### vovuh's blog

By vovuh, history, 19 months ago, translation, ,

I'm really sorry for the issue with the problem D difficulty, it was much harder than i expected, and there was a big difficulty gap between problems C and D. I hope in the next rounds it will never happen again.

UPD: I'd like to say a big thanks to kevinsogo for the great help with tutorials and the round preparation in general.

999A - Mishka and Contest

Tutorial
Solution (Vovuh)

999B - Reversing Encryption

Tutorial
Solution (Vovuh)

999C - Alphabetic Removals

Tutorial
Solution 1 (Vovuh)
Solution 2 (Vovuh)

999D - Equalize the Remainders

Tutorial
Solution (Vovuh)

999E - Reachability from the Capital

Tutorial
Solution (Vovuh)
Linear Solution (Vovuh)

999F - Cards and Joy

Tutorial
Solution (Vovuh)

• +72

 » 19 months ago, # |   +17 Thanks for the editorial, and interesting problems, surely we get to learn something!!
 » 19 months ago, # |   +2 Thanks for the editorial
 » 19 months ago, # |   0 My ideia on 999E - Reachability from the Capital is as follows: Build a new graph with all contracted SCCs in the original graph. In the new graph, for each vertex not reachable from scc[s], compute the in degree. Insert a pair composed of  < degree[i], i >  into a set and run one DFS per each element. Each DFS runned means one edge added. In the end, the numbers os DFS executed is the answer. Great contest! :) 39522069
•  » » 19 months ago, # ^ |   +6 You don't need DFS for each element in the set, the answer is the number of SCC with 0 in degree. All SCC with > 0 in degree will be reached by some vertex, so it is not good to add edges between the capital and some vertex in a SCC with > 0 in degree.
•  » » 18 months ago, # ^ |   0 whats the tc of your algo
 » 19 months ago, # |   0 What is D complexity? I think O(2*m + n)
•  » » 19 months ago, # ^ |   0 O, i wanted to write the time complexity for each of the described solutions and forget about it :( I will do it now :)
•  » » » 19 months ago, # ^ |   0 vovuh, in D if the problem were adjusted so that we can increase/decrease the numbers, can it still have a greedy solution or even DP, It can be solved by Max-flow min-cut but with the constraints adjusted, any idea?
•  » » » » 19 months ago, # ^ |   +3 I thought about it right now, with increase\decrease the elements this problem can be solved at least by mincost maxflow (but this is most stupid idea, i will think about it more time and may be i will find better solution)
•  » » » » » 19 months ago, # ^ |   0 vovuh please feel free to update me here once you've arrived at something.
•  » » » 19 months ago, # ^ | ← Rev. 2 →   0 i tried a different approach which led to the same min no. of changes...but diff array.. but I think the judge is checking only for a particular array...whereas in the question it says print any array...
 » 19 months ago, # |   +1 would anyone provide binary search solution for D as mentioned in tags
•  » » 19 months ago, # ^ |   0 I think this would be proper, I'm using std::set though to find the nearest number that has less than n / m numbers to fill from the current number I'm at with numbers > n / m 39484134Mostly the tags include concepts that can be used, it doesn't mean that their obvious strict implementation is intended but rather the way they work in a certain condition, hope this is clear!
 » 19 months ago, # |   +1 how should i start practicing for problem D, i am not able to solve them ?? (please don't go with my color).
