766A - Mahmoud and Longest Uncommon Subsequence

If the strings are the same, Any subsequence of *a* is indeed a subsequence of *b* so the answer is "-1", Otherwise the longer string can't be a subsequence of the other (If they are equal in length and aren't the same, No one can be a subsequence of the other) so the answer is maximum of their lengths.

Code : http://pastebin.com/aJbeTTjw

Time complexity : *O*(|*a*| + |*b*|).

Problem author : me.

Solution author : me.

Testers : me and mahmoudbadawy.

#### First solution :-

Let *x*, *y* and *z* be the lengths of 3 line segments such that *x* ≤ *y* ≤ *z*, If they can't form a non-degenerate triangle, Line segments of lengths *x* - 1, *y* and *z* or *x*, *y* and *z* + 1 can't form a non-degenerate triangle, So we don't need to try all the combinations, If we try *y* as the middle one, We need to try the maximum *x* that is less than or equal to *y* and the minimum *z* that is greater than or equal to *y*, The easiest way to do so is to sort the line segments and try every consecutive 3.

Code : http://pastebin.com/NsCkbQFS

Time complexity : *O*(*nlog*(*n*)).

#### Second solution :-

Depending on the note from the first solution, If we try to generate a sequence such that after sorting, Every consecutive 3 line segments will form a degenerate triangle, It will be 1 1 2 3 5 8 13 ... which is Fibonacci sequence, Fibonacci is a fast growing sequence, *fib*(45) = 1134903170, Notice that Fibonacci makes maximum *n* with "NO" as the answer, That means the answer is indeed "YES" for *n* ≥ 45, For *n* < 45, You can do the naive *O*(*n* ^{3}) solution or the first solution.

Code : http://pastebin.com/82XcJfgp

Let *x* be the number that satisfies these inequalities:-

*fib*(*x*) ≤ *maxAi*.

*fib*(*x* + 1) > *maxAi*.

Time complexity : *O*(*x* ^{3}) or *O*(*xlog*(*x*)).

Problem author : me.

Solutions author : me.

Testers : me and mahmoudbadawy.

Let *dp*[*i*] be the number of ways to split the prefix of *s* ending at index *i* into substrings that fulfills the conditions. Let it be 1-indexed. Our base case is *dp*[0] = 1. Our answer is *dp*[*n*]. Now let's calculate it for every *i*. Let *l* be the minimum possible index such that the substring from *l* to *i* satisfies the condition, Let *x* be a moving pointer, At the beginning *x* = *i* - 1 and it decreases, Every time we decrease *x*, We calculate the new value of *l* depending on the current character like that, *l* = *max*(*l*, *i* - *a* _{ s[x]}). While *x* is greater than or equal to *l* we add *dp*[*x*] to *dp*[*i*], To find the longest substring, Find maximum *i* - *x*, To find the minimum number of substrings, there is an easy greedy solution, Find the longest valid prefix and delete it and do the same again until the string is empty, The number of times this operation is repeated is our answer, Or see the dynamic programming solution in the code.

Code : http://pastebin.com/4JiXSwfU

Time complexity : *O*(*n* ^{2}).

Try to find an *O*(*n*) solution(I'll post a hard version of some problems on this blog soon).

Problem authors : me and mahmoudbadawy.

Solution authors : me and mahmoudbadawy.

Testers : me and mahmoudbadawy.

766D - Mahmoud and a Dictionary

Let's build a graph containing the words, For every relation in the input add a new edge with the weight of 0 if they are equal and 1 if they are opposites, If adding the edge doesn't make the graph cyclic, Our relation is valid, Otherwise it may be valid or invalid so we'll answer them offline. Check if adding that edge will make the graph cyclic or not using union-find like Kruskal's algorithm. Suspend answering relations that will make the graph cyclic, Now we have a forest of trees, Let *cum*[*i*] be the xor of the weights on the edges in the path from the root of the component of node *i* to node *i*. Calculate it using dfs. To find the relation between 2 words *u* and *v*, Check if they are in the same component using union-find, If they aren't, The answer is 3 otherwise the answer is , Now to answer suspended relations, Find the relation between the 2 words and check if it's the same as the input relation, Then answer the queries.

