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By rivalq, 12 months ago, Hint
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# 1682F — MCMF?

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Solution Tutorial of Codeforces Round 793 (Div. 2) Comments (96)
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 » :holyfrick: That was lightining fast .. Thnx for the hinted editorial !! helps a lot in upsolving :)
 » great problems, fast editorial, quick rating changes, and becoming specialist. thanks alot :D
•  » » Congratulations! I have become specialist too... again)))
•  » » » Thanks, Congrats on that :)
 » 12 months ago, # | ← Rev. 2 →   Missed C by just "+1", took single/2 instead of (single+1)/2. :(
 » I wish authors who put an anti-hash test in C a very pleasant evening.
•  » » Did authors include anti-hash test, or is it somebody's hack?
•  » » » authors did
•  » » » » How can you be so confident? lol
•  » » » » 12 months ago, # ^ | ← Rev. 2 →   No, we are not that cruel (At some point, I was also an expert).
•  » » » » » oh, sorry, than it must have been a very lucky coincidence for me lol
•  » » » »
•  » » » » » 11 months ago, # ^ | ← Rev. 2 →   .
 » 12 months ago, # | ← Rev. 2 →   Rating change is faster than editorial :D thanks
 » Is E related to something like toposort(finding indegree = 0)?Can someone give hint please
•  » » 12 months ago, # ^ | ← Rev. 4 →   hintmake the swaps in permutation into trees, slove the problem in it, and then using toposort.
•  » » » Can you elloborate on the idea ? I can't get how to construct a tree form permutation Thanks!
•  » » » » 12 months ago, # ^ | ← Rev. 2 →   If we connect $i$ and $p_i$ in a graph, you will see that the graph is some loops. So for one loop with length $k$, the minimal number of swap must be $k-1$ and the swap must connect all node in the loop. So you will see that if you connet all swap $(a_i,b_i)$, it will look like a forest.
•  » » » » » Yes, but I have no idea how to solve it on the tree
•  » » » » » » 12 months ago, # ^ | ← Rev. 3 →   OK. For an element in position $i$, we need to move it to position $p_i$. We will see that we can only use a series of swaps to move $p_i$, and two node have only one path in a tree. So if the edges in the path from $i$ to $p_i$ is represented as $x_1,x_2,\dots$, they must be used in such an order in the final. So we can make a new graph to describe the order for the swaps, and then use a topo sort can slove it.
•  » » » » » » » 12 months ago, # ^ | ← Rev. 2 →   But it seems that the number of edges is $O(n^2)$..?
•  » » » » » » » » 12 months ago, # ^ | ← Rev. 4 →   Oh, I forgot it. A good observation in this problem is that one edge may appear at most twice in all paths. Because A swap edge can only change two position, so it can be proved that if three or more paths through an edge, there must have no solution. So there are at most $2n$ visit for all the edges. Finally, the time will be $O(n)$.
•  » » » » » » » » » got it, thx
•  » » » » » » » » » It is unbelievable that I send all the solution in the discussion.
•  » » » » » » » » » lol
•  » » » » » 12 months ago, # ^ | ← Rev. 2 →   I only have an $O(n^2)$ algorithm. For each loop, choose one node as the starting point, then judge whether the dfs order on the tree is same as the loop's order.
•  » » » » » » You can see my explanation in the front.
•  » » » » » » » Awesome! Thank you very much
 » Awesome round. Fast Editorial. Quick Rating Changes. rivalq, CoderAnshu supremacy.
 » Can anyone tell me what is wrong with my code idea for B https://codeforces.com/contest/1682/my
•  » » I just edited one line on your code!
 » Just gonna share my construction for D here (no clue whether it's the same as the editorial): Spoiler Red crosses are odd degree vertices. The black lines trace out edges between even degree vertices. Blue lines are edges between odd degree vertices. Also, imagine I drew a circle.Submission: 158082919 (the code isn't very neat)
•  » » What do you do when there are two ones at the end?
•  » » » If the upper or lower segments of even degree vertices do not exist, it is fine to connect them to the next available point.You may be thinking that the previous 1 will have even degree, but it will be compensated by a blue edge.
•  » » » » 12 months ago, # ^ | ← Rev. 3 →   For example, this is what I get when I try to follow the intuition from the picture to construct the tree on string 10001100 (crosses are vertices with odd degrees, circles — vertices with even degrees) If I follow the intuition from the picture, I end up with a 1 with an even degree.