•  » » 19 months ago, # ^ | ← Rev. 4 →   +3 How can I judge the Kyubi??Let's start with the simple facts:1) numbers themselves aren't important, we only care about their modulus to m(i.e. )2) count the numbers that have the same and store their indices in the original array3) now iterate from 0 to m - 1(i.e. all possible mods under m) and if you have a mod that has numbers more than k = n / m store them somewhere call is Large, also store the numbers than have less than k = n / m in a set(or any container that supports finding the nearest number) call it Small4)now for each number in Large in increasing order, call X = count(large[i]) - n / m and you already know the mod of large[i] call it Z, now try to find the nearest mod in Small that is larger than Z in a cyclic order(i.e. if Z = m — 1, then search in[0, Z - 1]) call what you found V and replace the index of this number with any number that have mod = V, do this for all X numbers in large[i].5) now to the proving part, since the problem tells you that you can only increase the numbers you have a greedy solution, why? because assume you have at index i numbers than are larger than group size k, these numbers have to go to some modulus other than i (preferably near) in the direction from [i + 1, m - 1] then [0, i - 1], so you find the nearest modulus that have count < group size(k) and replace, this way you chose the best for index i now go on with the next and so on, to prove this is optimal, suppose there's another index larger than i with count > group size(k) and you decided to go with it first before finding the answer for i first, now suppose you found the answer for that other index and replaced, now i will still have to move much more steps to reach a valid number with count < group size(k), thus our choice of larger index first was wrong, hence the optimal way is to use modulus in increasing cyclic order.submission: 39484134
•  » » » 19 months ago, # ^ |   +3 Thanks for your proof. But I still have a small quetion. XDIn step 5), why i have to move much more steps to reach a valid number with count < group size(k) if we found the answer for another index larger than i with count > group size(k) and replaced ? Could you please give me some counterexamples ? :)
•  » » » » 19 months ago, # ^ |   +1 N = 8, M = 4, K = 2 0 4 12 1 5 8 9 13 Now numbers wih respect to mod M, are [4, 4, 0, 0] If you solved for first you'll have to move more steps to solve for
•  » » » » » 19 months ago, # ^ |   +3 Thanks for your example. XDIn your example, if we sloved for i mod m = 1 first, numbers with respect to mod M are [2, 2, 2(from 1), 2(from 0)], which the total steps are 2 * (2 - 1) + 2 * (3 - 0) = 8. If we sloved for i mod m = 0 first, numbers with respect to mod M are [2, 2, 2(from 0), 2(from 1)], which the total steps are 2 * (2 - 0) + 2 * (3 - 1) = 8, too.Does this prove that we are correct in using modulus in either increasing or decreasing cyclic order? :)Looking forward to your reply!
•  » » » » » » 19 months ago, # ^ | ← Rev. 3 →   +1 Ops! Perhaps this example was a bit weak to prove my point, and you can process mods in decreasing direction, as long as you're finding the nearest valid choice in increasing order a(i.e. a mod higher than the one you're currently processing in cyclic order of course), that last example was a proof for that.Here's a modified version of the counter example The mods array [4, 0, 4, 0] you can process i = 2, 0 in either increasing or decreasing cyclic order, but you cannot replace i = 1, with i = 2, because you now have to go through [2, m - 1] then to [0, 1] which is clearly not optimal.
•  » » » » » » » 19 months ago, # ^ |   +3 Got it. Thanks for your detailed explanation! :)
•  » » » 19 months ago, # ^ |   +3 thank you for your detailed solution .from kyubi.
 » 19 months ago, # |   0 For Question E , Why wouldn't the following solution work?(I'm getting WA):Answer is the No. of connected components in all the vertices that aren't reachable from s.First find all the unreachable vertices from s. Find the no. of strongly connected components among the vertices (using all the edges).
•  » » 19 months ago, # ^ |   0 You should find only SCC-s with indegree equal to 0.
•  » » » 19 months ago, # ^ |   0 Indegree of a Strongly Connected component? Can you explain a little further? Help is appreciated.
