YeuDieuNhieu's blog

By YeuDieuNhieu, history, 5 weeks ago, In English,

[](https://open.kattis.com/contests/asiasg18-prelim-open/standings) How to solve D and E, I tried but failed.

 
 
 
 
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5 weeks ago, # |
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D: Find all bridge in the graph, then DFS to see how many vertices reachable from vertex 0 after removing all the bridge.

E: Binary search the answer. Suppose we need to check whether it is possible to find a combination of N numbers with the minimum not less than X. Build a bipartite graph where vertices on the left represent row, and vertices on the right represent column. For each cell ai, j, if ai, j ≥ X, connect an edge from i-th vertices on the left to j-th vertices on the right. If the maximum matching of the bipartite graph consist N edges, it is possible to find a combination of N numbers satisfying the condition above (we choose all cell (i, j) such that the edge connecting i-th vertices on the left and j-th vertices on the right is in the maximum matching).

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5 weeks ago, # |
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Anyway, can anyone explain G and J?

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    5 weeks ago, # ^ |
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    G: precalculate all N such that Bob will win. There are only less than 400 such values. You need an efficient enough precalculate program. Ours runs in less than 10 minutes.

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      5 weeks ago, # ^ |
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      How did you precalculate these N efficiently?

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        5 weeks ago, # ^ |
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        For each i<=10^9, determine if there is any N found so far such that i-N+1 is prime. If none of them is, i is a new N we found.

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          5 weeks ago, # ^ |
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          Can you prove an upper bound on how many i from 1 to N that Bob will win?

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            5 weeks ago, # ^ |
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            No, I guess it's related to distribution of primes which is really complicated. In contest, I discovered there are really few N that Bob will win only by experiment.

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          5 weeks ago, # ^ |
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          If someone is interested.
          My offline sol to find all possible nos.
          And AC soln with all no hardcoded.
          But I as of now I don't have any proof on the upper bound.

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          5 weeks ago, # ^ |
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          Unable to solve A with DFS? I tried but got TLE, so I changed to an algorithm O(9*n^4).

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            5 weeks ago, # ^ |
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            My Soln of A.
            Can you share yours?

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              5 weeks ago, # ^ |
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                5 weeks ago, # ^ |
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                Thanks. You can reduce your time complexity to O(9*8*n^2). By using this piece of code.

                const lli r[10]={-2,-2,-1,-1,1,1,2,2};
                const lli c[10]={-1,1,-2,2,-2,2,-1,1};
                
                for (int i=1;i<=n;i++)
                for (int j=1;j<=n;j++)
                for (int k=0;k<8;k++)
                {
                p=x+r[k];
                q=y+c[k];
                if (1<=p&&p<=n&&1<=q&&q<=n&&a[i][j]=='I' && a[p][q]=='C')
                dd[p][q]=1;
                }
                
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    5 weeks ago, # ^ |
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    J:

    1. Enumerate all the possible cycles
    2. For a cycle, contract all the vertices into one vertex
    3. then, we can calculate the number of spanning tree after contracting with Kirchhoff theorem

    The first part can be computed by a O(2VV2) TSP DP. The overall time complexity is O(2VV3) since you need to compute at most 2V times Gaussian elimination.

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      5 weeks ago, # ^ |
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      Do you know any tutorial on how to compute determinant of a N * N matrix in O(N3)?

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        5 weeks ago, # ^ |
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        After applying Gaussian elimination in O(N3), the determinant of the matrix is equal to the product of all the element on the main diagonal.

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How to solve B?

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    5 weeks ago, # ^ |
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    B can be solved by using DSU or DFS to check whether (i,n-i+1) is in the same connected conponents. ( i<=1<=(n+1)/2). Here is my sol (using DSU): Here

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how to solve F?

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    5 weeks ago, # ^ |
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    My soln for F -

    Chose a const SIZE ~ sqrt(n). (500 gives AC, and 1500 gives TLE on last TC) Now divide all B into two half one less than SIZE and other greater than or equal to SIZE.

    For Type 1 queries - if B >  = SIZE. Directly update all values. //O(n/SIZE) if B < SIZE, //O(1) Keep an array inc[SIZE][SIZE] i.e. (B,A)
    And update as inc[b][a] +  = cst

    For Type 2 queries -

    cnt=a[D];
    for(i=1;i<SIZE;++i) //O(SIZE)
    cnt+=inc[i][D%i];
    

    Time Complexity — O(Q*SIZE) in worst cases.
    But unfortunately, it gives AC.

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5 weeks ago, # |
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How to solve C, H, I?

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    5 weeks ago, # ^ |
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    C: Dynamic Programming. Consider building the big string (I'll call it T) character by character. Because it has to be a palindrome, when we assign T[0], we also assign T[2N - 1]. Keep a pointer to the head and tail of S. When we assign positions in T, we may need to move the head and tail pointers of S. When S is a subsequence of the prefix / suffix of T that we've built so far, just multiply by 26remaining. The state is just how many characters are remaining and the positions of the pointers in S for a O(n3) solution.

    H: Just BFS from outside the grid and count how many walls you hit. If you enter an "active" cell don't continue the BFS from there.

    I: It looks like finding the coefficient of xk in (1 + x + x2 + ... + xm)n. No idea how to do that efficiently.

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      5 weeks ago, # ^ |
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      I is a pretty standard problem. For example, see 451E - Devu and Flowers for the application of the same idea.

      Firstly, consider the case when m (the number of objects of each type) is infinite. In this case the answer is simply (you need to count ways of splitting k in n non-negative integer summands; that is the same as splitting n + k in n positive integer summands; that is the same as splitting n + k balls lying in a row in n non-empty groups by placing n - 1 "walls" that separate the groups in n + k - 1 positions (between each pair of balls)).

      But we overcounted something: we counted the partitions of k in n non-negative summands where some summands are larger than m, though we should have not done that. Let's use inclusion-exclusion to handle the overcounting.

      How many partitions of k in n non-negative summands there are, if we are required to make some i fixed summands larger than m? Let's note that there are as many of them as there are partitions of k - (m + 1)i in n non-negative summands (simply subtract m + 1 from each "definitely overflowing" summand, after that there will be no restrictions on said summand).

      In the end, by inclusion-exclusion, the answer is equal to (we need to choose i definitely overflowing summands out of n and count the partitions of k into n summands such that i of them are definitely overflowing).

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        5 weeks ago, # ^ |
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        Does there any way to solve this problem with polynomial multiplication?

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      5 weeks ago, # ^ |
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      For I, we can rewrite the expression as (1 - xm + 1)n(1 - x) - n. Just use the binomial expansion on both and you have to evaluate the following:

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      5 weeks ago, # ^ |
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      I do not really understand the solution for C.

      "Because it has to be a palindrome, when we assign T[0], we also assign T[0]." Do you mean T[0] and T[2*N-1]?

      How do you handle S being in the middle (i.e. it does not lie entirely in the first N letters or the last N letters) for the DP state? I do not see how the transitions work for this case.

      EDIT: I misread the question as substring instead of subsequence, so the solution makes a lot more sense now...

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how to solve problem C An alphabetical string is a string consisting of 0 or more capital letters (i.e. [‘A’..‘Z’]). Given an alphabetical string S[1..N], determine the number of palindromic alphabetical strings of length 2N that contains S as a subsequence (not necessarily contiguous). A string is palindromic if it is equal to its reverse.