Wild_Hamster's blog

By Wild_Hamster, history, 3 years ago, In English,

677A - Vanya and Fence

For each friend we can check, if his height is more than h. If it is, then his width is 2, else 1.

Complexity O(n).

Code

677B - Vanya and Food Processor

The solution, that does same thing, as in the problem statement will fail with TL, because if the height of each piece of potato will be 109 and smashing speed will be 1, then for each piece we will do 109 operations.

With each new piece of potato ai we will smash the potato till a[i] MOD k, so we will waste a[i] / k seconds on it. If we can not put this piece of potato after that, we will waste 1 more second to smash everything, that inside, else just put this piece. We will get an answer same as we could get with actions from the statement.

Complexity O(n).

Code

677C - Vanya and Label

We can transform our word in binary notation, we can do it easily, because 64 = 26. Move through the bits of this number: if bit is equal to 0, then we can have 3 different optinos of this bit in our pair of words: 0&1, 1&0, 0&0, else we can have only one option: 1&1. So the result will be 3nullbits, where nullbits — is amount of zero bits.

Complexity O(|s|).

Code

677D - Vanya and Treasure

We can make dynamic programming dp[col][row], where dp[col][row] is minimal time, that we have waste to open the chest in the cell (col, row). For the cells of color 1: dp[x][y] = x + y. For each next color color we can look over all cells of color color - 1 and all cells of color color, then for each cell of color color with coordinates (x1, y1) and for each cell with color color - 1 and coordinates (x2, y2) dp[x1][y1] = dp[x2][y2] + abs(x1 - x2) + abs(y1 - y2).

But complexity of this solution is O(n2·m2), what is not enough.

We can do such improvement: let cnt[x] be the amount of cells of color x, then when cnt[colorcnt[color - 1] ≥ n·m, we can do bfs from cells of color color - 1 to cells of color color.

Then we will have complexity O(n·m·sqrt(n·m)).

Proof

Code

There also exists solution with 2D segment tree:

Code

677E - Vanya and Balloons

For each cell (x, y) take the maximum possible cross with center in this cell, that doesn't contains zeros. To do it fast, we can make arrays of partial sums for all possible 8 directions, in which each cell will contain the number of non-zero balloons in each direction. For example, if we want to know, how many non-zero balloons are right to cell (x, y), we can create an array p[x][y], where p[x][y] = p[x][y - 1] + 1 if a[x][y]! = 0 else p[x][y] = 0

So now we can for each cell (x, y) we can find the maximum size of cross with the centre in this cell, that will not contain zeros.

We can compare product for crosses with centers in the cells (x, y) and radius r using logarythms. For example, if we need to compare 2 crosses with values x1·x2·...·xn and y1·y2·...·ym, we can compare log(x1·x2·...·xn) and log(y1·y2·...·yn), what will be equivalent to comparing log(x1) + log(x2) + ... + log(xn) and log(y1) + log(y2) + ... + log(ym).

We can also use partial sum arrays to find value log(x1) + log(x2) + ... + log(xn) for each cross, so we can find the product of the values in each cross for O(1) time.

Complexity O(n2).

Code
 
 
 
 
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3 years ago, # |
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Human drinks a Vampire's blood. Out of curiosity, the Vampire asks what it tastes like.

"It's irony."

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How is this possible to prove time complexity in 677D - Vanya and Treasure ?

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I finished the problem C ten seconds after the contest... :'(

I enjoyed this round... Thanks.

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There's a bit of Russian in the English version of the description for C:

"В таком случае результатом будет 3nullbits, где nullbits — количество нулевых битов."

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Someone Please Post the code for all the problems.Please!!!!

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That editorial has been blazingly fast!

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how to solve third 677D

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Thanks for fast editorial.

I liked 677D - Vanya and Treasure the most and I want to share my approach, also in O(n3).

Let's iterate over value v from 1 to p - 1. Let's say that for each cell with this value we know how fast can we get here and open this cell. Now, we want to calculate the same for cells with value v + 1.