•  » » » » 19 months ago, # ^ | ← Rev. 2 →   0 I ment indegree of SCC in the condensed graph(if you don't know the concept fill free to ask). Indegree is number of incoming edges.To find indegrees of SCC-s we can go through all edges(u->v). If u and v belong to the same SCC then we do nothing. Else we increase indegree of the SCC containing v by 1.
•  » » » » » 19 months ago, # ^ | ← Rev. 3 →   0 Emil_PhysMath What is the meaning of graph condensation? What are its uses?Thanks in advance..
•  » » » » » » 19 months ago, # ^ |   +1 Strongly connected component(SCC) is subset of vertices such that any two vertices of this subset are reachable from each other. It is obvious that any vertice belongs to exactly 1 SCC. Thus we can divide the graph into SCC-s. Condensation graph is graph containing every SCC as one vertex. There is an oriented edge between two vertices SCCi and SCCj of the condensation graph if and only if there are two vertices u∈SCCi,v∈SCCj such that there is an edge(u;v) in initial graph. The most important property of the condensation graph is that it is acyclic.You can read about SCC and condensation graph here.You are welcome. If you have a question, feel free to ask. :)
•  » » » » » » » 19 months ago, # ^ |   0 sir i am still unclear with the idea of using SCC , what is wrong in that , if we find the reachable citiesfrom capital using dfs and then connect other from capital and print the count . i mean just find connected component . i know , i am missing one or two things ,please help thanks in advance !
•  » » » » » » » » 19 months ago, # ^ |   +4 My solutionAt first we build the condensation graph(which is acyclic). Now we can connect the capital to all SCC-s with indegree equal to 0(but we will not connect capitals SCC to itself). Note that if we can get to vertex u in some SCC, then we can get to all other vertexes in that SCC. Why is this optimal???Because if indegree of a SCC is equal to 0, we can't get there from the capital(it is obvious). So if their are k SCC-s with indegree equal to 0, we'll have to build at least k roads. Why does this work???Let's assume that we can't get to SCC u. Now if its indegree equals to 0 then we will build a road between it and capital and live happy. Else there is an edge (u1->u). We'll do the same with u1. As condensation graph is acyclic we will never get to the same vertex=>at last we will get to the capital ot to SCC with indegree equal to 0 and live happy. My codeAre you sure?I have warned you39490721 Remove the comments, the code itself isn't very complicated ;)
•  » » » » » » » » 19 months ago, # ^ |   +2 But I can't anderstand your approach.i mean just find connected component you mean to find SCC-s?P.S. You are welcome :)
•  » » » » » » » » » 19 months ago, # ^ |   0 I mean connected component , not scc . I have learnt about scc , just now while solving this problem , but I don't understand why are we using this in this problem Thanks for reply:)
•  » » » » » » » » 19 months ago, # ^ | ← Rev. 2 →   +3 Connected component of a directed graph has 2 definitions.1)SCC.2)It is weakly connected if it would be connected by ignoring the direction of edges.If you find number of connected components of type 2 you will not get the right answer. counter example4 2 4 1 2 3 2 Your answer would be 1, while the true answer is 2.Their is no other definition of connected component, if you used some other concept as connected component, please, tell about it in comments.
•  » » » » » » » » » 19 months ago, # ^ | ← Rev. 2 →   +1 sir can you tell me what is wrong with this approach? what i am trying to do is, run a dfs from the capital city and mark all the reachable nodes visited. Now , we again run dfs on those nodes which are not reachable from the capital city but we will not mark this node as visited. Example, if 2 is not reachable from capital city then we will run a dfs from 2 and will mark all the cities reachable from 2 as visited except for 2. And later, lets say 3 is also not reachable from capital city, so we run a dfs on 3 as well and mark all nodes reachable from it as visited except for 3.If 2 ,which was not marked visited initially , is reachable from 3 then 2 will be marked as visited. Finally we run a loop for all nodes and if a node is not visited just increment the final count. below is my solution 39578066
