### doreshnikov's blog

By doreshnikov, 3 years ago, translation,

1547A - Shortest Path with Obstacle

Idea: MikeMirzayanov

Tutorial
Solution

1547B - Alphabetical Strings

Idea: MikeMirzayanov

Tutorial
Solution

1547C - Pair Programming

Tutorial
Solution

1547D - Co-growing Sequence

Idea: doreshnikov

Tutorial
Solution

1547E - Air Conditioners

Tutorial
Solution

1547F - Array Stabilization (GCD version)

Idea: doreshnikov

Tutorial
Solution

1547G - How Many Paths?

Idea: MikeMirzayanov

Tutorial
Solution
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 » 3 years ago, # |   +25 E was a wonderful problem to solve, love it!
•  » » 3 years ago, # ^ |   0 Can u explain e?
•  » » » 3 years ago, # ^ | ← Rev. 2 →   +4 My solution differs from the editorial.Let's start with the following idea: if we shift a cell righter by 1, we get closer to the conditioners on the right side and get further from the conditioners on the left side.What this means is that we can precount the values for the very first cell with the given formula, and then we will have to see where do the air conditioners belong for the current cell, either on the right side or the left one, and change their current distance by the same value.I hope you're familiar with priority queues or at least sets, where we will keep pairs of values: the first will be the temperature value(not the initial one, but the one we've precounted with the formula for the first cell) and the second will be the position of the air conditioner on the line; this will give us the minimum value instantly.We will have to use two sets: one is for the left side, the other is for the right side. At first, all the conditioners obviously belong on the right side, so you should fill the right set.The loop from 0 to n — 1 is where the magic happens. We should take the minimum conditioner off top the right set and see if it is an impostor and should have been already swapped to the left side, because its position is smaller than the position for our current cell; if it has to be swapped to the left side, then we should delete it from the right set, update the temperature value for its pair, which will become its initial value + 1 — i, and insert that to the left set(if you're struggling with how to get the initial value, you can use map, where the key would be the conditioner's position and the value would be the index in the array of initial values). After we've done that, we can be sure that the conditioners do belong to the right sets.As we've assumed that we get closer to the right conditioners and further from the left ones, this means that we can find the minimum value as min((*left_side.begin()).second + i, (*right_side.begin()).second - i); This would give us the complexity of O(n + k * log(k)). Here's my solution with priority queues.I probably explained it badly, you might also want to see Colin Galen's solution(1:42:52), which is the same as in the editorial and is quite good.
•  » » » » 3 years ago, # ^ | ← Rev. 2 →   0 Hello, I think I came up an idea like you after the contest end. But my code got TLE. Can you take a look? 122107507
•  » » » » » 3 years ago, # ^ |   +15 Fixed: 122137899 You probably messed up with a formula a bit. I'm not saying it is incorrect because it was rather hard to understand the logic behind (n — i), but using just plain -i and +i looks easier for me. You don't check whether tmp1 does actually exist, i.e tmp1 != a.end(). Therefore *tmp1 is the reason of your TLE. You do the lower_bound() through the array a, but have you sorted it previously? Its elements can be a total mess and the function would return random stuff depending on the order of the input elements(we're not guaranteed that the input is sorted).
 » 3 years ago, # |   +54 F and G were absolute gold. Thanks for the set!
•  » » 3 years ago, # ^ |   0 yep, G was also my fav and most painful too haha.
 » 3 years ago, # |   +78 The number of accepted solutions increased gradual way.Proves that Level of question increased in a very uniform way. Loved this !!!
•  » » 3 years ago, # ^ |   0 do you have discord server or group that you discuss i want to connect with you and want to learn