For each row y if there is at least one cell in this row with value v, then let's iterate over cells in this row: once from left to right, once from right to left. We want to consider moving from one of cells with value v in this row, to any cell with value v + 1, not necessarily in this row. So, let's keep a variable best as we iterate over cells in the row — how quickly can we get to the current cell (with assumption that we already were in a cell with v in this row).

To update best we must: 1) increase it by 1 when we move to the next cell in the row, and 2) if the current cell has type v then do best = min(best, dp[cell]) (where dp[cell] means how fast we can get here from the start).

And we must use best to find dp values for cells of type v + 1. As we iterate of cells in the row y, for each x we must iterate over all cells of type v + 1 in column x and for each of them update dp using best, i.e. dp[cell] = min(dp[cell], best + abs(y - cell.y)). By doing it we consider moving from any cell of type p somewhere on the left to any cell in column x. Remember that you need to iterate second time from right to left (first iteration was from left to right).

code: 18192113

the same code with some comments added, should be easier to read: 18203850

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    3 years ago, # ^ |
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    Thanks for posting your solution, I'm trying to understand it.

    1. Did you mix up rows and columns? x usually goes for rows and y for columns, in the comments it says columns where it's actually rows and visa versa.
    2. Could you please explain why is it O(n3) but your code at some point contains 4 nested loops?

    Thanks in advance.

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      3 years ago, # ^ |
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      1. For me the x axis goes from left to right. Sorry if it confused you. So, x denotes column for me.

      2. You must think about the amortized complexity. For each cell in vector of cells odw[v] (v denotes type) how many times it can appear (how many times it can be processed). Turns out that we process it once for each row y when we consider going from type v to type v + 1. There are n rows and thus we process each cell there at most O(n) times. So, the amortized complexity is O(n3).

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        3 years ago, # ^ |
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        Thanks for sharing, amazing solution. But I still don't understand why the complexity is O(n^3). when going from type v to type v+1, it does visit O(n) times cells of type v+1 and visit only once cells of type v, but for each row y, each cell of this row of other types is visited too.

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          3 years ago, # ^ |
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          I understand it. Since each row has O(m) different types, each row will be scanned O(m) times, so there is O(mnm) best update operations and O(nmn) dp update operations.

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    3 years ago, # ^ |
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    Does the use of the 'visited' array change the complexity? If not, how/why did you come with such heuristic, if I may call?

    By the way, pretty neat. How do you come with such solutions?

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      3 years ago, # ^ |
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      Yes, it does. Consider any particular cell arr[i][j] in the matrix. How many times will it be processed? O(n) times, where n = number of rows. And there are n*m total cells. So complexity comes out to be O(n*n*m). But I am not sure about this. Please let me know if this is incorrect.

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      3 years ago, # ^ |
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      Think about a test where all cells have different types. Then, without vis[] we have O(n4) because for each of n2 types we will iterate over x and then over y. But, with using vis[] we can get to the last loop only once for each cell, so it's amortized O(n3).

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        3 years ago, # ^ |
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        I see... I have an intuitive feeling that this is right, but I can't think properly about the amortized complexity. I will spend some more time thinking on this. Thanks for the help!

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    3 years ago, # ^ |
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    To update dp[cell] of v+1, I think the formule should be

    dp[cell] = min(dp[cell], best + abs(y - cell.y))

    since x and cell.x are equal.

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      3 years ago, # ^ |
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      You are right, thank you very much. I've just corrected my comment, it's ok now.

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My solution for D is O(nmlog2(nm)). The basic idea is the following: define dp[r][c] analogously to the above definition.

Let's fix some chest at (r, c), and suppose that the previous chest we visited is at some position (r', c') with r' ≤ r and c' ≤ c, then dp[r][c] = dp[r'][c'] + (r - r') + (c - c'). Notice that we can split this into r + c and dp[r'][c'] - r' - c'. Similarly, suppose that we had r' ≥ r instead, then dp[r][c] = dp[r'][c'] + (r' - r) + (c - c') which splits into c - r and dp[r'][c'] + r' - c'.

