### Kniaz's blog

By Kniaz, history, 6 years ago, translation,

I'm sorry for a delay with publishing the editorial.

Problem author: MikeMirzayanov.

Problem author: MikeMirzayanov, Kniaz.

Problem author: MikeMirzayanov.

Problem author: MikeMirzayanov, Kniaz.

Автор задачи: Kniaz.

Автор задачи: MikeMirzayanov.

I want to thank Mikhail Piklyaev (awoo) for translation of tutorial!

• +70

 » 6 years ago, # |   +2 What about the problems E and F?
•  » » 6 years ago, # ^ |   +2 I think they should be published soon. You can also ask me if you would like some tips!
•  » » » 6 years ago, # ^ |   0 can you help me with implementations of C and D ?
 » 6 years ago, # |   0 Can anyone explain C and D in a simpler language? That would be a great help. Thanks in advance!
•  » » 6 years ago, # ^ |   0 I am not sure about the approach of C in the editorial, but here's my O(n) approach:Observation: a[] can be converted into b[] by grouping a[] in non-overlapping segments.If we start from the left, it is obvious that the answer exists only if the prefix sum (the sum of the first few elements) is equal to b[0] since a[] only consists of positive numbers. This fact could also be inducted to other segments if we consider that previous segments were cut away from a[] and treat the next element as the start of the segment for b[1...n].Now the only thing we have to care is whether the monster could start the momentum by eating others. By greedy we will assume that we are starting the eating processes of each group from the largest monster (as there are no non-positive elements), if you can't start the eating process (note that the eating process is only needed when the size of the group is larger than 1).Remember to store the correct positions of the monster during the eating process as the editorial has mentioned.The editorial for D pretty much nailed it and I don't think my explanation could beat that. Perhaps you should take a look at other's code.
•  » » » 6 years ago, # ^ |   +1 For problem C:You should select one largest monster which isn't surrounded by two equal monsters.To avoid being bothered with changes of position, you may process from the last block to the first block.Check my code: 21929693
•  » » » » 6 years ago, # ^ |   0 Tha is exactly my approach. In each prefix which has the correct sum simply found the maximal element which has a neighbor element strictly less than itself. Having its index and the information about the position of the lesser element it's easy to print all the eating sequences for that specific monster. My code http://codeforces.com/contest/733/submission/22160409
•  » » » » 6 years ago, # ^ |   0 Assuming you have the largest monster, how do you know whether to delete to the left or to the right? And how many elements to delete?
•  » » » » » 5 years ago, # ^ |   0 You should check and store min neighbour for current giant monster. See "getNeigh" function:http://codeforces.com/contest/733/submission/27457231
•  » » » 21 month(s) ago, # ^ |   0 haleyk100198 can you tell me the mistake in this solution 93391656 of Problem C(DIV2)[contest:378][problem:Epidemic in Monstropolis] THNAKS IN ADVANCE.
 » 6 years ago, # |   0 How to solve C in O(n)- challenge ?
•  » » 6 years ago, # ^ |   0 Like this. 21968223
•  » » 6 years ago, # ^ | ← Rev. 3 →   0 Not very different than what is explained in the editorial. For a group S that needs to get reduced to one monster, find a monster which has maximum value and doesn't have an equal neighbour(if there is no such monster, then no solution). Once he eats this unequal neighbour, he can eat everything else in any order. So you'll either eat everything to the left and then everything to the right, or eat everything to the right and then everything to the left. And this way it's easy to keep track of his position in the final array, without simulating the transformations on the array.
•  » » 6 years ago, # ^ |   0 try to use two pointers one starting at first of 1st array and other at start of 2nd array.first time traverse whole b array to check if solution exists.if doesnt exist print NO and exitelse if yes traverse again second time through 2 nd array and print the solution.for printing the solution,follow a greedy approach finding the maximum in it and assume it eats all the others in that unionhere is the link to my solution of it:http://codeforces.com/contest/733/submission/21949174
 » 6 years ago, # | ← Rev. 3 →   +19 Ooops.. looks like my solution was incorrect. Took us a while to figure it. Sorry to everyone who wasted time trying to figure it out. I'll think more carefully next time.Check the last edit if you wanna read the incorrect solution.
