Supermagzzz's blog

Codeforces Round #693 (Div. 3) Editorial
By Supermagzzz, history, 4 years ago, translation, In English

All problems were invented MikeMirzayanov and developed by me Supermagzzz and Stepavly.

1472A - Cards for Friends

Editorial
Solution

1472B - Fair Division

Editorial
Solution

1472C - Long Jumps

Editorial
Solution

1472D - Even-Odd Game

Editorial
Solution

1472E - Correct Placement

Editorial
Solution

1472F - New Year's Puzzle

Editorial
Solution

1472G - Moving to the Capital

Editorial
Solution
  • Vote: I like it
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  • Vote: I do not like it

Comments Comments (129)
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4 years ago, # |
  Vote: I like it -65 Vote: I do not like it

How the editorial posted 2 years ago? If the contest is running.

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    4 years ago, # ^ |
      Vote: I like it +15 Vote: I do not like it

    2 "Hours" ago.

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    4 years ago, # ^ |
      Vote: I like it +7 Vote: I do not like it

    You can write blogs and not publish them. In this case, it is so the author can get a fast editorial out — by pre-writing the editorial but not making it public yet.

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    4 years ago, # ^ |
      Vote: I like it +3 Vote: I do not like it

    You wrote "years" instead of "hours":)

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    20 months ago, # ^ |
      Vote: I like it +8 Vote: I do not like it

    Now it's 2 years old 😹

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4 years ago, # |
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Can someone please explain why in 1472G - Moving to the Capital you can't go in first sample:

6->2->5->1
corresponding d[i] is
[2]->[1]->[2]->[1]

so there is exactly one step of kind 2.

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    4 years ago, # ^ |
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    To calc the distances you need to consider the direction of the edges.

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    4 years ago, # ^ |
      Vote: I like it +16 Vote: I do not like it

    You use road with d[2] -> d[1] twice. It's not allowed

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      4 years ago, # ^ |
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      I use different roads. Are you talking about same distances change?

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        4 years ago, # ^ |
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        Yes, you use road with this distances twice

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          4 years ago, # ^ |
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          Oh, I got where is my mistake. I mixed up two kinds. Actually [2]->[1] is second kind of move. I thought it is first kind. shame.

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    4 years ago, # ^ |
      Vote: I like it +5 Vote: I do not like it

    Your move from 6->2 is a move of the second kind (dist 2 to dist 1) The move from 5->1 is also a move of the second kind (dist 2 to dist 0).

    Therefore, you should stop at city 2 — a distance of 1 from the capital.

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    4 years ago, # ^ |
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    yeah,I have the same question.

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4 years ago, # |
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how can editorial be posted if hacking phase is still ongoing

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4 years ago, # |
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anyone please explain d

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    4 years ago, # ^ |
      Vote: I like it +3 Vote: I do not like it

    For D you only have to justify to yourself why taking the biggest number is always the best choice to make on your turn. The editorial actually does a pretty job of this, but think of it like this — either you take the biggest item left, and it matches your parity (i.e. add it to your score), or it doesn't (your opponent doesn't gain from that item on their turn — also good for you). It's simple to see that therefore a player should ALWAYS take the biggest item (i.e. sort the array and check each move).

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    4 years ago, # ^ |
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    Another way to look at it:

    At each point the player wants to maximise the distance between the scores of one another.

    You may be tempted to look for parity while picking the numbers, but if the goal is to simply maximise the distance then you increasing your score is equal to hindering the others progress.

    therefore at each point you pick The largest number which has a potential to reduce the difference between two scores.

    I hope this makes sense.

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4 years ago, # |
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Can someone please explain the approach for B (non DP approach)?

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    4 years ago, # ^ |
    ← Rev. 2   Vote: I like it 0 Vote: I do not like it

    Approach for problem B.
    Let the number of 1's is one & the number of 2's is two .
    Let total sum = one + 2 x two.