This gives us the following recurrence:

dp[r][c] = min(topleft, topright, bottomleft, bottomright)

Where:

topleft = r + c + minr' ≤ r, c' ≤ cdp[r'][c'] - r' - c'

topright =  - r + c + minr' ≥ r, c' ≤ cdp[r'][c'] + r' - c'

topleft = r - c + minr' ≤ r, c' ≥ cdp[r'][c'] - r' + c'

topleft =  - r - c + minr' ≥ r, c' ≥ cdp[r'][c'] + r' + c'

.. with the additional constraint that the chest at (r', c') is has a key one less than the key at (r, c). What you'll notice is that each of these four values can be phrased as a range minimum query in some square (depending on r and c) plus/minus r and c. We can thus store all values after the 'min' in four distinct 2D Fenwick trees (one for all four directions).

So in the first 2D Fenwick tree we'll store dp[r'][c'] - r' - c' at (r', c'), etc. Then we can simply query the answer for each (r, c).

Notice that each time you 'finish' a type of key, you would want to reset the Fenwick trees such that they don't contain the values of smaller types of keys. If we'd reset naïvely, each reset would take O(nm) time, but you can speed this up by only resetting the positions that contain a value (this probably sounds a bit odd, see my code for reference, i.e. the reset_around method).

Code: 18198603, ~200ms.

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    3 years ago, # ^ |
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    Considering p <= nm you may remove it from complexity

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    3 years ago, # ^ |
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    I did the same, but I fixed r' so I did not need to worry about abs(r' - r). Then you can simply keep 2 fenwick trees per row of the grid ordered by the c' values.
    The complexity --> O(N * M * N * logM)
    Code -> 18197729 (It passes in 1.3 seconds)

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    3 years ago, # ^ |
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    I think your approach can be also simplified a little (not everyone is familiar with 2D segment trees), and you get rid of one log factor:

    1. It's sufficient to write code that solves the top-left case, and then run it four times, rotating the board
    2. We can take all the "points" (previous color), and "queries" (current color) and sweep them in the order of ascending x coordinate. For each y we store the smallest number out of all the points that had this y coordinate and were already processed. When we see a query (x,y), then we just take the minimum of values on the prefix from 0 to y. We'll just need a 1D fenwick for that.
    3. Our tree has now size O(n), so I guess we no longer need the lazy reset.
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      3 years ago, # ^ |
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      The step from 1D to 2D Fenwick trees is nothing more than duplicating a for loop and renaming some variables, you don't really need to understand it, just realize you can do 2D range queries :)

      Having said that, I do like your simplification, thanks for sharing!

      EDIT: Also, just to be clear, there's no lazy reset or something like that going on, I just follow the query path for (r, c) and reset all values on the path, for a total amortized cost of O(nmlognm) for the entire solution.

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    3 years ago, # ^ |
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    ".. with the additional constraint that the chest at (r', c') is has a key one less than the key at (r, c)"

    How do you ensure this in your query?

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    20 months ago, # ^ |
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    Thank you for detailed answer :). I'll try to implement it.

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Some notes about D: submissions of only 33 official participants had passed sys.tests

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I wrote an overly complicated solution for problem D involving four 2D BIT's, one for each quadrant, each with its separate update and query functions.

  • D[i][j] - i - j for upper left quadrant
  • D[i][j] - i + j for upper right quadrant
  • D[i][j] + i - j for lower left quadrant
  • D[i][j] + i + j for lower right quadrant

Then I processed the cells in ascending order of key, updating the four BIT's and resetting them as necessary.

I think total complexity is O(N * M * log2(max(N, M))), which is slightly better than the model solution from this editorial.

Here's my submission for further clarity: 18195269

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One can solve D in O(N*M*log^2(N*M)). The idea is like the DP one from the editorial.

We start by iterating all points starting from the one with the smallest number. We also build four 2d minimum segment trees on the points from the previous type.

We have four cases (let our point be x, y):

dp[x][y] = dp[x'][y'] + x - x' + y - y' = (dp[x'][y'] - x' - y') + (x + y)

Now in the first 2d segment tree we have to update all points from the previous type (x', y') value to (dp[x'][y'] — x' — y'). So now for a point of the current type the answer is minimum in rectangle [1, 1] to [x; y] + x + y. This is a simple 2d segment tree with range query and point update.