•  » » » 3 years ago, # ^ |   +11 I am a pupil bruh, not much to teach anywayI can give you advice that after every contest, watch upsolving videos on youtube and if they use a concept you are not aware of, go watch tutorial series of that conceptAnd of course, Keep practicing
 » 3 years ago, # |   +8 Fast editorial!
 » 3 years ago, # |   0 Wow, I really liked this contest, not just because I am expected to gain 127 points. A -> well, it's problem A.B -> nice, introductory 2 pointerC -> this took me a while because I was chasing down the wrong rabbitholeD -> interesting bit problem E -> nice, but easy, problemF -> nice segment treeG -> couldn't solve
•  » » 3 years ago, # ^ |   0 Can you explain how to solve F using segment tree
•  » » » 3 years ago, # ^ | ← Rev. 4 →   +32 Let's imagine our array in stages. For convenience, I duplicated my array just to deal with the cyclic nature of the problem.$a_1, a_2, a_3, \dots a_n$$\gcd(a_1,a_2), \gcd(a_2, a_3), \dots \gcd(a_n, a_{n + 1} )$$\gcd(a_1,a_2, a_3), \gcd(a_2, a_3, a_4), \dots \gcd(a_n, a_{n + 1}, a_{n + 2})$etcSo let's binary search on the answer. Say, we are wondering if the answer of 3 is possible. Then, we want to ensure that $\gcd(a_1,a_2, a_3) = \gcd(a_2, a_3, a_4) = \dots = \gcd(a_n, a_{n + 1}, a_{n + 2})$. We can check this using a segment tree with range gcd queries. Since we do a binary search, we have a final of $\mathcal{O}(N \log N \log N \log (\max(A_i ))$ Solution#include #include #include #include #include #include #include using namespace std; struct solve { int n; vector arr; int gcd(int a, int b) { if (min(a, b) == 0) return max(a, b); return gcd(max(a, b) % min(a, b), min(a, b)); } int *st; int findGcd(int ss, int se, int qs, int qe, int si) { if (ss > qe || se < qs) return 0; if (qs <= ss && qe >= se) return st[si]; int mid = ss + (se - ss) / 2; return gcd(findGcd(ss, mid, qs, qe, si * 2 + 1), findGcd(mid + 1, se, qs, qe, si * 2 + 2)); } int findRangeGcd(int ss, int se, int n) { if (ss < 0 || se > n - 1 || ss > se) { cout << "Invalid Arguments" << "\n"; return -1; } return findGcd(0, n - 1, ss, se, 0); } int constructST(int ss, int se, int si) { if (ss == se) { st[si] = arr[ss]; return st[si]; } int mid = ss + (se - ss) / 2; st[si] = gcd(constructST(ss, mid, si * 2 + 1), constructST(mid + 1, se, si * 2 + 2)); return st[si]; } int *constructSegmentTree(int n) { int height = (int) (ceil(log2(n))); int size = 2 * (int) pow(2, height) - 1; st = new int[size]; constructST(0, n - 1, 0); return st; } bool valid(int x){ set s; for(int i = 0; i < n; i++){ int l = i; int r = i + x; s.insert(findRangeGcd(l, r, arr.size())); } return (s.size() == 1); } int binSearch(int l, int r){ //cout << l << " " << r << endl; int m = (l + r + 1)/2; if(l == r){ return l; } if(!valid(m)){ return binSearch(m, r); }else{ return binSearch(l, m - 1); } } void read() { cin >> n; arr.resize(n); int GCD = 0; set s; for (int i = 0; i < n; i++) { cin >> arr[i]; GCD = gcd(arr[i], GCD); s.insert(arr[i]); } if(s.size() == 1){ cout << 0 << endl; return; } for(int i = 0; i < n; i++){ arr.push_back(arr[i]); } constructSegmentTree(arr.size()); cout << binSearch(0, n - 1) + 1 << endl; } }; int main() { int t; cin >> t; while (t--) { solve s; s.read(); } } 
•  » » » » 3 years ago, # ^ |   0 cool
•  » » » » 3 years ago, # ^ |   +7 Actually the time complexity is O(N"LogN*LogN*Log(Max(a[i])) , N*LogN for segment tree , Log N for binary search , and Log Max(A[i]) for GCD function .
•  » » » » » 3 years ago, # ^ |   0 oops, yes, my bad
•  » » » » » 3 years ago, # ^ | ← Rev. 2 →   +11 Time complexity is $O(N\log{N}(\log{N} + \log{MAX}))$, only two log factors.If you take gcd $x$ times, time complexity is $O(x + \log{MAX})$.
•  » » » » » » 3 years ago, # ^ |   0 How did you get this equation? Logarithmic time complexity for the worst case of Euclid gcd algorithm is bounded by sequential Fibonacci numbers and it sounds impossible to guarantee a big count of gcd operations on them in segment tree. Is it that?
•  » » » » 3 years ago, # ^ |   +5 Why do you not try the Sparse Table?It can solve this problem in O(n log n)
•  » » » » » 3 years ago, # ^ |   0 Two logs, I suppose.One for Sparse Table, and one for gcd.
•  » » » » 3 years ago, # ^ |   +1 another approach could be to check for every number that after how many steps or how right we'll have to go so that the gcd is 1 (I divided all the numbers by the gcd of the whole array in the start) then we take the max value. we can use seg tree to find the gcd of i to m (mid in binary search) the net complexity would be n*log^2(n)