    Casework:

    Case 1

    Case 2

    Case 3

    Case 4.1

    Case 4.2

    You can also see my solution here : 103195164

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      8 months ago, # ^ |
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      Used the same case by case exhaustive enumeration, but instead I just focused on remainders after division by 2 (one_cnt % 2 == 0, two_cnt % 2 == 1).

      count(1) % 2 count(2) % 2
      Case 1 0 0 The number of ones and twos is even. Possible to distribute
      Case 2 0 1 Needs more attention, because we could match extra 2 with {1, 1}
      Case 3 1 1 Not possible, because of extra 1
      Case 4 1 1 Not possible, because of extra 1
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    4 years ago, # ^ |
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    Lets say s represents the sum of all elements, If s is odd then its totally impossible to split it into two equal halves

    Otherwise if s / 2 odd then you need to have atleast one 1 in the array to make the odd half

    so the answer is no if (you have a odd total) or (a even total but odd half and no ones with you), else answer is yes

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4 years ago, # |
← Rev. 2   Vote: I like it +4 Vote: I do not like it

Problem B is really similar to Div-2A Kitahara Haruki's Gift (https://mirror.codeforces.com/problemset/problem/433/A)

Great Contest Overall

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4 years ago, # |
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Can someone explain last testcase of Problem G.

For node 3 , shouldn't the ans be 1

we can go

3 (2) --> 4 (3) ---> 2 (1)

value in brackets is the corresponding d values

3--> 4 is of type 1

and

4--> 2 is of type 2

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    4 years ago, # ^ |
    ← Rev. 3   Vote: I like it +1 Vote: I do not like it

    Distance to 4 is equal to 2 by 1>5>4, not 3.

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    3 years ago, # ^ |
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    Can someone explain why 6->2->5->1 isn't a valid path for the first testcase of the sample input for problem G? I'm pretty confused cause we go 6(2) -> 2(1) -> 5(2) -> 1(0).

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4 years ago, # |
  Vote: I like it +20 Vote: I do not like it

A way to avoid repeating the algorithm in E, is to always have H < W for every i. So just swap H and W for some person, if H > W.

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    4 years ago, # ^ |
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    Could you explain a little bit more ? thanks in advance.

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      4 years ago, # ^ |
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      Now I got it. the condition for the person i to be placed in front of the person j can be rewritten like this : min(h_i, w_i) < min(h_j, w_j) and max(h_i, w_i) < max(h_j, w_j) This means you can compare like that.

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    3 years ago, # ^ |
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    this makes question very easy. thanks

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4 years ago, # |
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Simple solution — let's iterate through how many candies of weight 2 we will give to Alice, then the remaining weight should be filled by candies of weight 1. If there are enough of them, then we have found a way of division. How is this true ? If we take the case of 1 1 2 2 2, then Alice will get 2 2, while Bob will get 2 ,1 and 1.

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4 years ago, # |
  Vote: I like it +15 Vote: I do not like it

Here are my attempts to explain solutions in video form after messing almost everything up (this seems to happen more and more often recently...)

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4 years ago, # |
  Vote: I like it +3 Vote: I do not like it

on G,can someone explain why (dist[u] must <dist[v]),my code doesn't add this condition and i got WA. Thanks!

if (!used[v] && dist[u] < dist[v]) { dfs(v, g, dist, dp, used); } (In the DFS function)

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    4 years ago, # ^ |
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    Because you can use only the first option from the statement as much times, as you want. We build the new graph, with edges that satisfy this condition. The second option we check by DP.