•  » » » » » » » » » 19 months ago, # ^ |   +1 the_halfbloodprince, your program fails on test case below. input4 3 1 2 3 3 2 4 3  right answer1 your programs output2Here is your mistake: After _dfs-s from nodes 1,2,3 _dfs from node 4 is called. It goes to node 3 which is already visited=>doesn't go to node 2, so node 2 remains unvisited.I am thinking how to fix it. If I have an idea I'll type it in comments
•  » » » » » » » » » 19 months ago, # ^ | ← Rev. 3 →   +1 the_halfbloodprince, you need two arrays visited instead of one. In _dfs function you will use the second array, and before calling _dfs in main() you will have to fill it with value false. And after calling _dfs: if for node u visited2[u]==true you will assign true to visited1[u]: visited1[ind]=true. like thisfor(int i=0;i
•  » » » » » » » » » 19 months ago, # ^ |   0 39601152. Finally AC!!! :)
•  » » » » » » » » » 19 months ago, # ^ |   0 Emil_PhysMath Thanks for your answer sir.Now, i got it.
•  » » » » » » » » » 19 months ago, # ^ |   0 the_halfbloodprince, you are welcome!!!
 » 19 months ago, # |   0 Could you please tell a bit more about solving question E with O(n+m) complexity?
•  » » 19 months ago, # ^ |   0 If you are interested, in the third spoiler of the problem is a linear solution to it. You can read more about Strongly Connected Components (SCC) here. Another part of the solution is quite easy i think. If I am wrong feel free to ask me.
•  » » » 19 months ago, # ^ |   0 Wouldn't be complexity for this solution is O(n*m) ?? As you are running two loops:for (int v = 0; v < n; ++v) { for (auto to : g[v]) { if (comp[v] != comp[to]) { cg[comp[v]].push_back(comp[to]); ++indeg[comp[to]]; } } }At worst case wouldn't it be O(n*m) ?
•  » » » » 19 months ago, # ^ |   +1 No, it will not, because of we will iterate over each vertex and each edge of the given graph only once in this two cycles. So the complexity is O(n + m).
•  » » » » » 6 months ago, # ^ | ← Rev. 2 →   0 but why do we need 2 cycles for problem d, why 1 is not giving right answer?
 » 19 months ago, # | ← Rev. 2 →   +3 In problem E, could anyone show that the solution, explained in second para of the tutorial, will always give an optimal answer.
•  » » 19 months ago, # ^ |   0 If the number of edges added is less than the number of vertices with indegree 0 in the graph's condensation ( other than the vertex containing the capital), then we will still have a vertex in the condensation with indegree 0 and which does not contain the capital. This vertex is unreachable. On the other hand, if you add all edges from the capital to such vertices, we can show that any vertex in the condensation is reachable; just "backtrack". Say if you want to reach a vertex u in the condensation, find a vertex v1 such that v1-> u is an edge. Then find v2 such that v2->v1 is an edge. We must terminate somewhere because the graph is a DAG. It can terminate only at a vertex with indegree 0, and this vertex is the vertex containing the capital.
 » 19 months ago, # |   0 In problem E , why do we need to again run a dfs according to count? Why does this solution 39525886 gives answer 1 more than the correct answer for the test case no:4? Can someone help me in understanding the incorrectness of my solution? If possible , please explain with a sample graph.
 » 19 months ago, # | ← Rev. 2 →   0 My idea for the problem D is that store the modulo of each element, and sort base on thier modulos. I use 2 pointer, for the end and start of the array. Let R is the modulo that each elemenent is need to be same, and CNT to check if CNT == N / M to inc R,for each modulo of element, let say Ai, if Ai <= R then the increment is R- Ai, else is M+ R- Ai, do the same on the second pointer, choose the least one, update the total cost, and decrease the list until empty. I hope my solution can help you solve easily.