Code : http://pastebin.com/WqwduaYs

Time complexity : *O*((*n* + *m* + *q*)*log*(*n*) * *maxL*) where *maxL* is the length of the longest string considering that union-find works in constant time.

Problem author : mahmoudbadawy.

Solution author : me.

Testers : me and mahmoudbadawy.

Wait for a hard version of this problem.

If we have an array *ans*[*i*] which represents the number of paths that makes the *i* ^{ th} bit sit to 1, Our answer will be

Let *arr*[*i*][*x*] be the binary value of the *x* ^{ th} bit of the number attached to node *i*(just to make work easier).

There are 2 types of paths from node *u* to node *v* where *u* is less in depth than or equal to *v*, Paths going down which are paths with *lca*(*u*, *v*) = *u* and other paths, Let's root the tree at node 1 and dfs, let current node be *node*, Let *dp*[*i*][*x*][*j*] be the number of paths going down from node *i* that makes the *x* ^{ th} bit's value equal to *j*. A path going down from node *i* is a path going down from a child of *i* with node *i* concatenated to it so let's calculate our *dp*. A path that isn't going down is a concatenation of 2 paths which are going down from *lca*(*u*, *v*), Now we can calculate *ans*. See the code for formulas.

Code : http://pastebin.com/n2a3kijD

Time complexity : *O*(*nlog*(*a* _{ i})).

Problem author : me.

Solution author : me.

Tester : mahmoudbadawy.

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Thanks for really fast editorial.

Waiting for O(N) solution for C

It's just the same thing again. maintain prefix sums on dp and use a two pointer approach to find the minimum index where we can split.

Or, Maintain another dp such that dp[i][x] is the last index less than or equal to i such that character x appears, You can calculate l in O(1) with that dp and the sum in O(1) with a cumulative.

Very fast editorials...Thanks alot :-)

Problem CO(N) solution hint: You can use two pointers with monotonic queue range minimum and partial sums.Or just keep a boolean array of size 26, indicating the presence of each letter in the current segment. :p

This will be

O(N* Σ) where Σ is the alphabet size :PAs the alphabet size is considerate to be constant (in this task) the complexity is O(N)

I should know min(v[j]), l<=j<=I, where l is my current left pointer, i is the current index. How do i get it in O(1)?

Link

Thanks. I've found a solution using a similar idea that uses 2 stacks instead.

I tried to solve this question with the matrix multiplication method but getting a timeout. I have only just implemented a partial solution that calculates no of ways to split. Can you help me? Here is my solution...

typedef long long ll;

using namespace std;

ll n;

string str;

map <char,ll> m;

ll solve(ll l, ll r) { if (l>r) { //cout<<"l>r "<<l<<" "<<r<<endl; return 0; }

}

int main() { cin>>n;

}

Thnx in advance...

Fast and Fantastic editorial. Really nice and enjoyable contest. Thanks to you both. Wishing you guys best of luck :)

We can do D using dsu which works online

I though the same. Can you help me? I got MLE, and I don't know why... Here is my code: http://codeforces.com/contest/766/submission/24508577

Can you explain how?

Cleaned my ugly dsu solution and made it pretty: 24513551

My idea was the following. For each set the representative of the set stores the index of a value of the opposite set (or -1 if it has no opposite). When combining two sets

`x`

and`y`

, you additionally have to union the sets`opposite(x)`

and`opposite(y)`

and set the opposite links afterwards. And when you want to make the sets`x`

and`y`

opposite, you have to union the sets`x`

and`opposite(y)`

and the sets`opposite(x)`

and`y`

and also set the opposite links new.Thanks for the idea. It was quite time consuming to get everything right with the code.