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      4 years ago, # ^ |
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      Hi, This is my Dfs function. Where did I perform the second operation twice?
      mark : stores the final answer.
      d : stores the distance from root. 

       void Dfs (int t){
        vis[t] = true; 
        for(auto u: g[t]){
        if(!vis[u]) Dfs(u);
        if(mark[u] < d[u]){
        if(d[u] > d[t]) mark[t] = min(mark[t], mark[u]);
        else mark[t] = min(mark[t], d[u]);
        }
        else mark[t] = min(mark[t], d[u]);
        }
       }

      Adding that conditon you mentioned, gives me an AC. Can you please tell, where am I going wrong? Thanks

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      4 years ago, # ^ |
      ← Rev. 2   Vote: I like it +3 Vote: I do not like it

      I am stuck in the same situation. Even without dist[u] < dist[v] condition, we seem to be ensuring that 2nd condition is only counted once because

      if (dist[u] < dist[v]) { // Only count dp[v] if it's not bridge
   dp[u] = min(dp[u], dp[v]);
} else {
   dp[u] = min(dp[u], dist[v]);
}
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      4 years ago, # ^ |
      ← Rev. 2   Vote: I like it 0 Vote: I do not like it

      Thumbs up to the guys above. I also got stuck over this condition. For anyone interested, here's a graph exposing this bug:

      5 7
1 2
1 5
2 3
3 4
4 5
4 1
5 3

      In short, without the dist[u] < dist[v] condition, your DFS may update answer for some vertex x (dp[x]) by using dp[y] at a moment when y is still being processed by DFS somewhere up the tree. In this case, dp[y] isn't finalized, it's just not ready. In the graph above, x is 5 and y is 3.

      Takeaway: DFS doesn't like when you close loops; better let the old DFS finish and only then spawn the new DFS tree to process x.

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4 years ago, # |
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The editorial for G is very smooth. Great work Supermagzzz and Stepavly.

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4 years ago, # |
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I don't know how people solve problem like today's D so fast !! sed lyf :((

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    4 years ago, # ^ |
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    Donald trump complaining even on someone solving a problem faster. Not surprising.

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4 years ago, # |
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I wonder with B they did not include the alternate solution with DP

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4 years ago, # |
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Could anyone please help me to resolve MLE in this solution for Problem C

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    4 years ago, # ^ |
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    Your dp size is 20000 instead of 200000

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    4 years ago, # ^ |
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    you're passing the vector by value, so each time you call solve() a new copy is created. Pass it by reference or make it global.

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4 years ago, # |
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Why this dp solution gives WA on B.I thought it is equal sum partition. https://mirror.codeforces.com/contest/1472/submission/103310451

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4 years ago, # |
  Vote: I like it +3 Vote: I do not like it

I came across an ID in this round which was submitting solution with wrong answer for a particular corner case. FOR eg if n==123234 then print "NO" , this was done so that it could be hacked by another IDs. :p

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4 years ago, # |
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Can someone please help me why my code gets a WA(I get why it might get TLE but no WA): 103300499

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    4 years ago, # ^ |
    ← Rev. 2   Vote: I like it 0 Vote: I do not like it

    wrong answer jury found the answer, but participant didn't (test case 11)

    This is a comment of your code, so you can look for test case 11(it is visible) and see why it doesn't work. EDIT: you're only looking for elements with smaller X but a correct answer could also be element whose X is larger Y is smaller and so when it is laid sideways than it could fit in the element that you are checking. So your for loop J should go from 0 to n (i.e. check all elements) P.S. even if you did this you would get TLE

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4 years ago, # |
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Can anyone please explain why my solution is showing WA. My logic is obvious. One can easily get that after seeing my code. Submission

Thanks in advance for helping :)

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    4 years ago, # ^ |
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    I think its because you didn't sort the arrays. You should always take largest number, and even if you divide numbers in odd and even you should sort the array and look for larges numbers. I used a similar approach 103227720. Hope this helps.

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4 years ago, # |
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Can anyone explain problem G? Why do we have to change the graph? Why can't we apply dp on the graph with cycles?

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    4 years ago, # ^ |
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    I have applied dp only my friend. You can see my submission. Reply here if you didn't get the logic.

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4 years ago, # |
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I enjoyed the last dp problems of this contest :)

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4 years ago, # |
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Does anyone else feel that E was too much implementation heavy?