•  » » » » 3 years ago, # ^ |   0 Hey Olympia, I have tried using the same approach as you have mentioned above. I have done a binary search for each element in the array and then found the first point where the segment of the array will be equal to the gcd of the entire array. I have used range queries( segment tree) for doing so. Here is a link to my submission. Could you please guide me as to where I went wrong. Thank you.
•  » » » » 3 years ago, # ^ | ← Rev. 2 →   0 I am quite noob :((, I don't understand why gcd(a1,a2,a3) = gcd(a2,a3,a4) = ... gcd(a(n-2), a(n-1), a(n). Can you explain that.
•  » » » » » 3 years ago, # ^ |   0 He is explaining an example where after two steps every element in array will become equal.(a1,a2,a3) = gcd(a2,a3,a4) = ... gcd(a(n-2), a(n-1), a(n)is only true for the case when ans=2 or ans=0.ans= No. of steps.
•  » » » » 21 month(s) ago, # ^ |   0 Can you help why my binary search + segment tree solution is getting a TLE on test4 SUBMISSION :1547F Thank You.
•  » » 3 years ago, # ^ |   +20 My solution to G.You can do a basic DFS to find if the answer is 0 or 1 or 2 for every component.Use Tarjan's SCC algorithm to find all strongly connected components. Do a multisourced BFS from every SCC reachable from 1. Every node reachable by this BFS is a -1.This solution works in O(N).
•  » » » 3 years ago, # ^ |   0 Can I use Kosaraju's algo instead of tarjan's SCC algo?
•  » » » » 3 years ago, # ^ |   +11 yes of course they're essentially equivalent.
•  » » » 3 years ago, # ^ |   0 Any Code?
•  » » 3 years ago, # ^ |   0 how do you predict your rating change?
•  » » » 3 years ago, # ^ |   +12 Incase you're using chrome, this will help: CF-predictorIt's not very accurate, but good enough to make you happy or sad.
 » 3 years ago, # | ← Rev. 2 →   0 Please, fix the code of F.Fix this line: const unsigned int MAX_A = 1000'000;Thank you for D, E and F. Didn't see G.
•  » » 3 years ago, # ^ |   +5 The code itself is technically correct but it's incorrectly parsed by the code highlighter tool. Will fix it in a few seconds, thank you!
 » 3 years ago, # |   +9 Maybe it is just me who came up with bad solutions (dp for C, dijkstra for E, bfs-based toposort for G), but the implementation was quite painful for most of the problems today. Not something worth coming to Codeforces for :c
•  » » 3 years ago, # ^ | ← Rev. 2 →   +11 you did come up with bad solutions, on C you can do a greedy algorithm (always try to add more lines), on E it's a simple dp where you try to find the best temperature from 1 to i and from i to n (in the end you combine both dp) and on G you can do a dfs and not pass through a guy only if he has infinite paths (you also need to store which vertices you passed through in your dfs), making you only pass through each vertice at most twice
•  » » » 3 years ago, # ^ | ← Rev. 2 →   +11 I don't question the fact that my solutions are bad, and I have also read the editorials, but I still think that implementations are unpleasant.
•  » » 3 years ago, # ^ |   0 Isn't C greedy? You take from either one as long as you can (need enough 0s). if stuck, no answer.E can be solved greedily. Start from the coolest AC and update all the cells unless a cell already has a better AC planted on it. Keep doing that for all ACs sorted from coolest to hottest. It can be proved that no cell will be updated more than twice.Couldn't solve F or G.
•  » » » 3 years ago, # ^ |   0 Yes it is, but I didn't want to think and proof anything, so just implemented the first idea I got, so dumb
•  » » » » 3 years ago, # ^ | ← Rev. 2 →   +3 Well, my implementation to all questions are short imo. Well, I'm not too sure as I started CP recently.
•  » » 3 years ago, # ^ |   +3 i used bfs for E, welcome to the "overthinking" club
•  » » » 3 years ago, # ^ |   0 hi
•  » » 3 years ago, # ^ | ← Rev. 2 →   0 I used lazy segment tree on E....I knew it's pretty bad when I started to debug. Edit: I'm surprised that my segment tree solution runs as fast as the linear solution.
•  » » » 3 years ago, # ^ |   0 I'm not sure how to solve E using Lazy segment tree..But one can solve the problem using two normal segment tree's itself.My code: 121978627
