xyz2606's blog

By xyz2606, 3 years ago, In English
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3 years ago, # |
  Vote: I like it +19 Vote: I do not like it

Auto comment: topic has been updated by xyz2606 (previous revision, new revision, compare).

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3 years ago, # |
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constructiveforces

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3 years ago, # |
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Can you also provide codes for the above editorials?

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3 years ago, # |
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Problem D: If we use dynamic programming to compute the shortest path from u with length k how can we be sure that we will come back to u after k steps?

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    3 years ago, # ^ |
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    find the shortest path upto k/2 moves and then come back along the same path...

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      3 years ago, # ^ |
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      I did the same thing still get wa on test 4 :(

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      3 years ago, # ^ |
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      solved

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        3 years ago, # ^ |
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        what was the problem in the 4th test? Do you have a test case?

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          3 years ago, # ^ |
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          Basically I was miscalculating the edge cost from going right to left and down to up. But somehow that buggy code passed the first 3 test cases and not the 4th one

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          3 years ago, # ^ |
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          Did you find your bug ! I've implemented the solution explained in the editorial , still somehow getting WA on test 4 . 114377482

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            3 years ago, # ^ |
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            yes. I used m instead of n somewhere. I think you should just reread your code. Its not some special testcase apparently

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    3 years ago, # ^ |
    Rev. 2   Vote: I like it +2 Vote: I do not like it

    When k is even, we need to find the shortest path from starting position by performing only k/2 steps. And then traverse back along this path to get the required answer for k steps.

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    3 years ago, # ^ |
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    The shortest path is not with length k, but with length i. If k/i is even, we can go this path from start to end and then from end to start all the time. So finally we can return u.

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3 years ago, # |
Rev. 6   Vote: I like it 0 Vote: I do not like it

Can someone help me with this

Screenshot-20210423-234101

According to what i searched this error means that comparator is broken but i tried my best to include all possibilities in comparator can somebody please help me out in this

Language java Problem B Code https://mirror.codeforces.com/contest/1517/submission/114037412

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    3 years ago, # ^ |
      Vote: I like it +4 Vote: I do not like it

    Hey, your comparator isn't defining the equality condition, when will two pairs be equal? Try changing line 157 from "return 1;" to "return 0;", it will work :)

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    3 years ago, # ^ |
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    You are returning 1 for this case:

             if (a.row == b.row) {
              return 1;
             }
    

    Ideally it should return 0. What seems to be happening is, when a=b and b=c then a=c but due to above case, a=c is probably not holding true

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3 years ago, # |
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I understand it intuitively but can we rigorously prove that the graph is bipartite?

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    3 years ago, # ^ |
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    You can think about it like this. Consider the sum of the 2 coordinates. Every step you take changes the parity of this sum. To return to your starting point, your parity can only change an even number of times, but it changes exactly k times. Therefore, k must be even.

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    3 years ago, # ^ |
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    All the nodes a[i][j] will be connected only to the nodes a[k][l] if i+j and k+l has opposite parity.

    So the given graph can be divided into two parts, one with the parity of sum of indices as even and other as odd.

    Hence the given graph is bipartite.

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    3 years ago, # ^ |
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    Graph is always bipartite when it has no odd simple cycles. Let's take any cycle in this graph. Let's add some vertices to it which still form a cycle with the past ones. Their count will be 3..5..7 and so on. Besides, we must remove one edge, because it's needed to make the cycle simple.

    Picture (sorry for Paint):

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    3 years ago, # ^ |
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    Because any n * m grid can be converted to a chess board.

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Are there any ways to optimize dijkstra algo for problem D?

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    3 years ago, # ^ |
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    Can we solve D using Dijkstra? I was getting wrong ans, not sure if I'm missing anything... https://mirror.codeforces.com/contest/1517/submission/114047555

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      3 years ago, # ^ |
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      I've got ML5 with not optimized dijkstra

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      3 years ago, # ^ |
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      Consider a cell for which the distance is .......say number 'X1' and total number of edges to reach it is 'L1'. Lets say there is another path, length 'X2' and number of edges 'L2' such that L2 < L1 but X1 < x2 (remember the graph is weighted) . The problem with Dijkstra's is that it will always prioritise the path with lesser overall distance.