I thought about exactly this solution but with 30 mins remaining.
Then dropped the idea thinking that it might have some better segment tree solution instead of so much implementation. lol

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    4 years ago, # ^ |
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    Actually, it is not (at least for me).

    It could be implemented using a segment tree in about 30 ~ 40 lines of code.

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4 years ago, # |
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G was cool, E took me a while but it was fun too. Overall good problems!

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4 years ago, # |
  Vote: I like it +1 Vote: I do not like it

Are we allowed (yet) to ask inquiries regarding hacks for this round?

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4 years ago, # |
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Can anyone tell me how ans of this test case of ques d is TIE.

3 2 1

I guess here bob should win. First alice chooses 2, then bob goes with 3 . Then at end alice chooses 1 and removes it. So bob wins here(since score of bob is 3 and alice score is 2).

How is it draw?

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    4 years ago, # ^ |
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    IF Alica pick 3 first, then if Bob pick 1 Alica wiil win because in next turn she will pick 2(Bob_score = 1, Alica_score = 2), if Bob picked 2 insted of 1, Alica will be left with 1 (Bob_score = 0, because he piced even number, Alica_score = 0, because she picked odd number). Then answer is TIE.

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4 years ago, # |
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Can C be done by top-down approach?

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4 years ago, # |
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Why is this Solution got hacked? It is almost same as the Editorial. Please let me know as I don't wanna to repeat the same mistake again. Thanks in advance !

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    4 years ago, # ^ |
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    Yes, I was wondering the same thing as well. My solution (and many other Java solutions of a similar format) were also hacked (and ultimately received TLE), but I cannot find any apparent inefficiencies.

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      4 years ago, # ^ |
        Vote: I like it +2 Vote: I do not like it

      I heard Java sort uses quick sort. Quick sort in a worst case takes $$$O(N^2)$$$ complexity in an anti quick sort test.

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4 years ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

I have two implementations for E which I think are easier than the one in the editorial.

A : With comments, Without comments

B : With comments, Without comments

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4 years ago, # |
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DP Contest :)

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    4 years ago, # ^ |
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    There is only 2 problems I solved with dp and I think this is enough. Problems which have dp tag also have some other solutions this contest

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4 years ago, # |
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For D first test case my code is giving right output in my pc whereas on codeforces it gives wrong output 103274282

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    4 years ago, # ^ |
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    else if (o[j]>=e[i] && !fl) { sumb+=**o[i];** --j; fl=!fl; }

    Maybe this is causing some problem

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4 years ago, # |
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I was given similar problem as D in one of my interviews and I couldn't think of greedy approach, could only come with DP approach and was rejected. Now I solved it with so much ease in comfortable environment. Hate interview pressure.

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4 years ago, # |
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Can someone please explain E?

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    4 years ago, # ^ |
    ← Rev. 2   Vote: I like it +1 Vote: I do not like it
    1. Ask everyone to stand (make $$$h_i \leq w_i$$$ for every $$$i$$$).
    2. Sort them based on $$$h_i$$$ then $$$w_i$$$.
    3. Now we know that the $$$i$$$-th person is at least as tall as all of the people who come before him ($$$h_i \geq h_j$$$ for every $$$j < i$$$).
    4. Iterate through everyone and find the largest $$$j$$$ such that $$$j$$$-th person is strictly shorter than the $$$i$$$-th person ($$$h_i > h_j$$$). As they are already sorted, now we know that the $$$i$$$-th person is also taller than all of the people who stand in front of $$$j$$$-th person.
    5. Now we have $$$j$$$ possible candidates that can be placed in front of $$$i$$$-th person. We no longer have to check whether they are shorter than $$$i$$$-th person or not, because we already know that from the previous step.
    6. We need to find among the $$$j$$$ possible candidates, is there a person who is thinner than the $$$i$$$-th person. There could be more than one person who fits the criteria, so we will just greedily pick the thinnest person among those $$$j$$$ people and check if the thinnest person is thinner than $$$i$$$-th person.
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4 years ago, # |
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In problem D (language used : JAVA 11) , I am getting TLE in testcase 10 with this snippet of code

while(t-->0){ int n = ni();

       int arr[] = new int[n];
       for(int i=0; i<n; i++){
         arr[i] = ni();
       }

