### gritukan's blog

By gritukan, history, 9 months ago, ,

(Idea — sender , developing — timgaripov)

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(Idea — glebushka98, developing — gritukan)

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 » 9 months ago, # |   0 Auto comment: topic has been updated by gritukan (previous revision, new revision, compare).
 » 9 months ago, # | ← Rev. 3 →   +20 I have a simple solution for D. It has complexity O(NlogNlogX), but can be applied when (1000, 2000) bigger, and changing bonus function.
•  » » 9 months ago, # ^ |   +10 Could you please explain the logic behind your solution?
•  » » » 9 months ago, # ^ |   +51 Assume we need to check if X is enough. Iterate a the number of whole cash paying for 1000. Clearly, we should pay b = (X - 1000 * a) / 2000 times for 2000. And We should do it for first a 1000 and first b 2000, each position we set values to 100/200. The remaining position are filled with their negative cost. Finally, add X - 1000 * a - 2000 * b to first position. If the minimal prefix is non-negative we know X is enough.
 » 9 months ago, # |   +4 Can someone link to the classical problem referenced in D? I can't seem to easily find it on Google.
 » 9 months ago, # | ← Rev. 2 →   0 In Div2-A, when n is even, won't the answer be the opposite of what you have written? If n is divided by 4, the answer should be n/2-1 and n/2+1.
 » 9 months ago, # |   0 Div2 C can be solved with DSU which is faster
•  » » 9 months ago, # ^ |   +1 Your solution isn't faster, you sort the numbers and that's the same complexity as the proposed solution.
•  » » » 9 months ago, # ^ |   +1 Oh you are right but since numbers are up to 1e7 you can sort them in linear time
 » 9 months ago, # |   0 Thanks for a fast and well explained Editorial....
 » 9 months ago, # |   +1 How does Div2C greedy approach work? We can create a bipartite graph with the left vertices as initial times and the right vertices as final times and an edge with the weight of the cost. It is a classic assignment problem. What condition in this question reduces the problem complexity making the greedy solution optimal?
•  » » 9 months ago, # ^ |   0 Call q is a heap contain planes can depart at time t (k + 1 <= t <= n) then the plane depart at time t will be the plane with the largest c in curent heap. And go on... we will have an optimal solution. Sorry for my bad English.
•  » » 9 months ago, # ^ |   +1 Let me try it. Consider we have a set of planes yet to leave. Now, if we choose ith plane to stay from our current set, we incur a penality of Ci. And we want this penality to be minimum. Now think about it and you will understand.
•  » » » 9 months ago, # ^ |   +1 Let's say at some point we have two planes x and y with cx = 2, cy = 5. Intially x is supposed to depart at 4 and y is supposed to depart at 3. Currently we are at 5th time unit. The total cost if we choose x first will be 2 * (5 — 4) + 5 * (6 — 3) equals 17. If we choose y, it will be 5 * (5 — 3) + 2 * (6 — 4) = 14. But the minimal penalty is given by cx at 5th time unit. That is not leading to the optimal solution. Correct me if I am wrong.
•  » » » » 9 months ago, # ^ |   0 I don't understand you. What we want is minimum cost and we will get it by taking cy cost first because it is bigger, and as you say this is optimal.
•  » » » » » 9 months ago, # ^ |   0 "We will show that following greedy is correct: let's for each moment of time use a plane, which can depart in this moment of time (and didn't depart earlier, of course) with minimal cost of delay." Does this not mean that the plane x gives us the minimal cost?
•  » » » » » » 9 months ago, # ^ |   0 I am having trouble understanding the editorial. But the solution i mentioned in first comment is correct. You can refer to my solution if it helps.
 » 9 months ago, # | ← Rev. 2 →   0 I think there is a mistake in the second solution of A. In line 2, it should be b=(n/b)+1; not b=(n/b);
•  » » 9 months ago, # ^ |   0 If you look carefully there is a "ceiling" sign. As n is a odd number, n/2 will always be x.5, here x is (n-1)/2. And ceiling of x.5 is x+1.