•  » » 19 months ago, # ^ |   0 Cant understand the use of r?
•  » » » 19 months ago, # ^ |   0 R is used to check the modulo is lying on, sorry for my bad description :( , you can check my code 39490011
 » 19 months ago, # |   0 Can someone explain to me how and WHY this dp solution for F works? I really don't understand O(n^2*k^2) for loop, but I don't understand also ans=(1,10^5)∑ i=dp[f[i]][c[i]] part. I think this way: if I let's say use c[i]=2 cards and k is equal to 3. I need to use my 3rd card that is not my favorite but that instantly means that some other card won't have same amount of c[j]. What am I missing?
 » 19 months ago, # | ← Rev. 2 →   0 can someone check this code: 39528268. Fails on test 8EDIT: nvm. got it.
 » 19 months ago, # |   0 Formal proof for O(n*m) in E? I have intuitive proof but I want to see mathematically precise proof.
 » 19 months ago, # |   0 for D i tried a different approach which led to the same min no. of changes...but diff array.. but I think the judge is checking only for a particular array...whereas in the question it says print any array..my code
•  » » 19 months ago, # ^ |   0 No, the minimum changes in testcase 1 is 3 whereas your code outputs 4 which is wrong
 » 19 months ago, # |   +1 I was wondering if there can be a solution for E with DSU We will count number of Disjoint Unit for ans..?
•  » » 19 months ago, # ^ |   +1 I think it can work with dsu ,but when we unite them we need dfs too.
•  » » 19 months ago, # ^ |   0 like this 39531065
 » 19 months ago, # |   0 Thanks a lot for nice tutorials:)
 » 19 months ago, # | ← Rev. 5 →   0 Need help for D problem: I am doing the same approach as in editorial but it is giving wrong answer in testcase 8. Please help. I think the problem is in updating the array. What I am doing is maintaining a pointer p1(representing remainder) which initially starts from free[0]%m and then moves forward in circular fashion. Whenever I see that p1%m has count < k, I change value free[i] such that free[i]%m = p1%m. Please I cannot find my mistake. Need help. Code
•  » » 19 months ago, # ^ |   0 I'm getting a wrong answer on the same test and I applied the approach as in the editorial. did it work out with you?
•  » » » 19 months ago, # ^ |   0 No man, I couldn't figure it out :(
 » 19 months ago, # |   0 Help needed understanding problem F, thanks
•  » » 19 months ago, # ^ | ← Rev. 2 →   0 Try to solve this problem first :Given N persons and M cards, where each person can have atmost K cards, what is the maximum possible value of total joy levels ? The joy level of a person having x cards is H[x].Example : N = 3 , M = 4 , K = 3 H[0] = 0, H[1] = 2 , H[2] = 6, H[3] = 7Maximum value of total joy level can be obtained by giving 2 cards to person 1 , 2 cards to person 2 and none to person 3. The ans is 12 (H[2] + H[2] + H[0]) in this case. Note that there exists other possible distribution of cards such that total joy level is 12. Above is just one way.
 » 19 months ago, # |   0 Can someone please explain problem F ? Thanks.
 » 19 months ago, # |   0 A very nice contest. In problem E, my approach is the following: I used the concept of Disjoint Set Union. I say 1. whenever (in the input) vertex v (as input) is the capital. OR 2. the root of vertex v (as input) is the capital I don't perform union operation. The answer are the number of vertices whose: root(vertex)==vertex && vertex!=capitalThe approach works fine for the sample test cases, I got wrong answer in 4th test case. Please provide some small test cases to try my approach on.
•  » » 19 months ago, # ^ |   0 Your program fails on this test case. input4 5 1 1 2 2 1 3 4 4 3 2 4  right answer0 your programs output1Good luck :)
 » 19 months ago, # |   0 Please someone explain why I got TLE in problem C Problem C Submission
•  » » 12 months ago, # ^ |   0 n