       Arrays.sort(arr);
       int k=0;
       long a=0, b=0;
       for(int i=n-1; i>=0; i--){
         k++;
         if(k%2 !=0 ){
          if(arr[i]%2==0)
              a+=arr[i];
         }
         else{
          if(arr[i]%2!=0)
              b+=arr[i];
         }
       }
       if(a==b)
         pn("Tie");
       else if(a>b)
         pn("Alice");
       else
         pn("Bob");

    }

and getting AC with this snippet of code

while(t-->0){ int n = ni();

       Integer arr[] = new Integer[n];
       for(int i=0; i<n; i++){
         arr[i] = ni();
       }

       Arrays.sort(arr, Collections.reverseOrder());
       long a=0, b=0;
       for(int i=0; i<n; i++){
         if(i%2 ==0 ){
          if(arr[i]%2==0)
              a+=arr[i];
         }
         else{
          if(arr[i]%2!=0)
              b+=arr[i];
         }
       }
       if(a==b)
         pn("Tie");
       else if(a>b)
         pn("Alice");
       else
         pn("Bob");

    }

Can anyone help me to know what is the cause?

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    4 years ago, # ^ |
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    Exactly what you are not able to understand?

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      4 years ago, # ^ |
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      In first code I sort the array in non decreasing order and I traverse the array from last for chosing the number for alice and bob

      and in second I sort the array in non increasing order and I traverse the array from starting for chosing the number for alice and bob

      Both are about same approaches nearly while first one is getting TLE and second one is AC

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4 years ago, # |
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https://mirror.codeforces.com/contest/1472/submission/103236341 this is my solution of D main idea is what person must bring max(his parity, opposite parity(it means not give this element to enemy))

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4 years ago, # |
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Can problem E be solved with monotone stack? Think the array to be circular, then for each element find the next element so that (h1 < h2 and w1 < w2) or (w1 < h2 and h1 < w2)

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4 years ago, # |
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anyone can provide better sol for Problem E or any video

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4 years ago, # |
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Hi, I am facing some problems with D. Even-Odd Game. I have read the editorial as well as the solution provided. I think I am doing the exact same thing given in the solution: always selecting the biggest available number for either player and adding or subtracting it from a global sum depending on who the player is and the parity. However, I am getting wrong answer on test case 3. ~~~~~

include <bits/stdc++.h>

using namespace std;

int main() { int t, n; cin >> t;

for(int i = 0; i < t; ++i)
{
    cin >> n;

    int a[n];

    for(int j = 0; j < n; ++j)
    {
       cin >> a[j];
    }

    int size = sizeof(a) / sizeof(a[0]);

    //sort array in ascending order
    sort(a, a + size);

    int score = 0;

    int num;

    int turn = 0;

    for(int j = n-1; j >= 0; --j)
    {
       num = a[j];

       //Alice Turn
       if(turn % 2 == 0)
       {

         if(num % 2 == 0)
          score += num;
       }
       //Bob Turn
       else
       {
         if(num % 2 == 1)
          score -= num;
       }

       ++turn;
    }


    if(score > 0)
       cout << "Alice\n";
    else  if(score < 0)
       cout << "Bob\n";
    else
       cout << "Tie\n";

}

return 0;

} ~~~~~

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4 years ago, # |
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Thanks for this round. I like problem G very much, it is the classic Codeforces' style task. Quality of this round was much better than previous five or even more.

