Codeforces Round #562 — Editorial

6 лет назад, # |

← Rev. 2 →

+21

I first implemented 1168D - Anagram Paths in $$$O(n \log n)$$$ with segment trees and only then saw that it doesn't say YES/NO correctly. But if the answer is YES, then I can correctly find the value to print (max number of each letter). But I later needed $$$O(n \sqrt n)$$$ anyway to check YES/NO.

So, how to find value of the answer in $$$O(n \log n)$$$? Run DFS to get preorder and postorder of every vertex. Create $$$26$$$ segment trees linearly, one for each letter. If there is a character on an edge, look at segment tree for this letter, and add $$$+1$$$ to segment corresponding to the subtree below the path. For each letter, you need a path to some leaf that has biggest number of occurrences of this letter — this is max in a segment tree.

This solution doesn't actually rely on the fact that there are at most two children. Maybe it's possible to extend it to get YES/NO too?

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300iq

6 лет назад, # ^ |

+16

It is very smart!

But I don't think that it is possible to extend it, honestly. Problems about dynamic tree DP are solved with some spooky HLD solutions :(

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AsleepAdhyyan

6 лет назад, # ^ |

Why doesn't this sol, say YES/No prorpely. I can't think of a reason. Do you have a test case or any intuition?

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6 лет назад, # ^ |

+11

Other than checking depths to be equal, the answer is NO if there exists a vertex such that the height of its subtree is smaller than the number of fixed characters below — for every letter, we take a path (to a leaf) with max number of occurrences of this letter, and we sum these $$$26$$$ values. The issue with my solution is that I only check the root for this inequality, while I should check all vertices with two children.

The counter test is: $$$N = 4$$$, edges: $$$(1-2, ?)$$$, $$$(2-3, a)$$$, $$$(2-4, b)$$$. The answer is NO because we need at least one a and at least one b below the vertex $$$2$$$.

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djq_cpp

6 лет назад, # ^ |

← Rev. 3 →

Point it out if I am wrong:

Let sum[v] denote the sum of maximum occurrences in a link for each character in the subtree v.

After replacing a '?' with a letter, what we need to do is a link addition.

Fix sum[v] + dep[v] and its minimum(which is required to answer "Shi" or "Fou" queries), which can be done in O(n log^2 n) (HLD) or O(n log n) (LCTs with some extensive information).

Another issue: how to find the length of the link we're going to update. Fix the 26 segment trees as you've mentioned above and do a binary search on the one that we need. Then, simply climb the tree through HLD or binary encoding.

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6 лет назад, # ^ |

What is LCT?

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ko_osaga

6 лет назад, # ^ |

link cut tree

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FizzyDavid

6 лет назад, # ^ |

+27

I implemented this idea and it works correctly. But it is slower than bruteforce because of large constant factors.

My submission: 54747584

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deepak_097

6 лет назад, # |

Can anyone please explain their solution of Problem C (Increasing by modulo) ?

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Malomalomalomalo

6 лет назад, # ^ |

+30

Sure.

First we notice that the number of overall operations is equal to the maximum number of operations used on a single index.

Next we notice that, if we can solve the problem in <= X operations, then we can also solve it in <= X + 1 operations. This is a monotonic sequence, so we can solve for the minimum value of X with binary search.

The question is, how do we check for <= X? Well, notice that it is never worse to have the leftmost index to be lower. However, we can only increase-- until it loops around. Thus we either don't change the value at x, or we set it to zero. (we check if the required number of operations to do this is <= X). Next, we move on to the next index, this time setting the "base" value equal to the previous index's value to ensure our sequence is nondecreasing. If at any point our current index is less than the base and X operations on it are not enough to make up for this, then we need to raise our value on X.

Now we have O(log m) searches and O(n) checking, so we have an overall O(n log m) solution.

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soccerstar10110

6 лет назад, # ^ |

Excuse me what do you mean by "However, we can only increase-- until it loops around. Thus we either don't change the value at x, or we set it to zero. (we check if the required number of operations to do this is <= X)."

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Malomalomalomalo

6 лет назад, # ^ |

lets say the first value int the array is 3, and m is 6.

after each operation, we get the following values: 3, 4, 5, 0, 1, 2

If we increase 3 through the operation, we don't want to make it 4 or 5, as this just makes getting a nondecreasing sequence harder, we want it to become less than 3.

However, to make it 1 or 2 we need to go past zero. Zero is better for nondecreasing sequences, and it is also easier to get to.

Therefore, our only choices for the first one are zero or the initial value.

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soccerstar10110

6 лет назад, # ^ |

Right! I just got it an hour ago after looking at your submission :) thanks again!

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6 лет назад, # ^ |

thanks for great explanation.. can u give some question related to C i.e. increase by modulo

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The_Kings_Gambit

6 лет назад, # ^ |

← Rev. 2 →

Your explanation is awesome! Thanks a lot!