 » 3 years ago, # |   +6 Just to confirm, for F, using Sparse Table shouldn't TLE right?
•  » » 3 years ago, # ^ |   +12 It shouldn't because even Segment trees are passing
•  » » 3 years ago, # ^ | ← Rev. 2 →   0 Not at all, I have used sparse table and my sol passed within 400ms. You can have a look at my solution if u want.
 » 3 years ago, # |   0 This round made me want to commit not alive
•  » » 3 years ago, # ^ |   +1 not sure why, the problems were pretty good
•  » » » 3 years ago, # ^ |   0 I was being dumb and then got hacked
•  » » » » 3 years ago, # ^ |   0 oh, pretty fair then
 » 3 years ago, # |   0 Almost greedyforces, upto E. F was a nice problem.
 » 3 years ago, # |   0 At the end of time I just know how to solve it but too late. Anyway I thing F is great. Great contest. Love!
 » 3 years ago, # |   +4 I used some kind of Dijkstra Algorithm (multi source) to solve E
•  » » 3 years ago, # ^ |   0 how ? can you explain
•  » » » 3 years ago, # ^ |   +12 So I push the initial temp to a priority queue, and start processing by the lowest one, supposed it's T. Then I push T+1 to its left and right, and put them in the priority queue
•  » » » » 3 years ago, # ^ | ← Rev. 2 →   +3 I found a very cool approach in which I first iterated from left to right and store the minimum temperature in the temp list and then from right to left and did the same at last the minimum temperature is stored in the respective position.Well fonmagnus can you please suggest me how can I do good in contests. It's been 1 and half year since I started and till now I am in Green. I don't know what is happening but eveyone else is just better than me ! What should I do ?
 » 3 years ago, # |   +44 During the contest, I think problem G is a wonderful problem since it's exactly a proper problem for those who just learn strongly connected components to practise.However, seeing the editorial, I think the G is more wonderful than I have thinked.
 » 3 years ago, # | ← Rev. 2 →   -10 please, somebody, tell me... why my solution is wrong for problem B and which test cases it is failing... 121938514there is a minute error... i.g. some very minute error is present... it failed the 5002nd case which is not visible.
•  » » 3 years ago, # ^ | ← Rev. 2 →   +1 The error seems to come on the string "abcdefghijklmnopqrstuvwxyz", your code gives No as output. I checked your code and it's because of the array declaration int alphabets[25], it should have a size of 26 not 25.
•  » » » 3 years ago, # ^ | ← Rev. 2 →   -41 checked this for my code ... It is giving correct output
•  » » » » 3 years ago, # ^ | ← Rev. 3 →   +2 First, completely unecessary to call him a dickhead for helping you.Second, you are wrong. In your computer it works because even though you are accessing a higher index than you allocated it doesn't give an error, which can happen sometimes when you're in your local computer, but it will give an error on codeforces and can give an error on some other computer (like brobat's).The worst part is that that's not the only reason why the code is wrong because you got wrong answer instead of runtime error, so your idea is also wrong
•  » » » » » 3 years ago, # ^ |   0 That's not necessary that the idea is wrong, accessing elements out of array bounds can return unexpected values and not RE — and that can cause WA.
•  » » » » » » 3 years ago, # ^ |   +1 As far as i know, in codeforces they actually go out of their way to check if you're accessing the array in bounds and not only give you a runtime error but they also highlight the line where this is happening.And the idea is wrong, they check where 'a' is and sees if the values increase as they go to the sides. This is not sufficient as "zax" would work but isn't actually a valid string. Also i believe the array they use to track frequencies can hold garbage data but im not sure
 » 3 years ago, # |   +1 I used binary search+ sliding window(with map/unordered_map) but why this code is giving TLE? complexity seems like O(n*factors[a[i]]*log(n*fact[a[i]]) (including map factor)[submission:link]
 » 3 years ago, # | ← Rev. 2 →   +2 There is no need to find the position of 'a' in problem B, we can start checking the values from endpoints of string as well...