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4 years ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

In problem D i have used erase for deleting the element and it has failed, but i see solution with same logic but using pop_back() instead of erase getting their solution accepted. I want to know that whether the time complexity of erase more than pop_back() or are they same in c++ ?

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4 years ago, # |
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I have a different solution for F, where I keep a offset between the top and bottom pointer, and use some case analysis. Time is O(mlogm).

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4 years ago, # |
  Vote: I like it +5 Vote: I do not like it

As the problem D has a trival solution, I have faced a false plagarism accusion on it.

"Your solution 103217882 for the problem 1472D significantly coincides with solutions lit2019039/103212550, whohet/103217882."

This question has a really trival solution and hence is clearly a coincidence. Please help and remove this plagarism mark. This is literally a coincidence, as the problem really have same solution. My solution link https://mirror.codeforces.com/contest/1472/submission/103217882

Match solution link https://mirror.codeforces.com/contest/1472/submission/103212550

Please have a look as problem really have a trival solution

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4 years ago, # |
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How does the code which was giving RE on testcase 4 with a slight modification in the comparator function (used for sorting) gives AC(aka happy new year). The change simply is using "<" instead of "<=". any input in this regard is highly appreciated

code with AC VERDICT:https://mirror.codeforces.com/contest/1472/submission/103375747

code with RE on test4 :https://mirror.codeforces.com/contest/1472/submission/103375782.

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    4 years ago, # ^ |
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    I bealive comparator function should be unique. Meaning if A > B then B < A. In your case with <= you are saying that at that if A == B then a could be before or after B which should not be possible because then sorting is not unique. I.E. if your functions returns A < B it should also return !(B < A)

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      4 years ago, # ^ |
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      I believe when you say "A<=B" that would mean that if A and B are equal then A should be before B , Although I understand what you are trying to infer, and I think that it could be right but that would depend on the sorting algorithm, furthermore if it was the case that A and B are repetitively compared with each other, shouldn't it result in a TLE instead of RE?

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        4 years ago, # ^ |
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        I bealive it works that if you compare A and B, than comparator automaticly compares B with A, and it should yield same results. In your case it gives A before B in first, and B before A in second which is contradictory. And so it gives an error.

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4 years ago, # |
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I felt E was harder than F. I spent wayy too long on E. Shouls have attempted F first.

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4 years ago, # |
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Why is there a "Tie" in the second test at problem D?

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    4 years ago, # ^ |
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    Each player takes the largest number so their opponent can't take it. So Alice takes 3, then Bob takes 2, than Alice takes 1, after that score is 0:0. This is optimal for both players. If Alice takes 2 in first turn then Bob will take 3 and win, so she has to take 3 herself. Same logic aplies after first move.

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4 years ago, # |
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Can someone please help me out for question D. My submission is showing runtime error on testcase 1 but it works fine on my ide

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    4 years ago, # ^ |
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    You pass odd.size() - 1 as a parameter — size() (in many implementations) returns an unsigned 64-bit value, so if it's 0 the result after subtracting will be the highest representable value. However, that is converted to a signed long long, and since the value is greater than can be represented in a long long, the value is implementation-defined.

    You can cast the result from size() to a signed type first.

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4 years ago, # |
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Is there any way that I can uphack my solution to G. It's linked here 103279910. I have tested it on custom invocation on CF and I know there is definitely a case that makes it TLE. I waited out the hack phase since I wanted to keep my rating, but now I don't see the option to hack it anymore.

EDIT: I think you have to be 1900+ to uphack. Here is my generator that would cause my solution to TLE. https://pastebin.com/uYxwMnYz

Any1 who is eligible to uphack, go for it :)

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4 years ago, # |
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Link1 Link2

Problem E.

Link1 solution is accepted wherease link2 is not. The only diff between the 2 solutions is that for pair(hi,wi) whose hi is least is being calculated differently:-

In link1: let the main loop take care of everything

In link2: comparing all (hj,wj) with (hi,wi) for second condition [wj < hi && hj < wi = > ans[i] = j] [Since first condition can never be satisfied]

Why does link2 solution fail?