 » 19 months ago, # | ← Rev. 2 →   0 For problem E, I got AC by pretty simple implementation, without taking reverse graph, without counting in degree: 39539472Is my solution correct or the test is weak? Is there any case when my solution will fail ?My approach is: 1. Mark all the nodes which are reachable from s 2. Run dfs in the original graph and push the nodes in stack according to dfs finishing time. 3. Pop each node from stack and if it is still unvisited run a dfs from here and increase answer by 1. My code:#include using namespace std; vector g[5002]; bool vis[5002]; void dfs(int u) { vis[u]=1; for(int v:g[u]) if(!vis[v]) dfs(v); } stack stk; void dfs2(int u) { vis[u]=1; for(int v:g[u]) if(!vis[v]) dfs(v); stk.push(u); } int main() { ios_base::sync_with_stdio(0);cin.tie(nullptr); //freopen("in.txt", "r", stdin); int n, e, s; cin>>n>>e>>s; for(int i = 1;i<=e;i++) { int u, v; cin>>u>>v; g[u].push_back(v); } dfs(s); for(int i = 1;i<=n;i++) if(vis[i]==0) dfs2(i); memset(vis, 0, sizeof vis); int cnt = 0; while(!stk.empty()) { int u = stk.top(); stk.pop(); if(!vis[u]) { cnt++; dfs(u); } } cout << cnt; return 0; } 
•  » » 19 months ago, # ^ |   0 Vovuh, can you please take a look ?
•  » » » 19 months ago, # ^ | ← Rev. 3 →   +1 Your approach is also correct, I have the same idea as yours and BledDest has proved this solution correctness
•  » » » » 19 months ago, # ^ |   0 Thanks for your reply. What's the proof of the correctness of such solution?
•  » » » » » » 19 months ago, # ^ |   0 Thanks for proof :)
 » 19 months ago, # |   0 for the problem E why do we need to sort the array which stores the cnt?
•  » » 19 months ago, # ^ |   0 Sorting is required because we have to connect the components of the graph greedily!So we need to find the source with maximum connected nodes!
•  » » 19 months ago, # ^ |   0 To find the source from which maximum nodes can be reached. This is important because we have to connect the graph greedily!
•  » » » 19 months ago, # ^ |   0 Isn't it possible that the nodes in sorted order may visit the similar nodes ?
•  » » » » 19 months ago, # ^ |   0 Yes it is .That's why after sorting the nodes we have to run the DFS and convert all the bad nodes into good ones.
•  » » » 19 months ago, # ^ |   0 What i was asking is let's say if node 1 in sorted order visit 4,5,6 and node 2 visits 5,6,7 and node 3 visits 2, 7. Then the 2nd node will be selected while it is not required. Am i missing something ?
•  » » » » 19 months ago, # ^ | ← Rev. 3 →   0 Node 2 won't be selected because according to your example node 3 visits node 2 that means you will run DFS from node 3 before node 2 . DFS from node 3 will convert bad node 2 into good node hence avoid revisit .
•  » » » » » 19 months ago, # ^ |   0 for (auto it : val) { if (!ok[it.second]) { ++ans; dfs1(it.second); } } In sorted list, node 2 will come before node 3 as its count is 3. Can you plz explain how node 3 is selected before node 2 ? "according to your example node 3 visits node 2 that mean you will run DFS from node 3 before node 2". How ? I hope you are considering all 1, 2 and 3 as bad nodes.
•  » » » » » » 19 months ago, # ^ |   0 You need to sort node according to the amount of node you can visit from that node ..in non increasing order In your example we can reach 4 nodes from node 3 but only 3 nodes from node 2 .That means 3 will come before 2.
•  » » » » » » » 19 months ago, # ^ |   0 Oh My bad.I got it. Thanks Buddy.
•  » » » » » » » » 19 months ago, # ^ |   0 Yeah :)
•  » » » 18 months ago, # ^ | ← Rev. 2 →   0 What is Greedily? And what happens if we don't sort? (I think I understand, but cant delete this message)
•  » » » » 18 months ago, # ^ |   0 Cool!
 » 19 months ago, # |   0 Hello, Could someone tell me what is the formal proof that the solution of problem D is optimal? I know someone has already written some proof but I really can't understand that one. Thanks,