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    20 months ago, # ^ |
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    Sorry for replying to such an old comment. Found a testcase which might help anyone facing the same problem in future.

    Spoiler
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3 years ago, # |
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In problem D, my solution was if I am at $$$(i, j)$$$ then I can go to $$$(i, j+1)$$$ using an edge of weight $$$W$$$ (let's say), and find the shortest path of length $$$k-2$$$ that starts at $$$(i, j+1)$$$ and also ends at $$$(i, j+1)$$$ and then finally return to $$$(i, j)$$$ using the same edge of weight $$$W$$$. So,

$$$dp[i][j][k] = 2*W + dp[i][j+1][k-2]$$$.

Do the same for all of the adjacent vertices, the minimum is the answer.

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    3 years ago, # ^ |
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    But how is this always minimum? What if we go out through a particular edge but return through a different edge?

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      3 years ago, # ^ |
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      It's easy to prove that it's always ideal to go half way in one direction and then just return along that path.

      Consider a node in the graph V. Let's say you have found a random path of length k that begins and ends at V, and this path enters and leaves V from two different edges.

      This would roughly look something like a loop. Now since the length of this path is even, we can divide it into two halves.

      There can be two cases —

      1. The costs of the two halves are unequal.

      2. The costs of the two halves are equal.

      If the costs of the two halves are unequal, we can reduce the total cost of the path by just going forwards and backwards on the cheaper half.

      If the costs are equal, going forwards and backwards on any half will not change the total cost.

      Thus, it is always ideal to go half way in one direction and then just return along that path.

      Hope this helps.

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3 years ago, # |
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Problem E: How can the case $$$(C/P)CC...CPCPC...PCPP...P(C/P)$$$ be calculated in $$$O(N)$$$? For me the definition of the state looks like this:

  • We have the C/P at the beginning or the end. This is a constant factor.
  • Then we have the CCCCC-Part, the PCPCPC-Part and the PPPPP-Part. To check the sum-condition for all cases we need to iterate over all 3 lengths. the 3rd lengths follows from the 1st and second since their sum is constant.

But that sounds like iterating two lengths and it amounts to $$$O(N^2)$$$? What is the observation that I am missing to make it $$$O(N)$$$?

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    3 years ago, # ^ |
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    You iterate over length of $$$CC...C$$$ and the minimum acceptable length of $$$PP...P$$$ can be found using binary search or two pointers, because all the numbers in the array are positive, and function $$$\sum_P - \sum_C$$$ is monotonous.

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    3 years ago, # ^ |
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    If you have some kind of alternating prefix sum array to quickly calculate the sum of P if some range is PCPCPC(let's call it $$$pr$$$ and if the range is between the $$$i+1$$$-th and the $$$j$$$-th element then the sum is $$$pr_j-pr_i$$$), when fixing the point where you switch from PCPCPC to PPPPPP(let's call it $$$j$$$), if the point where you switch from CCCCCC to PCPCPC is $$$i$$$, then the condition is $$$pr_j-pr_i+sum(j+1..n) \geq \left \lfloor{\frac{sum(array)}{2}+1}\right \rfloor$$$. You can find the maximum $$$i$$$ for which that condition holds using either binary search or 2 pointers.

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      3 years ago, # ^ |
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      Where is $$$\left \lfloor{\frac{n}{2}+1}\right \rfloor$$$ from? I would assume there had to be $$$sum(1 ... i-1)$$$ instead? Maybe I misunderstand your formula.

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    3 years ago, # ^ |
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    My solution is similar to dorijanlendvaj's for the part with alternate partial sum, but I used a data structure (e.g. fenwick tree, segment tree, treap) to maintain the number of i satisfying the inequality. For stupid people like me who can't spot the obvious property of monotonicity, it is a lot more intuitive. The code can also be pretty short if you are familiar with data structures, check my code here: 114032392.

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      3 years ago, # ^ |
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      Could you please explain how we concluded that:

      "There can't be two i's such that ci−c(i−1)>2, or pi−p(i−1) won't be non-increasing. And there can't be two i's such that pi−pi−1>2, or ci−c(i−1) won't be non-decreasing. So for any 2≤i<m, ci−ci−1≤2, and for any 2≤i<k, pi−pi−1≤2."