•  » » 6 years ago, # ^ |   0 Can you prove its correctness? I'm not sure about this.
•  » » » 6 years ago, # ^ | ← Rev. 3 →   0 I presume you are suspicious that the Kruskal is incorrect, now that edges change their initial values?Indeed, I see that I wasn't rigorous enough about it. Looking over the Kruskal proof, you can make it still hold if edges that you haven't added yet to the MST can only increase in cost during the algorithm(which is what happens in my algorithm. The edges of category 2 increase in value until they reach category 1, but we fix their cost if we add them to the MST before that).Later Edit: No, it doesn't work.
•  » » » » 6 years ago, # ^ | ← Rev. 3 →   0 Hi Mikester, I understood the following.For a tree with n — 1 edges, the dissatisfaction will be Sum(w(i)) — S/c where c = minimal weight of edges in the tree.Now, if I fix c, which is equal to weight of some edge, I need to minimize sum(w(i)), that is find the minimal spanning tree containing edge with the weight c and all other edges have weight >= c.Can you explain after that point? I mean how do I optimize it?Hmm, looks like I am close to the solutionhttp://codeforces.com/blog/entry/9570
•  » » 6 years ago, # ^ |   0 I can't understand what exactly are you representing with extra[i] array.Can you please elaborate a bit?
•  » » » 6 years ago, # ^ |   +3 extra[i] is in this instance of the problem -S/c[i]
•  » » 6 years ago, # ^ | ← Rev. 2 →   +5 It's incorrect. consider following case4 40 0 1 10100 100 5 11 21 32 32 410 your solution gives sum 1 and selects edge 3, 1, 4, but the optimal solution should give sum 0 and select 1 2 4.
•  » » » 6 years ago, # ^ |   0 Wow, you're right.. Good job! I thought I had a cool insight about Kruskal too. What I said about it was incorrect. My apologies.
 » 6 years ago, # |   +8 Implementing E was really painful, and had so many errors. Finally felt peaceful after getting AC. After seeing many people using binary search, I thought that I was making the problem very hard for myself but it seems that the editorial explanation is exactly what I have done(not sure). I considered all the cases separately in query function, and always made sure that when I am calculating the answer for s[i], it is always U. Otherwise I find the answer in the inverse string, where it would be U.
•  » » 6 years ago, # ^ |   0 What's the approach using binary search?
 » 6 years ago, # |   +5 I have a (possibly dumb) question about Problem F: Why would it be optimal, given a fixed selection of main roads, to reduce the dissatisfaction value of just a single road? Would that still hold if the budget is not divisible by the unit repair cost of the selected road? In that case, would it be more optimal to distribute the budget over multiple roads whose total repair cost could make better use of the budget?
•  » » 6 years ago, # ^ |   -11 That's a dumb question indeed :P . What you are saying is like this .. 19/3 can be greater than 15/5 + 4/4 . just because you are using all the money it doesn't provide you with more number of dissatisfaction units(because the cost is also high). Mathematically it goes like this.. a*x + (a-1) = Sigma(Bi*Yi). where each Bi is greater than 'a' implies Sigma(Yi) <= x. It's easy to prove :) . However my submission was giving wrong answer on test 75. can anyone please check ? what's special about testcase 75? here's my submission- [submission:21965336]
•  » » 6 years ago, # ^ | ← Rev. 2 →   0 Even if S % c[i] != 0, floor(S/c[i]) is maximized when c[i] is minimal. There's a simple argument: Assume you allocate some budget to road j such that c[j] > c[i]. Note that this budget could be moved to road i at no loss (indeed, possibly at a gain).
•  » » 6 years ago, # ^ |   0 Thanks @Artemis_Fowl and @OMGTallMonster for explaining! That was indeed dumb.