Anybody who's still struggling, here's further explanation —

synonyms must be in the same tree of the dsu structure. There can be many dsu trees. The whole input will represent a dsu forest.

only the representative element of a particular dsu tree will hold information about the antonym tree's representative element. and that antonym tree's rep. element will hold information about the former tree's representative elements.

the antonym value of each dictionary string will be = "none" (I used -1, because "none" can be in the dictionary) if it's not the rep. element of it's own tree.(this was important in my implementation, otherwise, there will be too many pointers between trees and they can't be tracked to the values we need)

the antonym value of rep. elements which don't have an antonym yet will also have antonym set to none.

to summarize,

the forest will have multiple dsu trees, and a few undirected antonym edges, such that each tree is associated with at most one antonym edge.

if a particular tree has more than one antonym edge( from rep. element only, as other vertices of a tree have antonym value set to null ) then each antonym tree of this particular tree can be joined into a single tree, therefore restoring the "each tree has at most one antonym edge" property.

We must now appropriately set some antonym values to "none", after joining the antonym trees, and make sure that the antonym edge that still remains points both ways.

hello. I need help. I think this problem (Uva-10158 War) is very much similar to D. so I tried to do exactly what the author says to solve that problem.It matches all the i/o i found in the net.But still getting TLE. Don't understand why. this is my code: (code) thanks in advance :)

I did it and it doesn't work. http://codeforces.com/contest/766/submission/24809940 No matter what I do it won't pass test 10.

Here's a minimal test case that doesn't work:

Thanks, here is my newest and clearest version and it passes this minimal test case, but not test 10. I really can't find the error here. http://codeforces.com/contest/766/submission/24810822

Try this test:

The error happens during the input

`2 A B`

in the following code:`a = "A"`

and`b = "B"`

. In the first line you union`"B"`

and`find_ant("A") = "C"`

.`"C"`

gets the new parent of the combined set. Then you update`a`

and`b`

to their parents, so`b`

becomes`"C"`

. But by doing this, you lost the information that`"B"`

is an antonymy of`"D"`

. So the`"A"`

and`"D"`

don't get unioned.Thank you so much, you really helped me a lot. I spent hours trying to find the mistake. How did you find it? Did you look into the code or did you find a counterexample?

Mostly by looking at the code and trying some examples on paper. But it also took me quite a while to find the error.

You always keep words that are equal in the same set. Furthermore for every word you keep a reference to a word of the antonymy set if it exists.

At the start every word is in its own set and does not have an antonymy. When adding a synonym you check if the antonym of one word is in the same set as the other word. (Then this relation is invalid). Otherwise you use Union to put them in the same set. You also have to join the antonyms of the words. When adding a antonym you check if both words are in the same set, then the relatioin is invalid. Otherwise you join the antonym of word a with the set of word b and vice versa.

I think I may have done some redundant stuff, but it works: http://codeforces.com/contest/766/submission/24510075

well i did exactly as you told but my solution is giving runtime error ie eof reached at test case 5. i tried using different online compiler like codechef or codetable but all give different errors and different outputs. even my pc gives different result.

can you please look into the code.

http://codeforces.com/contest/766/submission/24611187

For every question you have to print exactly one line. In the case where you check if(rb==root(c).first) you don't print anything in the else case.

well your advice worked but i had some really serious bugs. i have been debugging this since that day but finally i am getting TLE. can you tell me how did you optimize your solution. it would help me a lot

http://codeforces.com/contest/766/submission/24679609

I did exactly what you did and no matter how hard I try it won't pass test 10. I fixed everything I could, still the same. http://codeforces.com/contest/766/submission/24809940

yep! 24703668

I used some of the ideas shown here but here are my drawings of all online cases

sorry I didn't put when to mark the solution as invalid, In my code it is shown.

I hope my image helps somebody to reason this approach.

Variable naming is awesome.

Tried to implement C during contest in

O(N^{3}), but couldn't finish debugging.O(N^{3}) passed system tests, indeed 24514033.I should say u just passed.(1950ms) that's a thin margin.

Could someone explain why the dp (for the number of ways) works for C? I can't seem to understand why the base case is 1 or why we add dp[x] when x is also equal to L.