•  » » » » » Well actually the answer is NO for this testcase since the number of 1s is odd.
 » the round was lucky for me; I just became specialist ;)
 » Another easy to see proof for A: Number of element equal to the middle element has the same parity as that of N. So decrease N by 1, the parity of the middle elements should be changed
 » 12 months ago, # | ← Rev. 2 →   Here's a pictorial solution to D which is very similar to the one in the editorial. SpoilerAssume that the red dots are odd guys and there are lots of even guys on the boundaries. The blue curves denote the edges. The green one is a special case to handle some leftover even guys. The one in the editorial just picks a pivot adjacent to a red dot so that the green curve looks like a blue one. Observe that the parity of the degree of the pivot is correct irrespective of its choice.
•  » » 12 months ago, # ^ | ← Rev. 2 →   ありがとう
 » 12 months ago, # | ← Rev. 2 →   I think C solution is not correct, as it doesn't contemplate all possible cases. The one concretely isn't ok on the pretests 2 is the input num 22:1 1 2 3 3Where it says answer is 3. If it was the 1 or the 3 the number that had a single appearance it would be right, as it could be put in the middle of the array, but when is the 2 there is no way to order it that gives result 3. I made this awful code just to test it and it seems that the max increasing substring is in fact 2 // C++ program to display all permutations // of an array using STL in C++ #include using namespace std; int maxSubsequence(int a[], int n); // Function to display the array void display(int a[], int n) { for (int i = 0; i < n; i++) { cout << a[i] << " "; } cout << endl; cout << maxSubsequence(a, n) << "\n"; } int maxSubsequence(int a[], int n){ int rl = 1; int lr = 1; int aux = 1; for(int i = 0; i < n; i++){ aux = 1; int j = i + 1; while(j < n && a[j-1] < a[j]){ aux++; j++; } rl = max(aux, rl); } for(int i = n-1; i >= 0 ; i--){ aux = 1; int j = i; while(j > 0 && a[j-1] > a[j]){ aux++; j--; } lr = max(aux, lr); } return min(rl, lr); } // Function to find the permutations void findPermutations(int a[], int n) { // Sort the given array sort(a, a + n); // Find all possible permutations cout << "Possible permutations are:\n"; do { display(a, n); } while (next_permutation(a, a + n)); } // Driver code int main() { int a[] = { 1, 1, 2, 3, 3 }; int n = sizeof(a) / sizeof(a); findPermutations(a, n); return 0; } Am I missing something?
•  » » The answer is 1 3 2 3 1. Your testing code is just not corrected in finding LIS.
 » I liked problem F, but I feel like it would've been better to not add the artificial complexity from the bipartite flow graph stuff.I'd prefer something like "there is a wall of stacked 1 by 1 blocks along the number line. for most of the infinite number line the height is average, but unfortunately there are some positions where there are more or fewer than average blocks in a single column. can you figure out how many steps to the left and right in total the boxes need to be moved? The non-average heights are given in this format with queries..."
•  » » I second that, it'd be better this way
 » A shameful round for me only solved A-C ): :(
 » Great Round, enjoyed it a lot!!!
 » in problem B, with the input:180 1 2 7 4 5 3 6it seems like the answer is not correct, if im wrong, pls point out my mistakes.
•  » » what is your ans?
 » Man the solution for D blew my mind :D Great Round!
 » please add the problem rating in the tags of the problems.
 » 12 months ago, # | ← Rev. 2 →   unordered_map users : unordered_map is faster than map... (Le) Author : anti hash function...; " Does map faster than unordered_map " XD
 » It was not intuitive or provable in the contest that maximum X would be the AND of all misplaced elements of the array. Bad day for me:)
 » Changing just one line in my problem C code gave AC after the contest :(
 » For problem B, why the upper bound of the answer is the bitwise AND of all elements which are not at their correct positions.
•  » » Because X must be a submask of all such elements and bitwise AND is maximum of those X.