•  » » 19 months ago, # ^ |   0 By applying the approach in the Editorial, and as you are iterating over the modulos from 0 to m-1, then the last item in the vector called free is the nearest to you. then all you have to do is free.back() and then to pop_back() the element you used and increased. I can provide a more detailed explanation if you didn't get what I said.
•  » » » 19 months ago, # ^ |   0 Please provide a bit more explaination
 » 19 months ago, # |   0 My solution for problem E is as follows: Run a dfs and mark all the nodes visited through s. Run another dfs and store all the nodes not visited through first dfs in a stack in topological order. Iterate over the stack and after poping out each node check if visited or not(don't forgot to reinitialize the visited array as 0), if not visited run a dfs and add 1 to answer otherwise continue. Time Complexity: LinearSolution
 » 19 months ago, # |   0 would anyone provide better explanation for F . ps: provide recursive solution if possible
•  » » 19 months ago, # ^ |   0
•  » » » 19 months ago, # ^ |   0 plz explain the logic unable to get from tutorial
 » 19 months ago, # | ← Rev. 2 →   0 In E problem is that correct? Count SCC number with 0 indegree and substract 1 from answear if S lies in SCC with 0 indegree?
•  » » 19 months ago, # ^ |   0 Yes that works ... informally, when you connect a vertex to an SCC with in-degree 0 (let's call it C), you will be able to reach the maximum number of nodes, because the rest (unreachable from C) is completely disconnected from this SCC, so you'd have to add a new edge anyway, and if you choose to add an edge to an SCC reacahble from C, you won't reach C itself (if you find SCC so that none is a sub-graph of the other, e.g. with Kosaraju's algorithm), and will still have to add a new edge, which will make the previous one redundant.
 » 19 months ago, # | ← Rev. 10 →   0 Thanks for Editorial!
 » 19 months ago, # | ← Rev. 3 →   +3 Could somebody please point out to me why I get WA on this solution for 999D - Equalize the Remainders ? 39544753
•  » » 19 months ago, # ^ | ← Rev. 2 →   +1 Hi! Your programs fails for example on this test case. Spoiler6 63 5 3 5 2 8Right answer: Spoiler74 7 5 6 2 9I hope it will help you, good luck :)
•  » » » 19 months ago, # ^ |   0 Please look into this.This solution of mine passes your test case, still ending with WA on 8th tc.
•  » » » 19 months ago, # ^ | ← Rev. 2 →   0 Hey! Thanks for your time. I think I fixed this bug by not reversing the 'free' vector, but it still give me WA on test 8. I'm not sure if it actually got fixed for all the cases I don't quite understand the basis on which we find the closest element to increment(I don't understand why the editorial chose to do the adding and removing in the same loop instead of doing it in 2 loops. How does that make the solution optimal?). Here's the submission 39604949
 » 19 months ago, # | ← Rev. 2 →   0 Please look into this:This solution for D fails on TC 8.The approach is same as editorialist's : Link
 » 19 months ago, # | ← Rev. 2 →   0 Help needed in understanding Problem-F..can anyone tell how to arrive at that DP transition??
 » 19 months ago, # | ← Rev. 2 →   0 Can someone help me in problem F? In dp transition, the turorial doesn't avoid the illegal conditions. For example, if i'm considering x people, then i shouldn't consider more than k * x cards, thus the bound of the for y's loop shouldn't be more than k * x since we use current condition to update future condition, then for current condition dp[x][y], y should be <= x * k, and for dp[x + 1][y + i], i <= k, then y + i <= x * k + k == (x + 1) * k. What if for some i, c[i] is greater than k * f[i], which is a illegal condition? I suppose tutorial's solution doesn't avoid them, but treats them correctly. Can someone explain why?? Really appreciat it. :)
 » 19 months ago, # | ← Rev. 2 →   0 http://codeforces.com/contest/999/submission/39629449 I think the main idea is the same with the editorial.But I can't understand why my code is incorrect on test case #8. Iterating 10 times output is different from iterating twice.Would anyone explain why two output is different and my code is wrong. I'm sorry for my bad English.