Like for the first sample case when the string is "aab" and we get to letter b, L would end up equaling 1 (since i = 3 and a1 = 2) but why would we add the dp value at position 1 if the letter at position one can't be in the same substring as position 3?

did not get that too..confusing explanation

My DP solution is a little different. It is easy to dinamically calculate two functions: 1) f(i,j) = number of ways to split first i symbols of our string where rightmost word has length j, 2) g(i,j) = minimal number of words in splits of first i symbols of our string where rightmost word has length j. I can write recurrent equations if needed.

Interesting. I think I'll try and figure out the recurrent equations and probably solve this problem 2 ways. Thanks for the help!

Hey ! Yeah it seems a bit confusing but you can approach it like this as well ,

let dp[i] = number of partitions possible if the last string ends at index i .

for this you have to check for every substring ending at index i and check if it is a valid one .Let the string be j.....i .

If it is valid simply add dp[j-1] to dp[i] .i.e these much partitions can be made by having last string as s[j.....i] .

Coming to the base case :

You can have it as :

dp[0] = 1 where 0 indicates the first letter of the given string as there is only one way to split a single character .i.e the character itself.

and dp[n-1] will be the answer .

Can refer my submission : link

Thank you so much! This makes a lot more sense to me.

can someone explain the second solution for B?

Hi, for the question C,

For the first example test case, how come that "aab" (the whole string) can't be included into the number of possible ways?

Because from my understanding, the limit of a=2 and b=3 doesn't apply to the "aab" thus we can include it into the possible ways.

yes, we can include it. The solution splittings are — {a,a,b}, {a, ab}, {aa, b}

Why does {aab} don't get included into the splitting?

Because from the limit rules it said a1=2 and a2=3, but from {aab} it only has a=2 & b=1, so it's totally possible to include it into the splitting.

edit*: I assuming that a1 = limit of char A, a2 = limit of char B. Is my understanding is wrong?

It's wrong .. a1 = limit of the string size that we can write char A in it .. read the announcement about the problem B

I assumed the same during contest but it is wrong.

`That's because this magical paper doesn't allow character number i in the English alphabet to be written on it in a string of length more than ai.`

It means character i cannot be present in a substring of length more than ai.

OMG, this question really gives me a hard time to think. :(

I think maybe that I got confuse or not understand those text (your marked text) completely.

So, from the test case, we have a1=2 and a2=3, which means that the maximum of character 'a' and 'b' is both 2 and 3 respectively.

Thus, "aaa" isn't possible because it has three 'a's. The question said the maximum is a1=2, thus it's not possible.

Another example might be, "bbbb", also is not possible because it has four "b"s. The question said the maximum is a2=3, thus it's not possible.

Then why for "aab" it can't be included inside the splitting? I ask that because it only has a=2 and b=1, which totally fine with the rule a1=2 and a2=3.

That's because this magical paper doesn't allow character number i in the English alphabet to be written on it in a string of

lengthmore than ai.Suppose there is 'a' present in string then length of that string cannot be more than a[0].

Notice

lengthNOT frequency!So "aab" is not accepted because length of string is more than 2 but a[0]=2.

I finally understand the question! I misunderstood the question all the long way at competition until now. duhh.

Thanks a lot @dush1729

I think you misunderstood the problem.

Limit given for each characters is the max length of substring character can be a part of.

so if we include "aab" as one possible way then it violate the condition that a1 = 2 i.e 'a' cannot be part of substring more than length 2 (in case of "aab" length is 3)

I hope this helps.

Yeah this really help, I finally been able to understand the question!!

Thanks @alphaguy4

I think you misunderstood the problem.

Limit given for each characters is the max length of substring character can be a part of.

so if we include "aab" as one possible way then it violate the condition that a1 = 2 i.e 'a' cannot be part of substring more than length 2 (in case of "aab" length is 3)

I hope this helps.

Добрый день. Кто может написать решение 3 задачи нормальным русским языком? Заранее спасибо.

Задача А:

Задача АЕсли две строки одинаковые, нет там общый не-подпоследовательность — ответь -1. Если две строки не равны, тогда оба друг-друга не-подпоследовательность, а в задачи требует максимальный, ответь максимум длина этих строк.