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        3 years ago, # ^ |
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        Ah, so ci and pi is indexes of array "a", not the elements of "a". Now I got it.

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3 years ago, # |
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dpforces

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3 years ago, # |
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In D, is not not two dimension more complecated to think about the graph beeing bepartite, and hence does not allow paths of a given parity?

Instead, foreach step we do, we must do a step in the oposite direction. So the number is even.

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please provide cpde also

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3 years ago, # |
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In problem C could someone help in justifying the second construction ? How to prove that it will always work ?

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    3 years ago, # ^ |
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    My approach is also similar to editorial.

    Consider a rubber attached to the point i,i. We want to stretch the rubber p[i] times. So the optimal choice is to stretch the rubber in the following order

    1-Strecth the rubber up

    2-Strecth the rubber left

    3-Strecth the rubber down

    4-Strecth the rubber right

    This ordering will always ensure that we don't have any empty space left. See submission for more details:114016611

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      3 years ago, # ^ |
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      I was able to solve the question with this approach thnx a lot , but the point is how were u able to think of this??

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3 years ago, # |
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Video Editorial for problem C: https://youtu.be/RVE8Xprwsfo

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Hey guys, i found this very strange submission to problem D of this contest (114038234 belonging to the user pxsitivevxbess) and was wondering if anybody could explain to me the logic behind the excessive usage of "ashwet++".

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    3 years ago, # ^ |
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    I think I've seen it before, it's to evade the plagiarism checker. Maybe you could report that to MikeMirzayanov.

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3 years ago, # |
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Amazing editorial for problem G, the amount of details is simply astonishing

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    3 years ago, # ^ |
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    The problem has two types of forbidden pattern: the "square" and the "parallelogram". the square corresponds to 1-0-3-2-1 while parallelogram to 1-0-2-3-1. On the surface, it seems you have to add two kind of edges between rows: 0-3 1-2 (vertical ones) and 0->2 1->3 (diagonal ones)

    But the key observation is that you don't need to, you only need to add edges 0-3 and 1-2. The reason is because to check for parallelogram, you can actually just use the same check: whether a path 0-1-2-3 exists. If such a path exist, then it must either be a square or a parallelogram, e.g.:

    ...10...01...
    ...23...32...
    

    for squares and

    ...10...01...
    ..32.....23..
    

    for parallelograms

    Therefore, we consider the acyclic graph where all "0" nodes have edge to "1" node, all "1" node to "2" nodes, and all "2" nodes to "3" nodes. in this acyclic graph, we want to cut fewest weight nodes so that no path exists from "0" to "3". This can now be done easily using min cut.

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      3 years ago, # ^ |
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      What about a parallelogram that looks like

      ...10...

      .....32..

      It won't be covered by a 0-1-2-3 path, only a 1-0-3-2 path, so how do we take care of these cases?

      Ps. your second square case should be

      01

      32

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        3 years ago, # ^ |
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        These kind of parallelograms are not meant to be forbidden as important tents are represented by "1" and this parallelogram is not breaking the given condition ( $$$|x_2 - x_1| = 2$$$ but it should be $$$|x_2-x_1| \le 1$$$

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why in d you need k/2 only.suppose we go to next edge and its cost is 1 but after that costs are 10^6 so why shouldn't i go back and forth only to complete k steps rather than include 10^6 in my ans?

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    3 years ago, # ^ |
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    It is not possible.

    Supposition: The optimal cost to travel k/2 steps is C.

    If you move k/2 steps with cost C and if there exist a returning path with length k/2 that has C' < C.Then you can move in the later path to get a cost of C'.But this is not possible because of the supposition.

    Hence we can always travel k/2 steps and return with the same path.

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      3 years ago, # ^ |
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      i am asking if suppose i have next edge cost=1 and all edges after wards have cost of 10^6 then why should i go exact k/2 steps in forward direction why not jhust go b/w current vertic and next vertex whose edge has cost 1 and complete k steps in this manner ? why would u inlcude 10^6 in your ans?

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        3 years ago, # ^ |
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        You misunderstood my statement.