•  » » 19 months ago, # ^ |   0 I solved it. Thank you. (i--;)'s locations was reversed.
 » 19 months ago, # | ← Rev. 2 →   0 Can anyone please tell me what is wrong with my logic or implementation for problem E? what I am actually doing is- 1> dfs(capital) and mark all visited nodes as good. 2> now run a bfs(bad-nodes) and put them in priority_queue({degree,node}) 3> take nodes from priority_queue from top and run dfs(node) only if it is bad. here is my implementation
•  » » 19 months ago, # ^ |   +1 Your mistake is in bfs function for(ll u : adj[s]){ if(visited[s]) continue;//it should be v[u] if(v[s]) continue;//it should be v[u] nei++; v[s]=true;//it should be v[u] Q.push(u); } I fixed your code but it gave TLE on test 11. This is the code 39669926.
•  » » » 19 months ago, # ^ | ← Rev. 2 →   +3 thanks! But i think worst case time of my code is ~10^7 i.e n(n+m) + nlog(n) ~ n^2. can you please explain why i am getting TLE
•  » » » » 19 months ago, # ^ |   0 You are welcome! I fixed your code so it gets AC now(39677394). There are 3 main reasons why your code worked slowly.1)In your bfs function you had map v;. I changed it to bool v[5005];. map works in logarithmic time. Array works in O(1). Never use map, if you can use a simple array.2)Never use long long type if you don't need to. long long works slower than int.3)iostream works slowly. Use stdio.
 » 19 months ago, # |   0 For problem C i got this error. Can anyone tell me how to fix it? I'm stuckThis is the error:wrong answer Unexpected EOF in the participants outputThis is my submission: http://codeforces.com/contest/999/submission/39488231
•  » » 19 months ago, # ^ |   0 You have used variable temp without initialization. You have to add this line to your code. temp=0; But your solution is too slow(it gets TLE on test 4).
 » 19 months ago, # |   +1 In problem E, linear solution, what is the purpose of dfs3() ? In order to avoid source vertex containing SCC having indegree 0, just do indegree [comp[s]] = 1.
•  » » 19 months ago, # ^ |   0 Actually, neither the vector cg[N]; array is needed.
 » 19 months ago, # |   0 Can anybody help me with my solution.my solution Getting wa in test 8.
•  » » 19 months ago, # ^ |   0 When j == 1 (second iteration), the line abs(i - arr[rem.back()]%m) should be i + m - (arr[rem.back()] % m). Note the + m.
•  » » » 19 months ago, # ^ |   0 Got it , thanks
 » 19 months ago, # |   0 I have implemented top down approach for 999F,please have a look at my code and tell me where am i going wrong,as i am facing problem in clearing test case 3 http://codeforces.com/contest/999/submission/39853333
 » 12 months ago, # |   0 cnt++; please click *
 » 7 months ago, # |   0 why WA https://codeforces.com/contest/999/submission/55944188 problem D
 » 5 months ago, # |   0 My alternate solution to E: 1. Generate topological ordering on the graph, first calling dfs on s 3. for vertex v in topo: if (!reached[v]) ans++; dfs(v)(updating reached instead of explored); 4. Print ans
 » 3 months ago, # |   0 In 999E - Reachability from the Capital after we add an edge to a bad vertex shouldn't we be finding the order of bad vertices again. Can someone help me understand how the same order of bad vertices work even after adding an edge?
 » 3 months ago, # |   0 in problem 999E - Reachability from the Capital: i did a DfS from the capital to connect the good cities and for each bad city i started a DFS and if it reaches a new node or a node marked with 2 it changes it to 1 and when the DFS ends i mark it with 2 and the answer is the number of 2s but it is not running 63881015 can you please explain why it isn't working
 » 6 weeks ago, # |   0 In Question E, my solution is giving MLE. Could anyone please help me. Link to submission : https://codeforces.com/contest/999/submission/66542471