Задача В

Задача ВЗдесь предлагается две решение: 1) отсортируем всех числа , и будем проверять последовательно по трех чисел, если a[i] , a[i+1], a[i+2] -- можно построит треугольник ответь — YES, иначи нет. Почему ??? Чуть немножка подумай).

ИсходникВторое решение основано на это: Нельзя построит треугольник из множества чисел только если эти числа — числа Фиббоначи ( 1 1 2 3 5 8 13 21 34 ... ). Но факт числа Фиббоначчи очень быстро растет, значить 45-число Фиббоначи больше 10^9. ОТкуда, если n > 45 -- ответь всегда YES, иначи будем проверять все число с трем циклом.

Задача С.

Задача С1-вопрос количество разрез2-вопрос

Максимальный разрезМинимальный количество разрезWho knows features of Test 27 on problem D? For some reason my solution (Python) have an RE on it... Using onlene dsu like in this comment.

Can anyone explain why my code fails on testcase 10 problem D Code Is the problem hashing collision?

well i have some doubts in 766E

my algo is that i will apply dfs recursively and add values of path from that vertex to vertices coming after that. these will be used to determine other distances. but it is giving wrong answers. can somebody point out a mistake in this?

here is link to my code:

http://codeforces.com/contest/766/submission/24524352

Where does log(n) come from in the described solution for D? Assuming dsu works in constant time it takes linear time to construct the graph andlinear time to run 1 dfs, which is also linear, and then we do m+n queries at O(1). Have I missed anything?

You missed the map. Look at his code.

Thank you)

Although I believe hashmap is enough for this problem, so the logarythm is still not neccessary.

When I had read the problem E,I wrote a code using Centroid Decomposition without thinking twice....

It's been 3 years though, but if you can read the statement again, and help with, how to approach this with centroid decomposition? . Thanks

As a Chinese participant, I was surprised to find that 766D is very similar to a problem

Food Chainin China National Olympiad in Informatics in 2006....Can someone please explain Problem E in detail ???? .Particularly how do we come to such DP states and the transiton between the states .Thanks in advance :)

http://codeforces.com/contest/766/submission/24547430

The narrations should do the job

Very nice Code.

That's much better than editorials description.

hello, I am new in coding, can plz explain the solution in simpler words and what is the basic idea that is used ??

D becomes much simpler if while merging we copy each element of smaller set to larger set and change their color accordingly.Not faster, but easier to code.

The number of times we have to push an element in a larger set is at max logN time, because whenever we push a smaller set to larger set the size of smaller set becomes atleast double.How manny times can a set size becomes double? Log2(N). So Complexity is N*logN.

That's why DSU with only Union by rank has time complexity of N*LogN ,But when we do path compression along with Union by rank the amortized complexity becomes O(K*N) where K<=6 for practical values of N.

A nice solution for problem B(div.2) with random!! ;) Code: 24497457 The exception value is really good=)

Well, how you can be sure that it will work on every time? hmm

## include<bits/stdc++.h>

using namespace std;

## define ll long long int

## define pb push_back

## define mp make_pair

## define ff first

## define ss second

## define fr(i,a,b) for(ll i=a;i<b;i++)

## define frn(i,a,b) for(ll i=a;i<=b;i++)

## define frb(i,a,b) for(ll i=a;i>=b;i--)

## define ssi(a) scanf("%lld",&a)

## define si(a) scanf("%d",&a)

const int N=100007; int main() { ll n,a[N]={0},i,j,flag=0,mid,l,h,sum; ssi(n); fr(i,0,n) ssi(a[i]); sort(a,a+n); fr(i,0,n) { fr(j,i+1,n-1) { l=j+1; h=n-1; sum=a[i]+a[j]; while(l<=h) { mid=(l+h)/2; if(a[mid]<sum) { flag=1; break; } else l=mid+1; } if(flag==1) break; } if(flag) break; } if(flag) printf("YES\n"); else printf("NO\n"); return 0; }

In the above code i am getting TLE in test case 5 but the code given below is accepted. Can anyone point out the reason?