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4 years ago, # |
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What's the DP state/solution for D no. problem? I can just feel the greedy approach. But problem tag shows DP too.

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4 years ago, # |
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Another way to think about E:

We can think about the pairs $$$(h_i, w_i)$$$ and $$$(w_i, h_i)$$$ as points on a 2D plane.

Then the problem becomes: for each point $$$P(h_i, w_i)$$$, find if there is any point that lies completely inside the rectangle that is formed by the origin $$$O(0, 0)$$$ as its bottom left corner and $$$P$$$ as its upper right corner.

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4 years ago, # |
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I've uploaded my post-contest stream to YouTube: https://youtu.be/nCSLlTCDnDs

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4 years ago, # |
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For D how can this input be a tie? 3 2 1

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    4 years ago, # ^ |
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    Alice will pick 3 in his first move(because he picks 2, then bob will pick 3 and wins thereafter), Then bob will pick 2(same reason for 1). Then alice will pick 1 at end. Both will have 0pts at the end.

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4 years ago, # |
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Can someone help me find the complexity of this code:103390550? It just barely passed the 4s time limit.

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4 years ago, # |
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I cant stand the fact that the dude on problem E invite n friends 1<= n <= 2 * 10^5 to celebrate new year! Guys we are in corona virus pandemic we should be more carefull! Imagine what will happen after 14 Days if there was someone with corona in the party! Be more carefule stay home and code, stay safe :D

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4 years ago, # |
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Simpler solution for F: 103672111

  1. Sort the blocked cells by x, y
  2. Pair the adjacent(after sort) blocked cells in disjoint pairs.
  3. Between blocked ceils within the same pair: distance has to be odd(or opposite color in chess grid as described in editorial)
  4. Between the second cell from ith pair and first ceil from i+1th pair: they cannot be on the same column. Otherwise it's going to block the flow of white cells, which is going to create an odd number of white cells in both side, which is impossible to fill.
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4 years ago, # |
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i need help..! submission to D.

Please find what is wrong with this code Your text to link here.... i am not able to find any error.! it is giving runtime error.!

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4 years ago, # |
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Does anyone know a possible reason for MLE in Problem G?

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4 years ago, # |
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include <bits/stdc++.h>

using namespace std; typedef long long int lli; typedef long double lld;

define FOR(i,l,u) for(lli i=l;i<=u;i++)

int main() { lli t;cin>>t; FOR(k,1,t) { lli n;cin>>n; lli A[n+1]; FOR(j,1,n) { cin>>A[j]; } lli score[n+1]={0},maxi=0; FOR(j,1,n) { if(score[j]==0) { lli i=j; score[j]+=A[i]; i+=A[i]; while(i<=n) { score[i]=-1; score[j]+=A[i]; i+=A[i]; } if(score[j]>maxi) maxi=score[j]; } } cout<<maxi<<"\n";

}

}

why am I getting time limit exceeded? In the above code no element is traversed more than twice, so time complexity is-O(n) for each test case.

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10 months ago, # |
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Kind of late to the party, but I think that problem G could be solved in a lot easier way. Firstly, just as in the official editorial, find out using BFS all the shortest distances from node 1 to every other node. Then we create 2*N states for the dfs (vis[2][N] = false and ans[2][N] = INF). First state will represent, if we already took the "larger to smaller distance edge" or not. Now, we will iterate through every node (v) and if vis[1][v] is false, then we do the dfs (our dfs will be defined as dfs(took, v)).

In the dfs, for each neighbor (u) of a current node (v), we check if d[u] > d[v]. If yes, we can go there. Else we have to check if took is 1. If yes, we can go into u, but we have to update took to 0. The dfs will be returning the shortest path found yet. Also, if we encounter a visited vertex, we just answer with the solution that we precalculated (recall ans array). For better understanding you can check my solution.

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