 » 3 years ago, # |   +1 I think I have a very easy to understand solution for E: https://codeforces.com/contest/1547/submission/122007085Iterate over conditioners, and for each go left and right, while you can cool down cells, it gives you O(n+k).
 » 3 years ago, # |   0 Awesome contest. Where can I find more problems similar to D?
•  » » 3 years ago, # ^ |   0 bitmask problem from codeforce you can search in the problemset by tag
 » 3 years ago, # |   0 I was able to do A,C,D. But I got WA for B. I don't know where I did a mistake.Every test case I tried randomly seems to be fine.Can someone plz help me with it? https://codeforces.com/contest/1547/submission/121969497
•  » » 3 years ago, # ^ | ← Rev. 2 →   -9 The same happened to me, dude... I don't know where's the problem... you help me please 121938514 this is my solution... please please help me as soon as you get it
•  » » » 3 years ago, # ^ |   0 Yeah, sure
•  » » 3 years ago, # ^ |   0 i think you just read the problem wrong because i have no idea of what you're checking. Could you elaborate on your idea?
•  » » » 3 years ago, # ^ |   0 First I'm checking for duplicates and any alphabet whose value is greater than length of string(considering a's value to be 1.If string length is 3 only a,b,c can be there.) and storing the positions of characters.Now as we build from lowest alphabet, each character can go to right or left of string,i.e it can just be beside previous character or at other extreme(so the gap between to consecutive alphabets would be equal to all the previous letters before these 2 consecutive ones).( abs(mp[i]-mp[i-1])==1 || abs(mp[i]-mp[i-1])==i)As we're building and checking the gap from lowest pair of consecutive alphabets ,if for all alphabets until length of string ,every condition satisfies then it would be alphabetical string. Hope this helps :)
 » 3 years ago, # |   +8 Test Cases separated by a line helped a lot while debugging. Please continue doing so.
 » 3 years ago, # |   +10 Pizzeria Queries sends its regards.
•  » » 3 years ago, # ^ |   +3 Thanks for the Reference!
 » 3 years ago, # |   0 I used priority queue to solve E.121971913
 » 3 years ago, # | ← Rev. 3 →   +4 Am I the only one who used the minimum on the queue to solve the problem F? 121980439
•  » » 3 years ago, # ^ |   0 Can you explain your solution, please?
•  » » » 3 years ago, # ^ |   0 Yes, I usd struct similar to the one described here.Instead of the minimum, I implemented the gcd function.Using two pointers and this struct i found longest segment where gcd() not equal to one.Its length is the solution to the problem.
•  » » 3 years ago, # ^ |   0 Thanks for your sharing your solution. Learnt something new.
 » 3 years ago, # |   0 My own solutions for B, C and E:B. If we reverse the operations, we'll need to remove letters from the beginning or from the end of the given string. Two pointers (L and R) showed the part of the string that is not removed yet. So I iterated on letters from 'a'+s.length()-1 to 'a' and shifted one of the pointers. That allows to solve this task in O(n), not O(n^2) as in editorial.C. On each turn, we can take the minimum of a[i] and b[j] — if it's invalid, the other value is invalid as well. So, checks with a[i] == 0 and b[j] == 0 are not needed.E. Tried to add only "effective" conditioners — conditioners which cool some cells. I sorted the given conditioners by increase of their temperature and iterated over them, checking if some conditioners could be skipped. Then, the answer is affected only with conditioners near the cell (the nearest on left side and the nearest on right side).
 » 3 years ago, # |   0 121992524Please provide some hint on where this solution seems to be incorrect.
•  » » 3 years ago, # ^ |   0 What does && k!=0 mean in if((i
 » 3 years ago, # |   0 it was a realy good contest thanks for all setters and testers for make beautiful contest
 » 3 years ago, # |   0 D was the best and I really liked it =)
 » 3 years ago, # |   0 Problem: 1547F - Array Stabilization (GCD version) can be solved using two pointers stack trick, which can be learned through EDU "Two Pointer" section. Link: https://codeforces.com/edu/course/2/lesson/9/2, video is called "Segment with Small Spread", my solution: 122018965