## include<bits/stdc++.h>

using namespace std;

## define ll long long int

## define pb push_back

## define mp make_pair

## define ff first

## define ss second

## define fr(i,a,b) for(ll i=a;i<b;i++)

## define frn(i,a,b) for(ll i=a;i<=b;i++)

## define frb(i,a,b) for(ll i=a;i>=b;i--)

## define ssi(a) scanf("%lld",&a)

## define si(a) scanf("%d",&a)

const int N=100007; int main() { ll n,a[N]={0},i,j,flag=0,mid,l,h,sum; ssi(n); fr(i,0,n) ssi(a[i]); sort(a,a+n); fr(i,0,n) { fr(j,i+1,n-1) { l=j+1; h=n-1; sum=a[i]+a[j]; while(l<=h) { mid=(l+h)/2; if(a[mid]<sum) { flag=1; break; } else h=mid-1; } if(flag==1) break; } if(flag) break; } if(flag) printf("YES\n"); else printf("NO\n"); return 0; }

Please explain second solution of problem B, Mahmoud and Triangles

for a non-degenerate triangle(non-zero area) we have to satisfy the condition x+y>z. If we sort the length in increasing order, we only have to check for three consecutive lengths to be the sides of triangle (i-1, i, i+1) and check if(a[i+1]<(a[i]+a[i-1])).

If this condition is false for i-1, i, i+1 sides it will be false for any triples where largest sides is greater than i+1th value. Similarly we take i-1th side as smaller side if the condition is false to i-1, i, i+1 it will be false to any value smaller than i-1th side length.

So, in the best case for an ith length i-1th and i+1th value is the best case for being a triangle.

Thanks. But I understood this one. I was asking for explanation to the solution using Fibonacci sequence.

In Febo sequence 1, 1, 2, 3, 5.. observe all the consecutive elements just form degenerate triangle.

So, if difference of smallest and largest value is smaller than F(n) (n is the size of array 'a') there exist 3 consecutive elements such that it form a triangle.

F(45)=1134903170>(10^9) For n>=45 for any value dif(larg-sml)<F(45) so, triangle exist. For n<45 run a n^3 loop and check for all the triplet.

Great!

I need help. I think this problem (Uva-10158 War) is very much similar to D. so I tried to do exactly what the author says to solve that problem.It matches all the i/o i found in the net.But still getting TLE. Don't understand why. this is my code: (code) thanks in advance :)

You haven't added any weight to the trees on Union. If you always add as first element as parent it would lead to linear time in parent_finding();

Add a weight value to each node and take node with higher weight as parent. Weight would be the no. of element attached to it. Actually you have to take the element with larger height as parent but the weight would do as, for reaching 3-height atleast 5 nodes are required, and from there on the required nodes would double with each increment of height, for example 3-5, 4-10, 5-20, 6-40...

Hopes that helps

thank u. i implemented a rank based union system. in union what i do is if rank[u]<rank[v] then par[u]=par[v],rank[v]+=rank[u]; otherwise i did par[v]=u; ranking[u]+=ranking[v];

it helps me to run in time limit and get accepted :)

Can anyone please explain Div2 C? The solution is not clear to me.

It uses dp: Basically for a substring s[x...i] we determine if every character satisfies the condition of its occurence as: a[s[x]-'a']<=len (where len is i-x+1 ) same as 'if(f>x)' in authors solution if yes then: 1. we add the result of the sub-problem dp[x] to dp[i] 2. we also keep track of maximum valid string length by l = max(l,i-x); 3. dp2[] is for minimum substrings , or minimum cuts to be done in the string so that it can satisfy the condition. so for every valid x we add 1 and take the minimum of all. dp2[i]=min(dp2[i],dp2[x]+1)

Wrote DSU for D without reading the solution. My idea is exactly the same as others' but it won't pass test 10. No matter what I do, no matter how much I fix it this solution just won't pass test 10. Here is my cleanest and last version of the program. Still won't pass test 10. http://codeforces.com/contest/766/submission/24810631 What the hell is in test 10?

Can someone explain solution For B using Fibonacci Sequence? Thanks in Advance.