•  » » » 9 months ago, # ^ |   0 Thanks.
 » 9 months ago, # | ← Rev. 2 →   0 Can anyone please explain the solution of problem 853B - Jury Meeting more?
 » 9 months ago, # |   0 Can someone explain DIV2B.I didn't get the idea from the explanation in the editorial
 » 9 months ago, # |   0 Auto comment: topic has been updated by gritukan (previous revision, new revision, compare).
 » 9 months ago, # |   +2 In problem Div2D, you say that: "Obviously, each member of the jury needs to buy exactly two tickets — to the capital and back." Does it mean that two jury members can't fly in the same plane? o.O
•  » » 9 months ago, # ^ | ← Rev. 2 →   0 EDIT: we7d right, they obviously can't — silly me ^_^
•  » » » 9 months ago, # ^ |   +1 they can't because only one member in each city and flights are only from city x to city 0 and vice versa
•  » » » » 9 months ago, # ^ |   0 Thx for answer, now it's much simpler task.:D Previously I thought that flights like 1 10 8 5000 and so could exist. Silly me :(
 » 9 months ago, # |   0 My same solution which passed during system testing if I submit again now gives TLE. Is it that system testing servers are better or any other reason.
 » 9 months ago, # |   0 can someone explain the greedy proof in div2 C could not understand it from editorial like in this statement "Let x be plane with minimal cx, such ax ≠ bx. At any moment greedy algorithm takes avaliable plane with lowest cx, so ax < bx. " what is the proof for ax
•  » » 9 months ago, # ^ |   0 I think there is a better explanation. You can think about this, how to get the best answer? Supposing that we have an initial arrangement, how to optimize it. First of all, we will come up with an idea to swap some planes' departure time. If plane i departs at ti, plane j departs at tj, then the cost of them two is ci * (ti — i) + cj * (tj — j) = ci * ti + cj * tj — ci * i — cj * j, and ci * i — cj * j is a constant. So our goal is to minimize ci * ti + cj * tj, and it's obvious if ci > cj, ti > tj, we should swap ti and tj. So for every moment t, we should choose the highest cost from those who can take off at this time. And this is my code.http://codeforces.com/contest/853/submission/30136836
•  » » » 9 months ago, # ^ |   0 tnx ,good explanation.
•  » » » 9 months ago, # ^ |   0 nice explanation, thank you.
 » 9 months ago, # |   0 In Div1C : can anyone please elaborate how to find the number of points in the query rectangle efficiently?
•  » » 9 months ago, # ^ |   0 Try persistent segment tree, where root[i] represents the root of segment tree for all elements having a row <= i.
•  » » » 9 months ago, # ^ |   0 You can also solve the queries offline and you won't need persistent tree.
•  » » » » 9 months ago, # ^ |   0 can you please explain how we can answer the queries offline?
•  » » » » » 9 months ago, # ^ |   0 Sort the queries and marked squares from left to right(column),and use segment tree to maintain it.When adding a suqare(i,p_i),add 1 to p_i in the segment tree,and you can find the answer when asking a query.
 » 9 months ago, # | ← Rev. 3 →   +38 Div 1 B can be easily solved in linear time. Fix a window [l, r] of size K. Then all the jury must go to city 0 before time l and must return to their cities after time r. So let's find L[i] = minimum cost of making all jury travel to the capital at time i or less and R[i] = minimum cost of making all jury return to their cities at time i or later. Since all times are  ≤ 106, we can keep a list of flights for each time. First, go in increasing order, processing flights and keeping the minimum value for each jury as well as the total minimum cost. Then do the same, but in decreasing order of time. All this process is linear. Once we have calculated L and R, we simply try out all possible windows [l, r] of size K. The cost for a fixed window is L[l - 1] + R[r + 1]. Checking all windows is also linear.
•  » » 9 months ago, # ^ |   0 Can you provide a solution using this approach?
