that end somewhere in the subtree rooted in u and have length j (mod k). Hence, if b is a child of a, then the paths described above which end in a's subtree include those that end in b's subtree (because a's subtree includes b's subtree, for all children b of a).Thanks I understand it now

int remainder = subtract(i + j, 2 * depth); Can you explain this line to me? Why is the disance ((i + j) — 2 * depth) % k?

Can someone please try to rephrase the explanation for Div2-D Bear and Tree Jumps? I followed both the text description and the author's code.

I could not understand the part with the two imbricated for loops over k inside the DFS. The author states he's "computting distance modulo k" then proceeds to subtract 2*depth from i + j, which are remainders as far as I can tell. What distance is this? What are we actually trying to achieve here?

Any visual explanation or suggestion for how to make sense of this?

LE: Alright, I think I got it, given two subtrees rooted in vertices

uandvrespectively, and given:number of paths, as well asPufrom the (global) root, having lengthi (mod k), that end inu's subtreenumber of paths, we're trying to compute the lengths of the paths between those vertices inPvfrom the (global) root, having lengthj (mod k), that end inv's subtreeu's subtree where pathsPuend and those vertices inv's subtree where pathsPvend, hence we employ the LCA-like formula...As I_love_Lucy explained above, depth(u) + depth(v) — 2 * depth(LCA(u, v)).

If problem div2 D. Bear and Tree Jumps asked to print the sum of the number of jumps from every node, could it still be done using one dfs? Or two dfs must be applied in that case?

I guess we would need centroids then.

Yeah it could be done using two dfs. I used this exact approach to solve this problem. I calculated the number of jumps for each node to all other nodes, calculated their sum and print(sum/2) ! CODE Time complexity is O(n*k)

Can someone explain to me what's wrong with this submission please: http://codeforces.com/contest/791/submission/25671681

How to solve Bear and Tree Jumps? Can someone please explain in detail?

You can read: the editorial, comments under this blog, my code (provided in the editorial).

In problem 771D, what is the intuition behind coming up with that particular state ? I could not come up with anything like it when I thought of it on my own. My ideas revolved around having states for a n length string and relating it with answer for (n-1) length string but I could not come up with anything that worked. Is this (the one in the editorial) a common way of choosing states for these type of problems ? If not, could you suggest how to learn to come up with them during contests ?

Thinking about some particular moves and how the string looks like after a few moves turns out to be hard (I think you can't get a solution this way). Instead, sometimes it makes sense to think about the final string and this is a way to go in this problem. If you will try to create a dp state to know something about the final string, you should get the idea similar to the one described in the editorial.

I am confused with statement that is given in Bear and Tree jumps's editorial...

" We know the sum of L/k because it is S/k in total. "

it might be S/k ,,, and it might not be equal to S/k...

Even the first statement in the same paragraph says that answer is "AROUND" S/k, not exactly S/k... so am I understanding anything wrong there ???

S is defined as the sum of length (i.e. the sum of L), so S/k is equal to the sum of L/k.

And yeah, the answer is something different than S/k, because the answer is equal to the sum of values L/k rounded up.

okay :)

771A — Bear and Friendship Condition. Why do I get a runtime_error in 11 test? 25713930

It was a very interesting contest. It is sad that it is over

Does anyone have a proof of correctness for the solution for 771B — Bear and Different Names?

I can't seem to understand why the other groups of k consecutive names are not affected if we place a duplicate at the start and end of each window of size k.

I can see that the the duplicate element at the start of each group will be shifted out at the next iteration. But why doesn't the duplicate element at the end affect the other groups?

Start by proving the following. We choose

ndifferent names and we make some two namesiandi+k- 1 equal by replacings_{i + k - 1}=s_{i}. Do you agree that other groups ofkconsecutive names aren't affected?The second part is to notice that we can go from left to right and make that operation many times. The proof will still be correct. What matters is that last

kconsidered names are different, not that all names are different.We choose n different names and we make some two names i and i + k - 1 equal by replacing si + k - 1 = si. Do you agree that other groups of k consecutive names aren't affected?Sorry. But I don't understand why the other groups of k consecutive names aren't affected.

If group is affected it contains both this names (all other names are different), but it means that it contains all the elements in between too. So, it contains at least this k element, so no other elements(because it contains exactly k), but if it's this group, then it's not other group

Let me see if I understood you correctly.

I have drafted a proof by contradiction:

Since all n names are distinct, if a group is affected, that means that it will become a "NO" instead of "YES". For group i to affect another group j, that means that the two equal elements at position x and x + k — 1 in group i will also be in group j but that is not possible because all the names starting from from x + 1 are distinct, contradicting the fact that we will have the same 2 elements in group j.

For 771C — Bear and the Tree Jumps, how did the author manage to derive that the solution for the sum of path length between all pairs of nodes is S/k?

Also, I do not understand why f(L,k) contributes to the answer. How does it help us to minimize path length between nodes?

Please advise.

The sum of path length is

S— this is how I defined it. And note that we don't minimize anything. We usef(L,k) to get rid of the ceil symbol:Hi. Thank you for your quick reply. I understand your explanation but I do not understand why the answer is S/k. Also, if the solution does not minimize anything, why does it manage to compute minimum jumps between all pairs of vertices?

`Also, if the solution does not minimize anything, why does it manage to compute minimum jumps between all pairs of vertices?`

There is only one path between any two vertices in a tree.

Thanks! That clarifies half of my doubts. Do you know how the author got the expression S/k?

Read my answer to drexdelta1's comment (a few comments above).

I read the comment. Unfortunately, that was not the answer I was looking for. I am confused about how the expression S/k was derived in the first place?

Sis defined in the editorial. I divided it bykto get the expressionS/k. It you ask about something else, use some other words.Sorry if I didn't make my question clear. Why do you divide S by k?

Because the statement says that we should find the sum of ⌈

L/k⌉, so computing the sum ofL/kis a reasonable start.Oh. I see. Thank you so much!

who can explain exactly what algorithm in 771E — Bear and Rectangle Strips ????I can't understand much

https://www.hackerrank.com/contests/101hack22/challenges/sum-of-all-distances/editorial I find it helpful for div2 D

maybe we can get a solution for C by using segmenttree.

then we can

we can solve this by dp twice and using seg to tag

771A — Bear and Friendship Condition

-> Why is this below submission failing, anyone ? http://codeforces.com/contest/791/submission/26127531

Why this code does not pass the tests?

27540821

The solution is based on KAN's code.

25627860

Maybe python recursion limit?

In 771C - Bear and Tree Jumps author states "For each subtree we compute the

kvalues: for each remainder modulokhow many paths (starting from the root of this subtree) have this remainder." What does it mean that path starts in some vertex?"A path starts in vertex X" means "Vertex X is one of two ends of a path".

In problem E(Rectangular strips), can anyone explain how to do the update operations in the dp and still manage to get overall complexity of O(n) ?? I am stuck on this part. Thanks!

Hi KAN! In 771A's code 25627860, shouldn't the type of "cnt_edges" be "long long"?

Well this may sound stupid, but in the

Bear and Friendship conditionif we store the edges as directional then the condition, edges = (vertices)*(vertices-1)/2 doesn't work in one of the cases, can anyone help!I dont' know if this was a stupid question or not. Tutorial : "It's a clique if and only if edges = (vertices)*(vertices-1)/2" but in your solution you only check 'edges = (vertices)*(vertices-1)'. what's the difference ? And how could your solution work? Thank you very much!!

I count every edge $$$(a, b)$$$ twice: once from vertex $$$a$$$ and once from $$$b$$$.

Yah, I see. My bad. ^^. Thank you very much.