Блог пользователя chinesecoder

Help in this problem.
Автор chinesecoder, история, 6 лет назад, По-английски

The problem , 814c , my approach was , for each segment [l, r ] , check whether its possible to color it with at most m colors or not.

since there can be 26*n queries and n^2 segments , the solution will work in 26*n^3 which will give TLE .

The editorial mentions some prefix based approach but i couldn't understand the 2nd part , how its optimizing the solution .

How to reduce 26 * n ^3 to 26*n ^2 .

is there any dp solution possible to it . ?

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6 лет назад, # |
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let f(l) be the biggest r that you can color [l, r] with at most m colors. then it's clear that if l1 ≤ l2 then f(l1) ≤ f(l2). so you can find answer for all queries in O(n) with two-pointer.

I think this solution works too:

let DP[i][j][c] be maximum number of continuous character of c ending at i if we color exactly j characters. then DP[i][j][c] is equal to DP[i - 1][j][c] + 1 if s[i] = c and DP[i - 1][j - 1][c] otherwise.

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    6 лет назад, # ^ |
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    Thanks , two pointer method is quite tough to implement .

    in 2nd method do we have to add any more coding or they r sufficient.

    Sigyn

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      6 лет назад, # ^ |
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      For every query mi ci, you should get the maximum value for every 1 ≤ i ≤ n: DP[i][min(n, mi)][ci].

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6 лет назад, # |
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Here is my approach as requested.

Let f(x, r) represent the maximum length of substring of character x with EXACTLY r replacements. How do we calculate it for a given x ?

We visit every range [L, R]. The number of replacements in this range is the number of characters that are not x. 

    for(int ch = 0; ch < MAX_ALPHABETS; ch++)
    {
        for(int left = 0; left < length; left++)
        {
            int replacements = 0;

            for(int right = left; right < length; right++)
            {
                replacements += (S[right] != 'a' + ch);

                maximum_segment[ch][replacements] = max(maximum_segment[ch][replacements], right - left + 1);
            }
        }
    }

This takes O(σ N2)

Now let g(x, r) represent the maximum length of string if we do at most r replacements.

g(x, r) = max{f(x, 0), f(x, 1), f(x, 2), ... , f(x, r)}

    for(int ch = 0; ch < MAX_ALPHABETS; ch++)
    {
        for(int replacements = 1; replacements <= length; replacements++)
        {
            maximum_segment[ch][replacements] = max(maximum_segment[ch][replacements], maximum_segment[ch][replacements - 1]);
        }
    }
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