 » 3 years ago, # |   +4 Problem E was really interesting!But I think, there is a small mistake in Editorial. In my opinion, it should be $L_{i}=min(L_{i-1}+1,c_{i})$ instead of $L_{i}=min(L_{i+1}+1,c_{i})$Thanks for contest again!
 » 3 years ago, # | ← Rev. 2 →   0 For problem D, there is a solution using greedy with less thinking, but with a little worse time complexity.If we treat every integer as a 29 bits binary numbers. To make $y$ lexicographically minimal, start scan the sequence from position 1, and each time try make $y_i$ as minimum as possible.First of all, $y_1 = 0$.And then for $i > 1$, we can let $y_i = g(x_i)$ first, where $g(x_i)$ is bitwise-not of $x_i$. Since by doing this, $x_i \oplus y_i$ will be all 1's, and this must satisfy the requirement of growing.After this, we can enumerate every 1 in $y_i$ from higer position, and change every 1 to 0 if possible.
 » 3 years ago, # |   0 in problem G, one of the test cases is:3 2 3 3 2 2my answer was 1 0 0 expected answer: 1 -1 -1acc to me, there ate 0 ways to go from vertex 1 to vertex 2 or 3. but expected answer was that there are infinite solutions. Am i missing something? My Solution link
•  » » 3 years ago, # ^ |   +3 Check it in CF's custom test, your answer is "1 -1 -1 " to this test, which is wrong and the correct answer is "1 0 0". Maybe you've read "Answer" part instead of "Output" part.
•  » » » 3 years ago, # ^ |   +5 AAAh right. Thank you.
 » 3 years ago, # | ← Rev. 4 →   0 Problem E was easier than I thought...My solution was different, involving Segment Trees (This is an overkill, btw).I had solved a kind-of-similar problem on CSES. The problem: Pizzeria QueriesThe major difference being that there were updates in that problem. However, I went ahead and coded up that solution during the contest. After the contest did I realise how simple and elegant the code for E was.Anyways, incase someone is interested in the code: 121978627
•  » » 3 years ago, # ^ |   0 My solution was even more different — I used Treap. I've thought of Segment tree, but didn't understand, how to account elements only later than i in it.
•  » » » 3 years ago, # ^ | ← Rev. 2 →   0 USACO guide has explained the solution for Pizzeria Queries pretty well.The code I implemented is pretty similar to the guide's solution.Incase you want to understand how to solve it using segment trees, you can look into the solution linked above.
 » 3 years ago, # |   0 Can someone look at this solution for E. I used bfs for it, but for some reason I got TLE on test 12. Can someone spot a mistake that causes an infinite loop, because it should run in O(n)
•  » » 3 years ago, # ^ |   0 Sorry, turned out solution is just shit
 » 3 years ago, # |   0 F is heck of a problem!!!✨
 » 3 years ago, # |   0 Hmmm!
 » 3 years ago, # |   +1 A bit different approach to problem F:Notice, that the number of times a[i] changes is small (worst case is log(a[i]), I believe). So if we can do a brute-force solution where we only change the elements that need to be changed, we can get N*log(MaxValue) complexity.Note, that a[i] changes only if it's not a divisor of a[i+1]`. Let's keep track of segments of elements where $a[l] \mathrel{\vdots} a[l+1] \mathrel{\vdots} ... \mathrel{\vdots} a[r]$. Only the last element may change after one iteration. It's also pretty straightforward to update the segments after the iteration as well.Code: 122069028 ( I used std::set here, so the complexity is not optimal, you can rewrite it using std::vector).
 » 3 years ago, # | ← Rev. 2 →   0 If you're interested in a video, I explained all the problems on stream, and you can watch at https://youtu.be/QKW3_9RFxDI
 » 3 years ago, # |   0 is it possible to do binary search on the F sum? like by taking x over 0 to n steps ?
•  » » 3 years ago, # ^ |   0 Yeah, you can binary search on the number of operations. Basically, at $k$ th step $a_i$ becomes $gcd(a_i, ... , a_{(i+k)\%n})$. And to calculate gcd fast, you can use binary lifting. Look at my solution 121972800. This is $\mathcal{O}(n\cdot log^2 n)$ but you can make it faster by combining binary lifting and binary search. Similar to the idea mentioned here.
 » 3 years ago, # |   +1 Problem E was great. Can Problem E be solved using priority_queue ?
 » 2 years ago, # | ← Rev. 3 →   0 148276776 Can anyone please explain, why it is giving WA on test-2?