Endagorion's blog

Technocup 2017 Finals and Codeforces Round #403 Editorial
By Endagorion, history, 18 months ago, translation, In English,
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18 months ago, # |
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Auto comment: topic has been translated by Endagorion (original revision, translated revision, compare)

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18 months ago, # |
← Rev. 2   Vote: I like it +2 Vote: I do not like it

Why it shows Tutorial is not available ?

UPD: It's available now :)

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18 months ago, # |
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Can anyone explain how Div2 B can be solved using Ternary Search

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    18 months ago, # ^ |
    ← Rev. 4   Vote: I like it +20 Vote: I do not like it

    I have done this.

    We analyze total time in relation to position, I will call it p selected. Time of each point x with speed v is abs(x - p) / v. If we make p vary, as v and x are fixed we'll see an abs function

    Which is the total time? The worst of all those times. formally max(v1, v2, ..vn). If we make drawings we can see that if we plot the total time in function to the position selected, that graphic has no local maximum, and the left and right sides tend to positive infinite. So we observe that as the function is continuous we need to have a unique local minimum somewhere in the center of the graphic (otherwise the sides can't tend both to infinity). There can't be more than one minimum, because otherwise we would need a maximum to exist and we accepted it didn't exist. The left side of the minimum must be strictly decreasing (as it comes from positive infinity), and the right side strictly increasing (as it goes to positive infinitive), and so, the graphic is something like this

    Note that the left derivative is negative and the right one is positive. Which means we can make a ternary search on the point when it changes, from negative, to positive, that will be a local minimum, as the only one, the global one, hence the solution of the problem.

    Evaluating the function will be O(n), so total complexity the product of O(n) and your number of ternary search iterations (of logarithmic behavior)

    25253744

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      17 months ago, # ^ |
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      Great explanation, I have seen this problem on virtual contest and tried to solve with similar approach. Intuitively I felt that there should be only one point where the required time is minimum. But couldn't solve during the contest, after reading this comment everything clicked and got AC.

      By the way, I have checked your solution and why use mabs? cpp has its own fabs which does exactly same

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        17 months ago, # ^ |
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        By ignorance :) Sometimes, during a contest it is faster to write these simple functions rather than open google haha. Also sometimes I decide to avoid to use some c++ functions to reduce the risk of generate unexpected algorithm complexity, however this is not the case :)

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18 months ago, # |
  Vote: I like it +3 Vote: I do not like it

for Div.2 D (Div.1 B)
somebody says try all ai, if there is the conflict then try all bi. if these two cases both have conflict then the answer is NO.
like submission 25322923 and try the input: 

  4 
  ABC DDD 
  ABC EEE 
  QWE FFF 
  QWR FFF
get the output: NO but isn't 
 ABD, ABE, QWE, QWR
a solution?
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    18 months ago, # ^ |
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    Here we don't try all b_i when there is some conflicts for a_i. But only those b_i whose corresponding a_i conflict with each other.

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      18 months ago, # ^ |
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      I know but the somebody's code doesn't seems like that.

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18 months ago, # |
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Can someone please help me understand the editorial of problem 782B — The Meeting Place Cannot Be Changed. how to iterate for all t's and how to apply binary search.

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18 months ago, # |
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782B can be solved without binary search. Working time is much better comparing to binary search solutions. heuristic with enumeration all pairs & O(n*log(n)) sorting + O(n) scan

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18 months ago, # |
← Rev. 2   Vote: I like it +8 Vote: I do not like it

in div2 E, why is it guaranteed that the tour exists? and even if it doesn't which, why is writing the nodes in that way is guaranteed to generate the answer?

edit: i understood the solution, but why is it called euler tour?

and is it me, or mentioning that traversing an edge twice is fine had to be mentioned in the statement?

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    18 months ago, # ^ |
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    would someone tell me if we can visit the same edge twice? and if so, why is it called "euler tour" in the editorial?

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      18 months ago, # ^ |
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      Yeah, we can visit the same edge twice. Suppose in the dfs ordering of nodes, we backtrack all edges (in the dfs tree) once.

      -> Total edges traversed = 2*(n-1) -> If we divide into k parts, we get [2*(n-1)/k] <= [2n/k] -> We can always assign dfs (preorder) accordingly.

      I have no idea why it's called Euler tour, since that means we should be visiting each edge once, and end up on the beginning node.

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        18 months ago, # ^ |
        ← Rev. 3   Vote: I like it +10 Vote: I do not like it

        You should code simple recursive algorithm for Euler tour. But write vertice twice first time on visiting and another time after recursive return (on each return to the vertice). This will produce required sequence.

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18 months ago, # |
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In div1D, what is meant by "optimizing boolean multiplication with bitsets" and how do we achieve this optimization?

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18 months ago, # |
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Can you figure out what is wrong with my code in problem 781B (wrong in test 16)? 25333425

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18 months ago, # |
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what meaning "The route is a closed polygon line in the place, with all segments parallel to one of the axes" in Intranet of Buses? help on Intranet of Buses!

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18 months ago, # |
← Rev. 3   Vote: I like it +13 Vote: I do not like it

I'm getting 'Tutorial is not available'.

And tutorials for some other rounds are also not available currently see this

KAN MikeMirzayanov please fix this.

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18 months ago, # |
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In Div1 B (football league problem) what is the mistake in my algorithm? I really don't see the bug. http://codeforces.com/contest/780/submission/25485848

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18 months ago, # |
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can someone please find why am I getting runtime error in python2.7 in problem C div 2?

Here is my code, a big thanks in advance

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18 months ago, # |
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in task d(div 2) test 5 aaa b aaa c aaq z aab z aab o craches lots of solutions. The result is probably YES aab,aac,aaq,aaz,aao but it isn't)

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17 months ago, # |
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What's the idea behind giving 5 seconds for div 2 B if the intended solution is 46 ms 26456429 . Is it to allow for N^2 solution or to trick people into attempting N^2 solution?

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    17 months ago, # ^ |
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    I believe it was done to allow solutions using slower programming languages.

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      17 months ago, # ^ |
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      interesting. There is programming language 100x slower than c++? If so, what language is that?

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        17 months ago, # ^ |
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        I'll give you a hint: the name of this language starts with letter p.

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17 months ago, # |
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Can anyone please explain me why the complexity in DIV 2 B is O(n * log (eta inverse) ) and not O(n * log ( max (eta inverse,(h-l) ) ), here h=upper bound and l = lower bound of the binary search respectively. Thank you in Advance :)

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7 months ago, # |
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Endagorion That may seem too old, but for problem B div1 for the second part after having only distinct Ai, isn't maximum matching needed?

A case like this causes a lot of accepted solutions to fail:

4

ABC D

ABC E

ABF X

ABX E

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    6 months ago, # ^ |
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    I was wondering about the same thing. Both solutions 25258392 of V--o_o--V and 25257297 of OO0OOO00O0OOO0O00OOO0OO make use of maximum matching too.

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      6 months ago, # ^ |
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      Most of the solutions used greedy approach which is definitely not accurate that it fails on the mentioned case and similar cases.

      That would match the fact that N <= 1000, otherwise if greedy was enough N could have been <= 100000

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5 months ago, # |
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Can someone explain why my algorithm used in Div2 B The Meeting Place Cannot Be Changed is wrong. I assigned each friend the velocity plus the velocity of last friend(using the concept of relative velocity, last friend which is on the northest side has been given zero velocity while another friend will move towards him with a greater velocity) So minimum time I am getting is less than jury's answer. Here is the link to submission

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