Tutorial is loading...

Tutorial is loading...

Tutorial is loading...

Tutorial is loading...

Tutorial is loading...

Tutorial is loading...

Tutorial is loading...

Tutorial is loading...

Before contest

Manthan, Codefest 19 (open for everyone, rated, Div. 1 + Div. 2)

17:10:48

Register now »

Manthan, Codefest 19 (open for everyone, rated, Div. 1 + Div. 2)

17:10:48

Register now »

*has extra registration

# | User | Rating |
---|---|---|

1 | tourist | 3645 |

2 | Radewoosh | 3403 |

3 | Um_nik | 3348 |

4 | LHiC | 3336 |

5 | Benq | 3316 |

6 | V--o_o--V | 3275 |

7 | mnbvmar | 3241 |

8 | yutaka1999 | 3190 |

9 | ainta | 3180 |

10 | Petr | 3106 |

# | User | Contrib. |
---|---|---|

1 | Errichto | 191 |

2 | Radewoosh | 177 |

3 | PikMike | 165 |

4 | rng_58 | 164 |

5 | Vovuh | 160 |

6 | majk | 158 |

7 | antontrygubO_o | 154 |

7 | 300iq | 154 |

9 | Um_nik | 151 |

10 | kostka | 149 |

Tutorial is loading...

Tutorial is loading...

Tutorial is loading...

Tutorial is loading...

Tutorial is loading...

Tutorial is loading...

Tutorial is loading...

Tutorial is loading...

↑

↓

Codeforces (c) Copyright 2010-2019 Mike Mirzayanov

The only programming contests Web 2.0 platform

Server time: Aug/25/2019 00:24:12 (e2).

Desktop version, switch to mobile version.

Supported by

User lists

Name |
---|

Auto comment: topic has been translated by Endagorion (original revision, translated revision, compare)Why it shows

Tutorial is not available?UPD: It's available now :)

because it's not available yet

Can anyone explain how Div2 B can be solved using Ternary Search

I have done this.

We analyze total time in relation to position, I will call it

pselected. Time of each pointxwith speedvisabs(x-p) /v. If we make p vary, asvandxare fixed we'll see an abs functionWhich 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 thisNote 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 ofO(n) and your number of ternary search iterations (of logarithmic behavior)25253744

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

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 :)

for Div.2 D (Div.1 B)

get the output: NO but isn't a solution?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:

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.

I know but the somebody's code doesn't seems like that.

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.

we need to find the minimum time so that all participants can be in one place. that place need not be an integer, so it could be an integer but it doesn't have to be. so our time would be a long double variable.

lets assume that the time we picked is very large. the larger the time the better the odds of us find a point where every person can meet. why? because each person can then travel farther distances (north or south) which means the odds of persons crossing each other's paths is higher. if lets say there are only 2 people and the distance between them is large, as we give them more and more time to travel they will get closer and closer to each other and at some time t, they will cross each other. the same argument can be extended to more than 2 people...

so if we picked some time t1 and we somehow were able to figure out that everyone CAN meet at some point x within that time, what does that mean for all times t GREATER than this t1? it means that for any time greater than t1 all the people CAN meet at that point x. why? because the only thing that we do by increasing the time is that we give everyone a chance to travel even further north or south. so this point x is obviously within everyone's reach, even if we give everyone extra time. so basically there is no point of checking for times greater than t1, if at t1 everyone can meet at point x.

now, if we picked some time t1 and we somehow were able to figure that everyone CANNOT meet at some point x within that time, what does that mean for all times t LESS than this t1? it means that for any time less than t1 all people CANNOT meet at any point x. why? because reducing time to travel means everyone travels every a shorter distance than before, which means the odds of everyone meeting reduces further. so basicaly there is no point of checking for times less than t1, if everyone cannot meet at point x.

^ the above approach is binary search. so we pick a time t1, see if its possible for everyone to meet at some point x. if it is possible then we check for a time less than this t1, as t1 is already the best answer we have. if it is not possible then we check for a time greater than this t1...

now given a time t1 how do we figure out if it is possible for all people to meet at some point x? given a time t1, person P can travel anywhere between (positionP — speedP * t1) and (positionP + speedP * t1). speedP is a limiting value. the question says person P can travel at speeds no greater than speedP. so we have a range [x,y] where we know the person can travel to. we need to find [x,y] for every person and see if all these ranges have an intersection range. if it there is no such range that is common to all [x,y] then for this particular time t1, all people cannot meet at some point x. if there is a range that is common to all [x,y] then it means all people can meet at any point within this common range.

**here we will be doing binary search on long double variables, which is not as straight forward as it is with integers. we usually stop binarySearch(i,j) when i == j, but how can u say i == j when i and j are double values? i might be equal to j for a few decimal places but not necessarily for all. how do we tackle this?

as the question states the error should be within 10^-6 doing binary search at least 80 times would mean we have correctly covered 6 decimal places. if the error rate was maybe 10^-10 we would need to increase the number of times we do binary search maybe to 110 times... u can create a dummy binarySeachForDouble(x.0, y.0, steps) and keep doing binarySearchForDouble(x.0, (x.0 + y.0)/2, steps — 1) to see that x and y keep getting closer and closer....

refer to https://codeforces.com/contest/780/submission/46841304 for better understanding

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

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?

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

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.

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.

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

Can you figure out what is wrong with my code in problem 781B (wrong in test 16)? 25333425

Same problem, can't find the mistake.

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!

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.

Same here,what's the matter?

watch this legendary chess match while you wait

Woah

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

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

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)

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?

I believe it was done to allow solutions using slower programming languages.

interesting. There is programming language 100x slower than c++? If so, what language is that?

I'll give you a hint: the name of this language starts with letter p.

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 :)

Same question here. I think it might be an error.

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

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.

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

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