•  » » » Thanks:)
 » In question C , on using unordered_map , I am getting TLE whereas using map accepts the solution. Can anyone please tell why it is so? My both submissions : Accepted Solution using map [problem:C][TLE submission using unordered_map](https://codeforces.com/contest/1682/submission/158110821)
•  » » 12 months ago, # ^ | ← Rev. 2 →   unordered map answers its queries(add, get) in amortized O(1) time. The key word is amortized, which means that there are cases when it's not that fast. The worst case is O(n) in time for these operations, which you've probably dealt with in tc 28.
 » speedforces
 » can anyone please elaborate editorial B
•  » » From the given sequence of numbers, there is subsequence of numbers from the original sequence that are not in order and can be sorted by swapping among themselves with the help of a fixed value, X in this case, which is also part of the original sequence, when when we take bitwise and of all the numbers from the non-sorted subsequence we make sure the final answer(bitwise AND) is the value made out of all set bits of numbers part of the subsequence hence we can reorder this subsequence with this largest fixed number X. I hope this helps.
 » Video: A very interesting Multiset Hashing Solution to Problem EPosting this because the Editorial to problem E is not posted yet (and I talked to the problemsetters and their solution is different from mine)
•  » » This was the soln I implemented during testing. 158094993
 » For A, don't you mean $s_i = s_{n - i + 1}$ instead of $s_i = {n - i + 1}$?
 » In second problem ,AND SORTING i am confuse in why it is taking 'and(&)' of all elements present at not correct position, please help
•  » » We don't need to move the elements at the right places. Hence only the elements at wrong places should be eligible for swapping. Thus X should be & of all those values.
 » In part C https://codeforces.com/contest/1682/submission/158147851 if I don't write i
 » I'm afraid that I didn't catch the point of problem C.Why don't we just make the subsequence in a monotonically increasing order?Like this:[2 2 4 4 5 5].By this way,LIS(a')=0.
•  » » Oh, I made such a stupid mistake. :( Neglect it.
 » Good Problem E!Maybe Better for having a sooner editorial XD
 » Why not use Hash to solve the 1682A?And why do you Hack unordered_map in 1682C?
•  » » Why not use Hash to solve the 1682A? Because its an overkill. Simpler easy to implement solns exist. And why do you Hack unordered_map in 1682C? Because it's hackable.
 » I have added editorial of problem E. It's really long and detailed. Hope you would like it.
•  » » For problem E, how to understand the phrase "because we have to sort the permutation in a minimum number of moves which isn't possible if two cycles are merged at any instant".
•  » » » I have understood, thanks.
 » 12 months ago, # | ← Rev. 2 →   In problem 1682C — LIS or Reverse LIS?, how do we have the answer 3 for this input: 1 1 2 3 3?
•  » » 1 3 2 3 1
•  » » » Am I blind or what the ans in 13231 is still 2.Where r u getting an increasing sequence of 3
 » Can anyone explan the problem F?
 » Really good problem D and E! But why the editorial of problem F is really late......
•  » » We both are extremely busy in office, sorry for the delay. Will post it very soon.
•  » » » Thanks really much!
•  » » » Understand! Thanks very much!
 » Hey guys I have a question regarding C :Consider the test case : 1 3 1 2 2 Now, there are three arrangements of $[1, 2, 2]$ : $[1, 2, 2]$ : here, $LIS(a) = 2$ and $LIS(a') = 1$ so beauty is $1$. $[2, 1, 2]$ : here, $LIS(a) = 1$ and $LIS(a') = 1$ so beauty is $1$. $[2, 2, 1]$ : here, $LIS(a) = 1$ and $LIS(a') = 2$ so beauty is $1$. So beauty in every case is $1$. My code outputs $1$ as well. However, the output and the formula from the editorial suggest the answer should be $2$. Can anybody point out where the flaw in my reasoning is?
•  » » Nevermind I made a mistake in case $[2, 1, 2]$ got it
 » 12 months ago, # | ← Rev. 2 →   I'm getting TLE at testcase #28 in Problem C. Here's my submission. Can anyone help me why I'm getting TLE.
•  » » Unordered maps can be blown up easily. You can refer comments above for the same.
•  » » » LOL what made hash function slow here ? I'm new to this could you plz explain it.
•  » » » »
•  » » » » » thanks
 » hey, I am trying to understand why in problem C the solution is said to be NlogN? We simply add to the map and then traverse it, isn't it just N? what am I missing here?
•  » » Adding elements to std:map in C++ is log(size).