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dragonzurfer

6 лет назад, # ^ |

← Rev. 2 →

https://medium.com/@harryjobz/increasing-by-modulo-c057eadfa1f0 for a bit more explanation

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6 лет назад, # ^ |

how we come to conclusion that in first case we have make a[I] equal to prev if we have enough operation….. but it is okay to leave a[I] as it is as the sequence is still non decreasing... plz tell me the proof of this, as I cant understand it and thanks for great explanation.. sry for bad English.

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EbTech

6 лет назад, # |

← Rev. 5 →

+11

Are there many other programmers who compete in Rust?

My solution to 1168C seems similar to the editorial, but it times out and I don't know why.

Also, why isn't the i-j-k loop in the official solution n log^2 n?

Edit: locally, my code runs a lot faster if I use -O. What rustc compiler options does Codeforces use?

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The_Kings_Gambit

6 лет назад, # ^ |

+10

It's roughly 10^8 operations. Should be fine with C++. I've heard Rust is faster than C++?

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EbTech

5 лет назад, # ^ |

← Rev. 3 →

+19

I got word from Mike, we use: rustc -O -C link-args=/STACK:268435456 --cfg ONLINE_JUDGE %1

The inner loop should only run 300,000 * floor(19/2) * ceil(19/2) = 27 million times, so it's still puzzling that my solution is so slow.

Edit: I figured it out! Unfortunately, it turns out stdout is line-buffered, so it's slow if you print a lot of newlines, such as using println!() or print!("\n").

Here's a passing submission using writeln!()

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The_Kings_Gambit

5 лет назад, # ^ |

Interesting!

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RedStone

6 лет назад, # |

Is there anyone who has a proof for the lemma of problem D ? Why there are no strings without such x,k of length at least 9 ?

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DeliciousFlatChest

6 лет назад, # ^ |

Just try out all bitstrings of length 9? Is it that hard?

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MidoriFuse

6 лет назад, # ^ |

+27

Yes, there is a proof

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Pannu

6 лет назад, # ^ |

Hey there, can you provide more sites like above mentioned(mathematics and proofs).

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RedStone

6 лет назад, # ^ |

Thank you :)

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Ice_Cream

5 лет назад, # ^ |

Amazing site I had ever seen! I love it.

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motatoes

6 лет назад, # |

what do these lines do?

#ifdef ONPC
  mt19937 rnd(228);
#else
  mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
#endif

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HosseinYousefi

6 лет назад, # ^ |

+16

If running on local machine (ONPC!), use a constant seed for random (228), but when submitting to codeforces for instance, use the time as seed to avoid hacks. This will make debugging your random code easier, because the values produced by rnd(228) are always the same every time you run the code on your machine.

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xiaoshizhan

6 лет назад, # |

I checked your solution of problem D Good Triple. I don't understand why get the largest r. If we fix l, and we find the smallest possible r, and r of value from r to n-1 (inclusive) is ok. I think we should get the smallest r, and I changed your code by add a break if I get a smallest r.

    if (s[i] == s[i + k] && s[i + k] == s[i + 2 * k]) {
                vl[i] = i + 2 * k;
                break;
            }

And the code get AC. Is there any bug of the solution or the system test does not cover all cases?

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300iq

6 лет назад, # ^ |

for (int k = 1; i + 2 * k < vl[i]; k++) {

This part already ensures that you will make vl smaller, so if you will update vl, you will break because of the for condition.

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McDic

6 лет назад, # ^ |

Then the code works to find the smallest $$$r$$$, not the largest.

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ko_osaga

6 лет назад, # ^ |

If we find largest $$$r$$$, then it’s obviously n-1 or bust. So it’s a typo.

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priyansupulak

6 лет назад, # |

How to solve 1169B using graphs? Thanks.

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devanshsaxena16

6 лет назад, # ^ |

Actually its a question of determining whether the vertex cover set has atmost 2 elements or not

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Bojack

6 лет назад, # ^ |

Isn't that a NP Complete problem for general graph?

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devanshsaxena16

6 лет назад, # ^ |

yes it is np complete for general graph but that problem is finding minimum set of vertex with contains all edges of a graph,here the problem is different

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Noob27999

6 лет назад, # ^ |

Can you please explain further , possibly with a pseudo code .

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devanshsaxena16

6 лет назад, # ^ |

go to this link https://codeforces.me/blog/entry/67248

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__soli_deo_gloria

6 лет назад, # ^ |

hey can you point out why my code is wrong ? I swapped the x and y if x is smaller than y, then put it in a vector and continue to sort it, and then picking the last and the first only to check, why is this wrong ?

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6 лет назад, # ^ |

can us pzl explain little more or provide some link to related article

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priyansupulak

6 лет назад, # ^ |

← Rev. 2 →

@devanshsaxena16 Thanks for the reply. I got it now and I am a Dragon ball fan too.

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AriaB

6 лет назад, # |

← Rev. 2 →

Hi,

In B Question in the solution of author ==>

a--, b--;

Why the author use this line when his getting pairs?

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majdoor

6 лет назад, # ^ |

Because , he want to store them in the range between [0,n-1]. Many programmers are used to this notation .

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alphatrance

6 лет назад, # |

Can someone explain 1169E? I didn't get it from the editorial.