        Suppose you are initially at $$$A$$$ and after completing $$$k/2$$$ steps you are at position $$$C$$$. So in order to get the minimum answer you must retrace the same path that you have to taken to reach to $$$C$$$ to $$$A$$$.

        That's why we only have to check $$$k/2$$$ steps.

        And steps doesn't mean you have to always move forward. You can also move 1 step forward then backward then forward and so on until all $$$k/2$$$ steps are exhausted.

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implementForces :)

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In Problem D, multi-source Dijkstra with a priority queue gets TLE: 114052809.
But if the priority queue is replaced by a normal queue, it gets AC in 405 ms: 114052881.

I think the solution with the queue should get TLE too. But I couldn't come up with an appropriate case. Can anyone try to uphack the solution? Or am I missing something?

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    3 years ago, # ^ |
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    For Dijkstra's, the exit condition should be -

    new_code

    instead of

    old_code

    check out my submission, I just modified this line and it passses — https://mirror.codeforces.com/contest/1517/submission/114073092

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      3 years ago, # ^ |
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      Actually, both are performing the same purpose here.

      dist[u.r][u.c][u.lv] holds the minimum possible cost for [u.r][u.c][u.lv] till now. So, obviously dist[u.r][u.c][u.lv] != u.w means the value of u.w is greater than dist[u.r][u.c][u.lv].

      Edit: Your submission is passing because of the compiler. I've submitted with != operator using "GNU C++17" compiler and got AC in 1856 ms: 114108137.

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    3 years ago, # ^ |
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    By the way, can you explain your logic for Multi-Source Dijkstra's? Mostly why this holds true in this problem?

    Also, the queue solution should be incorrect, because Dijkstra's requires a priority queue for the same?

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      3 years ago, # ^ |
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      Logic:
      For a cell (r, c), we need to find the minimum cost after h steps if we start from the cell (r, c). As the graph is undirected, it is equivalent to reach the cell (r, c) from an arbitrary cell after h steps. So, we can do it by taking every cell as the source and calculate the dist[r][c][h] for each cell. It is normal Dijkstra. The rest calculation part for the minimum boredness is quite easy.

      Queue solution:
      Dijkstra needs a priority queue (min-heap) just to reduce the complexity. Taking queue won't make your solution incorrect. In normal graphs, it will increase your complexity.

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        3 years ago, # ^ |
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        So you are saying that -

        1. We can do Dijkstra's using queue OR
        2. Required implementation in this problem doesn't require a priority queue
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          3 years ago, # ^ |
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          1. Yes, we can do Dijkstra using a queue. But the complexity will be bigger. In the worst case, it will be $$$O(VE)$$$.

          2. I'm not sure actually. But most probably, yes.
            Firstly, the way I've implemented Dijkstra, using queue in it almost makes it SPFA.
            Secondly, For a cell $$$(r,c)$$$, you need to go at most $$$10$$$ steps. So, the number of cells needed to traverse from a single source reduces a lot. It is quite a large restriction. With this restriction, there's a chance that making the worst case for $$$SPFA$$$ is impossible.

          Edit: I'm saying both.

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            3 years ago, # ^ |
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            Thanks for the descriptive response :) I didn't know the queue solution looked like SPFA. That's nice observation.

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3 years ago, # |
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Can you explain the $$$n^2log(n)$$$ solution for problem F? Thanks.

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    3 years ago, # ^ |
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    I haven't yet tried, but I guess you can optimize the dp with some heavy-light decomposition trick. I guess, since the number of dp states at node v is of O(depth_of_subree_rooted_at_v), when dealing with node v, you can inherit dp states from the heavy child (child with the maximum subtree depth) and iterate the rest children. That way, you end up with a O(nlogn) dp per radius, then O(n^2 log n) for all radius.

