My solution should use 2 * M * K indexations of linear array in the worst case (100 000 000 for max M and K) but works about 0.5s, twice less than time limit, and it is Java!

Does it mean that so simple algorithm is enough or that tests are weak?

Edited by author 15.05.2007 02:27

I used greedy algo, pascal, 0.265sec works, O(2*M*K) too.

For such strong authors request of failed solver!
Give us short clarification of this great optimization
problem!
What is more simple prototipe of it?
What class of the problem? Graphs?DP? May be....

Timus Online Judge server is fast enough to perform more than 100000000 operations per second.

The problem is based on the Activity Selection problem.

I don't know.

I'd said simulation. And I think Alexander Kouprin's definition "greedy" is right.

The problem gets simpler if you change its model. You can allow driver to debus any boarded passenger at any stop < F[i] refunding the passenger all her money P. The driver in this case would get the same amound of money as if he would not board that debussed half-way passengers at all.

Edited by author 15.05.2007 14:09

The truth of this statement depends on operation kind. Recently I solved problem in which 50 000 000 operations * and % on 64-bit integers worked 2s.

Dmitry, do you know more precise numbers, how many operations of each kind the server can execute per second? For example linear array indexations - 180 bln, long multiplication - 100 bln and so on...

I'd said it's sorting problem. :)
You have segments of way: first point and last.
Your task is how to combinate maximum of segment in K lines.
This lines can be severed inside of itself and have big holes.

I used sorting too, but I think the main problem is to devise what to do after sorting.

Just write a kind of performance mark program, submit it and you will get exactly what you want.

Then you may post the results here and everyone will appreciate your work.

Is Activity Selection problem standard internet term?

I think it's a lecture hall assignment problem. One thing is that number of lecture halls here is limited by M...

Thank for "lecture hall assignment" brand!

I tried to solve it running Greedy-Activity-Selector M times. But WA at test 4. What is wrong with such approach?

Gready _Activity_Selector of course.
But with auxiliary subproblem:
Let we have a set S={[ai,bi]} intervals chosen to some
moment and according with greedy should include in S
new segment [c,d]. Can we do it without excess of M.
For, we must solve the problem of maximal overlapping
value. I used augmented red-black tree
as in Cormen but have very bad time 0.843 AC.
Intuition says that good times taken due better
ways of solving auxiliary problem.

Edited by author 29.09.2007 14:41

Thanks, I realised that. I simply store now all stations beginnings in an array (maximum K). I initialise it with M value for each element. Now, for each request I go from start to end and check if all elements are non-zero (not including the end). If that requirement is met, I add the request to the result and decrement all checked elements

By the way. Should we sort just time ending of the activity? In that case I get WA#4, If I sort beginnings in decreasing order in case of end equals, I get TLE #4

my algo is
O(N + K *log(M))  -> AC in 0.1

Mine too :)

i use segment tree and got AC in 0.109.
what's the best algo?
order of my algo is O((n+m)logn + klogk).
GOOD LUCK!!!