We will hold KEYENCE Programming Contest 2023 Autumn（AtCoder Beginner Contest 325）.
- Contest URL: https://atcoder.jp/contests/abc325
- Start Time: http://www.timeanddate.com/worldclock/fixedtime.html?iso=20231021T2100&p1=248
- Duration: 100 minutes
- Writer: yuto1115, cn449, ideas of all tasks are provided by KEYENCE
- Tester: PCTprobability, math957963
- Rated range: ~ 1999
- The point values: 100-250-300-450-450-500-575
We are looking forward to your participation!
250 for B will it be hard
don't have such preconceptions on a difficulty of a problem based on points
How to today's D? E was easier than D.
is E a DP problem ??
I also tried with dp first,but it was modified dijktras
code — https://atcoder.jp/contests/abc325/submissions/46824003
What's the complexity of dijkstra approach?? Just curious
it is ElogV where E=O(N^2) and V=O(N)
I solved D, but couldn't solve E.
can you explain your approach,E was modified djiktras. here is the code https://atcoder.jp/contests/abc325/submissions/46824003
For D, I put all eligible candidates in a priority queue and then pick one which is the earliest to finish: https://atcoder.jp/contests/abc325/submissions/46822856
I was not sure if greedy would work :/ thanks..
I tried greedy and got 2 WAs.
same -_-
For problem C, why does it not work if I do not use a visited set? It should be sufficient to mark it as empty "." right? Is there an edge case i am missing here?
https://atcoder.jp/contests/abc325/submissions/46818827
Maybe you referenced $$$grid[n][m]$$$ where $$$n\ge N$$$ or $$$m\ge M$$$?
if r<0 or r>=n or c<0 or c>=m or grid[r][c] == ".": return
I have this check in the dfs function to ensure I am not referencing that
Just checked your code here and there, and resubmitted with
sys.setrecursionlimit(500000). Got AC.Can someone tell me what is wrong in my solution for problem D?
can someone tell what wrong in my solution for D https://atcoder.jp/contests/abc325/submissions/46827017
You have to use a priority queue. Imagine when you are at time
t, and there are two segments which containt. At this point, it doesn't matter which of these segments begins first. All it matter is which of these segments finish first.the multiset contains the range of unassigned point so which case is missed?
Not sure, but the WAs in your submission are exactly the same as mine when I tried using greedy: https://atcoder.jp/contests/abc325/submissions/46811464
So, your approach has the same outputs as my greedy approach (which has 6 WAs).
So the correct approach is also greedy right or two pointer?
Right, greedy + PQ is correct. Greedy without PQ is incorrect.
my approach also passed just right sorting method was needed
can any one tell me what the topic of problem d?
scheduling trick: pick the first-ending task use cp-handbook greedy section to learn about it
Not exactly the same problem... if i recall the one in cph disallows choosing another task whilst performing the first one, so the proof for this problem is different from in CPH. However its true the greedy is the same.
Seems like the proof is a bit difficult... wonder why they omit it in the editorial :|
How to solve $$$F$$$? I thought of dp but couldn't think how we gonna do it!
$$$dp_{i, j} = \min$$$ number of type-2 sensors to cover first $$$i$$$ ranges if we also use $$$j$$$ type-1 sensors
Thanks!
tbh I misread constraints for $$$k$$$, assumed that $$$k \le 10^9$$$ :(
UPD: Never mind
Forbidden
REVEL_CSRF: tokens mismatch.
I think problem F is very inspiring, even though I didn't come up with that definition as the editorial mentions.
However, this reminds me of a problem that I have met before, which involves a similar idea https://mirror.codeforces.com/contest/745/problem/E.
When I first met some new ideas, I could keep that in mind, maybe for several days, and this time, it shows again that, I will forget it sooner or later. I think I should practice harder and more.
Can anyone help me why this code is wrong for problem G?
https://ideone.com/TZcGWP
Could someone expain me this sentence?
You can switch from company car to train, but not vice versa. You can do so without spending time, but only in a city.
Does that mean once I get on the train,I can't get on the car anymore,then I must get to the n city by train?
Yes
My solution in problem D is almost the same as abc214_e