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hey anyone who can find whats wring in my code of problem B

include<bits/stdc++.h>

using namespace std;

define ll long long int

define el "\n"

define mod 1000000000

int main() { ll t; cin>>t; while(t--){ ll n,m; cin>>n>>m; ll a[n][m]; ll b[n][m]; ll temp=0; vector<pair<ll,ll> > v(m);

for(ll i=0;i<n;i++){
    for(ll j=0;j<m;j++){
        cin>>a[i][j];
        b[i][j]=a[i][j];
    }
  }
  //v[0].first=3;

while(temp!=m){
        ll c=mod;
   for(ll i=0;i<n;i++){
    for(ll j=0;j<m;j++){

        if(a[i][j]<=c){

            c=a[i][j];

           v[temp].first=i;

        v[temp].second=j;
        }

    }
  }
  a[v[temp].first][v[temp].second]=mod;

temp++; }

for(ll j=0;j<m;j++){
   swap(b[v[j].first][v[j].second],b[v[j].first][j]);

}

// cout<<el; for(ll i=0;i<n;i++){ for(ll j=0;j<m;j++){ cout<<b[i][j]<<" "; } cout<<el;}

} return 0; }

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    3 years ago, # ^ |
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    Please use pastebin or use spoiler to share your code, directly pasting the code pollutes the blog :)

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Help me out with A folks I don't know how to find sum of its digits in decimal representation.

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    3 years ago, # ^ |
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    Well, you can prove to yourself that the last digit of "n" is always (n%10) = n modulo 10 (the rest when you devide n by 10. Now, you use a while and at every step you add n%10(last digit) and then devide n by 10 (that eliminates the last digit). You have to do this until n is equal to 0. When that happens you stop the iterration and that's the sum. Hope it was not a joke and I actually helped. :))

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I haven't understood the editorial of problem E, can anyone explain it in detail?

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    3 years ago, # ^ |
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    C and P are two sets of all indexes of "a" (together — "n" elements). Indexes of C is growing wider left-to-right, and P — right-to-left. If for some ci-c(i-1) > 2 than there should be two P's in a row between them. And this means there can't be no more P's to the left from them, and to the right(max — 1), because pi-p(i-1) become 1, and to jump over C it need to become 2 — this breaks monotonic rule. And so on.

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.

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Can someone explain Problem B?

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    3 years ago, # ^ |
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    Check this image out —

    click the image

    Basically, get the top m minimum numbers from the all the entries, and position them stategically such that the order of min elements is followed.

    As you can see the image, we want to permute each row, such that the min numbers(in red) are positioned one after the other.

    You can check my submission for more details — https://mirror.codeforces.com/contest/1517/submission/114011935

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      3 years ago, # ^ |
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      Hi, thanks for the reply. I understood the basic idea. Could you tell me what is wrong with my submission? https://mirror.codeforces.com/contest/1517/submission/114070887

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        3 years ago, # ^ |
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        You are modifying the modified elements in each row, you need to skip those positions, which have been modified.

        Like you might modify (0, 0) and then later you might check for an element(0, i) which matches at (0, 0), so might change the value.

        Keep a set of modified elements, which would want to skip, that might help

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Can anyone help can't figure out what's wrong with my problem D solution.

WA on Test 5 114051343

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3 years ago, # |
Rev. 3   Vote: I like it -18 Vote: I do not like it

I really liked the idea for the problem C & D.

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3 years ago, # |
Rev. 3   Vote: I like it -8 Vote: I do not like it

In prob-D, Can anybody tell the number of states we will have while calculating the shortest path of length K/2 from some vertex u? I think its around K^2 because starting from vertex 'u' ideally, we will traverse in square of K*K size

Update: Got the answer

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3 years ago, # |
  Vote: I like it 0 Vote: I do not like it

I did try to implement what you have said in B.But somehow its not passing all test cases. Can someone please tell what is wrong with mycode. https://mirror.codeforces.com/contest/1517/submission/114183214

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3 years ago, # |
  Vote: I like it 0 Vote: I do not like it

please give the codes as well

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3 years ago, # |
  Vote: I like it 0 Vote: I do not like it

The editorial for problem H says: 'We will explain the precision issues later', but that part seems to be missing. Can somebody please add that part? In my attempts to solve the problem, I struggled with precious issues for quite some time, and I added several hacks until I finally got AC (114778549).

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3 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Did anyone solve Problem D with Matrix exponentiation?

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3 months ago, # |
  Vote: I like it 0 Vote: I do not like it

following code can be referred for the Problem D for an easy implementation, the basic optimization is that we can reduce by 2 every time and by doing this we wouldn't have to reset the dp matrix for every (i,j) :

https://mirror.codeforces.com/contest/1517/submission/246684152