•  » » » 9 months ago, # ^ |   +2 Yes, here's my submission: 30179926
•  » » 9 months ago, # ^ |   0 The exact complexity of this solution is O(Max(di) + m)
•  » » 9 months ago, # ^ | ← Rev. 2 →   0 I'll provide my non-linear two-pointer version without multiset (actually, it's replaced by inverse of two mini-stacks) for clarity: 30313607
 » 9 months ago, # | ← Rev. 2 →   +5 In Div1-D, what is "next after i interesting day"? Did you mean "the interesting day next after i"?
 » 9 months ago, # |   +13 Spent two hours to debug 1C, just to find out that p[i] is supposed to be row p[i] column i instead of row i column p[i]...
•  » » 9 months ago, # ^ |   0 same here :(
•  » » 9 months ago, # ^ |   0 Same! Though I've even read this right but swapped rows and columns in queries in my mind.
 » 9 months ago, # |   +1 is there any explanation about "2d-tree" or whatever you use to reach O(q·logn) complexity in 1C/2E ?
•  » » 9 months ago, # ^ | ← Rev. 4 →   +6 Yes, actually with a 2D Tree, you get O(Q * log2N). To reach O(Q * logN), you can use an approach similar to sweep line supported by a BIT. The process would be as follows. Sort the queries by top row. Process queries from top to bottom row by row. For each row, do the following in order: Query the BIT for every query in this row, add column of this row to BIT. To query the BIT, consider we're at row i, and we're processing query (i, j1) -  > (i2, j2), then we'll query the BIT in the range (1, j1 - 1) (squares up and to the left of query rectangle) and in the range (j2 + 1, N) (squares up and to the right of query rectangle). Since we only add the column of a row once we've processed all queries of that row, and since we're processing rows from top to bottom, we know that all data in the BIT will be from rows above the current one. Sort the queries by bottom row now and do the same as in previous step but this time from bottom to top.
•  » » » 9 months ago, # ^ |   0 Is it even possible to get a 2D tree under the memory limit?
 » 9 months ago, # |   0 Can someone provide a formal proof that Div2C accepts a greedy solution? for instance, using a matroid?
 » 9 months ago, # | ← Rev. 3 →   0 in Div1-C I tried persistent segment tree to solve the queries in O(n*logn)but I failed at test 4 , any help or hint please codeUPD : problem solved , I had to swap rows and columns from the input ...
•  » » 9 months ago, # ^ |   +3 I solved it during competition using persistent segment tree. You can check my code 30147451
 » 9 months ago, # | ← Rev. 2 →   0 Why is Div1C offline search O(qlogn)?I think it should be O(qn)process time and O(1) query time.
•  » » 9 months ago, # ^ |   0 Sorry, I understand now.
 » 9 months ago, # | ← Rev. 2 →   0 http://codeforces.com/contest/854/submission/30213879 Please help,why runtime error on testcase 49,although allocated memory properly :(. Thanks
 » 9 months ago, # |   0 Can anyone please explain Div2-D/Div-1 B. I'm not able to understand much from the editorial.
 » 9 months ago, # |   0 I can't understand the proof for Div. 1 D. Why is it always optimal to have x_i <= 3000?
 » 9 months ago, # |   0 854 B Please Explain this part. Otherwise, if k > x, assume that apartments with indices 2, 5, 8 and so on are occupied, so any room has at least one inhabited room adjacent to it. Therefore number of good apartments is equal to number of vacant apartments and is equal to n - k.
•  » » 9 months ago, # ^ |   0 One neighbour can cover two adjacent apartments. If there are more than x occupied apartments, one can divide the apartments in groups of 3 and assign them to the middle of each group (hence the 2 in 123, 5 in 456, etc.)
 » 9 months ago, # |   0 Auto comment: topic has been updated by gritukan (previous revision, new revision, compare).
 » 9 months ago, # |   +5 Actually Div.1 C can be solved in O(q log^2 n) and get Accepted...
•  » » 9 months ago, # ^ |   0 haha you're right, I have just changed oset for a vector and got accepted