•  » » 6 years ago, # ^ | ← Rev. 4 →   0 I was also confused about that one. The OP says that fix an edge to be reduced and find MST corresponding to that edge.Now, it would be correct if the cost of that reducing edge was minimal. But what if there was an edge with smaller cost of reducing in that same spanning tree. Then we could reduce the dissatisfaction by reducing that edge. So I thought this, fix a road, find a MST containing that road, where all edges have cost of reduction >= c, where c is the cost of reducing dissatisfaction of the edge we initially chose. But it turns out I don't have to do actually that. What we need to do is that if there are multiple spanning trees of the same weight, then for all those spanning trees, we should choose the tree containing the edge with smallest reducing cost. So, we select an edge P, find MST corresponding to that edge. Suppose w(P) > w(Q) (w(p) = cost of reducing dissatisfaction of the edge p) for some other edge in the MST, then we know that MST(P) >= MST(Q)(either MST(P) = MST(Q), or at least one edge is different with smaller weight, so MST(Q) < MST(P) ), therefore we know that when we finally find MST corresponding to Q, the answer is obviously better and hence no need to find MST with weight >= c.
 » 6 years ago, # |   +5 Really liked the problem C.Need more problems like this.I didn't think that it would be that easy.
 » 6 years ago, # |   +3 Problem F should add this case:5 63 3 3 3 4 9100 100 100 100 2 11 22 33 44 51 51 57The answer is 10. But my solution got 11 and passed all system test cases.
•  » » 6 years ago, # ^ |   0 They should put more corner cases in the System test... We don't expect weak test cases from Codeforces
 » 6 years ago, # | ← Rev. 3 →   0 problem F : Precalc takes ? or for Binary lifting ?
•  » » 6 years ago, # ^ |   0 O(nlogn)
 » 6 years ago, # |   0 Thank you for such intersting problemset! The problems(especially E and F) worth thinking & solving :)
 » 6 years ago, # | ← Rev. 2 →   0 Can someone help me in finding the bug in my submission for problem C . I followed the same approach in the editorial but I'm getting a WA at test 64 . The checker comment is Wrong answer Can not eat not less . Since it's a large input I couldn't find the bug.I would be very thankful if someone could help me out.My submission with comments : LINKUPDATE : I managed to find the bug and solved it.
 » 6 years ago, # |   0 I think I'm missing something in the complexity argument for F. How do you add an edge to a tree and delete some edge in the cycle without paying O(m) to reverse the links in the path between the added edge and the deleted edge? (The links are directed, and thus must be reversed, based on the LCA binary lifting article elsewhere in the blogs.) Which would raise the total time complexity to O(n log n + m^2).Example: Tree rooted at A, with a child B. C is a descendant of B at depth O(m) [since this is an unbalanced tree]. Leaf D with depth O(m), not in B's subtree. Need to add an edge between C and D: LCA is A; suppose the heaviest edge in the cycle is (A,B). Then we need to reverse links on path B..C with total cost O(m).
•  » » 6 years ago, # ^ |   0 You don't have to do the whole operation. Instead, you just find the LCA to identify the unique cycle that will be created in log time, and then use binary lifting to identify the best edge to be removed.(When you create the binary lifting lookup table, you lift it in the way that the edges are directed to the parents, that's the way you want to do it anyways, you don't need to reverse it everytime you lookup the table then.)
•  » » » 6 years ago, # ^ |   0 I got all that, and now after typing a long reply explaining my question better I think I was actually confused about something else =)I thought that after you found the edge to delete, you actually needed to store and use the new MST with that edge "forced", and having new binary lifting tables for the new MST. That would mean throwing away all the precomputation work you did before because it's invalid; that's the parent/child reversals I was talking about. Obviously that would be O(m) precomputations which is no good.But I think I see now that you don't store the new MST (well, you might store the minimal one); you just have a single MST you reuse over and over with a "hypothetical" delete/insert at every step.I think part of my confusion was that the time bound given seems achievable with LC trees supporting path reversal ("evert" operation in Sleator and Tarjan), which would allow you to actually modify the tree over and over.
 » 5 years ago, # | ← Rev. 2 →   0 Sorry, maybe I don't understand problem B(div2) but in 3d sample which is 6/ 5 9/ 1 3/ 4 8/ 4 5/ 23 54/ 12 32/ "/" means new string We can change places of 23 and 54 and we'll have optimal answer(then suml=80 and sumr=80) So why answer that we can't do anything? Please help in understanding problem!!
•  » » 5 years ago, # ^ |   0 I'm so stupid.........
 » 5 years ago, # |   0 Can anyone prove the correctness of given solution for F ?
 » 7 months ago, # |   0 what about the 11th test case in problem A? I didn't understand the logic