because it's not available yet

Не могли бы вы объяснить каким же образом можно использовать стек?!

Заранее большое спасибо :)

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

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.

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://mirror.codeforces.com/contest/780/submission/46841304 for better understanding

Красным и синим рассматриваемые последовательности, серые линии без красных и синих участков не участвуют в рассмотрении и никак не влияют на результат.

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

Same problem, can't find the mistake.

Same here,what's the matter?

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

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

I was wondering about the same thing. Both solutions 25258392 of V--o_o--V and 25257297 of jqdai0815 make use of maximum matching too.

Answer is YES?