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meaningIess

6 лет назад, # |

Problem 1168E is just same to the problem in the Topcoder! Here is the address: https://community.topcoder.com/stat?c=problem_statement&pm=14159
The problem is in SRM686.

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advitiyabrijesh

6 лет назад, # |

← Rev. 2 →

-8

why this approach doesn't works for E — AND Reachability, for every bit i merge all numbers have that bit on and to answer the query if two numbers lie in same component then they are reachable otherwise not.

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Malomalomalomalo

6 лет назад, # ^ |

Its not guaranteed that two numbers that are reachable share a bit. (the and checks two numbers for adjacancy, not the whole sequence.)

For example: 8 12 4

We can reach the 4 from 8, but they don't share any bit.

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advitiyabrijesh

6 лет назад, # ^ |

what I'm trying to do is for each bit I'll merge every number whose bit is on so for your case 4&12 will get merged when we check 2nd bit then 12&8 will get merged when we check 3rd bit so every number lies in single component

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nks43

6 лет назад, # ^ |

Assume you need to find reachability from index 3 to index 4 and a3&a4 is 0 but a3&a1>0 and a4&a1>0, then your code will return reachable as a1, a3 and a4 share a component but since index 1 is not between 3 and 4 this route shouldn't be used.

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advitiyabrijesh

6 лет назад, # ^ |

but it's mentioned in the question we can use any permutation of indexes.

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poojanpatel101

6 лет назад, # ^ |

← Rev. 2 →

x=p1<p2<…<pk=y , and api&api+1>0 for all integers i such that 1≤i<k.

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zarrym911

6 лет назад, # |

Is there any other way to solve 1169C, instead of the one provided in the editorial. The question was really good BTW!

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https://codeforces.me/blog/entry/67241?#comment-513902

6 лет назад, # ^ |

see this comment for better explanation

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6 лет назад, # |

Can anyone please explain their solution of Problem 1169E (And Reachability) ?

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mohamedeltair

6 лет назад, # |

+52

Complexity of Div2 E is $$$n(log(n)^2)$$$ not $$$nlog(n)$$$

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zarif_2002

6 лет назад, # |

please anyone explain problem D proof.

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_abcd_

6 лет назад, # ^ |

← Rev. 2 →

I am not able to prove it any formal way but i can do this in informal way.
if x=1,k=1, positions are 1,2,3
so if we want to find upper bound of k it is better not to put triplets of same numbers together.

0011001100110011

Now consider x=1,
k=1->[1,2,3],2->[1,3,5],3->[1,4,7],4->[1,5,9]
for k=1,2,3 condition does not satisfy but for k=4 this conditions satisfied.so our maximum position will be atmost 9.

  for (int i = n - 1; i >= 0; i--) {
    vl[i] = vl[i + 1];
    for (int k = 1; i + 2 * k <n; k++) {
      if (s[i] == s[i + k] && s[i + k] == s[i + 2 * k]) {
        vl[i] =min(vl[i], i + 2 * k);
      }
    }
    ans += n - vl[i];
}

Here we iterate from backward find the minimum possible r for this l.

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Advancingxq

5 лет назад, # ^ |

You can use the "pigeonhole principle" to understand this lemma. For a 01 string of length 9, it means at least five zeros or five ones. Assuming there are five ones, then no matter where 0 is, there will be a k satisfying condition.

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5 лет назад, # |

can someone explain me the div2 B solution using graph..? Thanks.

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devarshi09

5 лет назад, # |

+12

Div 2E:AND Reachability is a beautiful problem with a beautiful solution!

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5 лет назад, # ^ |

Please Can you explain me how to solve it. I cann't understand the editorial.

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5 лет назад, # ^ |

I got it nvm.

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solimm4sks

5 лет назад, # |

← Rev. 2 →

Why is Div2B tagged as "Graphs"?

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Xellos

5 лет назад, # |

+18

Huh. So I solved E with machine learning. I thought for some time, didn't prove anything useful except the form of the answer if the input is a permutation as well and $$$p = I$$$, so I just took that as the starting point and did some random swaps until I got the target sequence. It should be slow theoretically and doesn't have to terminate, but it worked pretty damn fast after some speedups.

Specifically, I keep $$$p = I$$$ and only care about the number of occurrences $$$c_i$$$ of each number in the target (input) sequence compared to the number of occurrences in my current sequence $$$p \oplus q$$$. I calculate the loss function $$$L = \sum |c_i - c^*_i|$$$ and perform only swaps in $$$q$$$ that don't increase $$$L$$$. I go through the whole permutation $$$q$$$ multiple times in random order, ignore some elements that probably won't decrease the loss when swapped and try all swaps otherwise. Swaps that strictly decrease $$$L$$$ are made immediately. If there are currently no such swaps and $$$L > 0$$$, I also make swaps that keep $$$L$$$ constant.

It's really just hill climbing — small incremental improvements to minimise loss, with some "epochs" that start by shuffling and a bit of extra swapping after each epoch to increase entropy.

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