We do four cases of this type:

X < x'; Y < y'
X > x'; Y < y'
X < x'; Y > y'
X > x'; Y > y'

We do N*M point updates and N*M range queris so the complexity is O(N*M*log^2(N*M)). In fact this can be done with not a 2d segment tree but with a 2d minimum BIT, because all of our queries are of a prefix/suffix rectangle.

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    3 years ago, # ^ |
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    That's exactly what I did! I thought I was the only weirdo doing that xD!

    I believe the complexity is actually O(N * M * log2(max(N, M))).

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      3 years ago, # ^ |
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      O(log(nm)) and O(log(max(n, m)) is one thing, actually

      As well as O(log(n + m))

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        3 years ago, # ^ |
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        What do you mean?

        For example, if N = M = 1000, then log(N * M) = 20, while log(max(N, M)) = 10. In general, if N = M, then log(N * M) = 2 * log(max(N, M)).

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          3 years ago, # ^ |
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          Are you familiar with O-notation?

          log(max(n, m)) <= log(mn) = O(log(mn))
          log(mn) <= log(max(m,n)^2) = 2 log(max(m, n)) = O(log(max(m, n))

          If f(n) = O(g(n)) and g(n) = O(f(n)) then h(n) = O(f(n)) <=> h(n) = O(g(n))

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      3 years ago, # ^ |
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      Well I also did think it's weird and I was 100% sure I overkilled it :D

      About the complexity, I think it should actually be O(N*M*logN*logM).

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Could Someone plz explain Div2 C in detail. if our word is "0125" then how will we write it's binary representation and why is the answer 3^(total zero bits)

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    3 years ago, # ^ |
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    Firstly, you should understand, that for each character in string there is a number. So "0125" is actually the same 0 1 2 5, now we should translate it to 2 base system. So it will be 000000 000001 000010 000101 . You should remember to add zeros till length is 6, cause max number in 2cc has 6 numbers. Now we should mention that 1 appears from pair 1&1 , however 0 there are 3 pairs: 0&0, 0&1 , 1&0. So now we just need to count how many zeros are there and print 3^cnt. P.S.Sorry for my English

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      3 years ago, # ^ |
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      Thank you so much. It's totally clear now and my solution also got accepted.

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      hey i have one doubt. lets say initial string is 1 only then binary of this will be 000001. Now if initial string has 1 as a bit then the other string should also have 1 at the same position and if initial string has 0 as bit then the other string can be 0 and 1. i am not able to understand how 1&0 and 0&1 are same because if initial string has 1 then other string must contain 1 but if initial string is 0 then other string can only have 1 or 0. Please explain me this.

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I think my solution for C is more natural.

I precalculated values A[i], which means how many pairs of characters are there such that their AND is equal to i. We can do this by just trying all 64 * 64 possible pairs. Then we just multiplicate all these values for every character of S and we're done.

Here's my submission: 18190366

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    3 years ago, # ^ |
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    sorry to ask a foolish question but could you plz explain the meaning of the the question with example.Although i understood your solution but still i wanted to understand the question clearly. Thanks

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How about solving D by shortest path algorithms, such as Dijkstra? Just model the palace as a graph, and cells with color C will be connected to cells with C+1. We set a dummy starting point (0,0) and it can be connected to itself if (0,0) has color 1. I haven't started to practice on graph problems so I don't know how to implement this algorithm. But I understand the theory and I think Dijkstra can solve it. The complexity is VlogV+E, where V = n*m. E is also n*m since each cell only have at most one edge out. So total complexity is (n*m)log(n*m). Please correct me if you find this is impossible.

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    3 years ago, # ^ |
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    If you want to connect cells of color C with cells of color C+1, then you will get (n·m)2 edges.

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My solution which I failed only because the limits in To was improperly set.

I am checking only 2 Chests for every row and every column for updation of dp array of chest with key x, the reason being every other Chest with number x+1 will be useless as at least one of these points give the same result .

I don't know why won't CF give me Purple.

18203988

Link to Image

More Explanation: As In normal dp for a chest with number x ,

we had to relax for every chest with number x , every chest number with x+1.

Now my observation for bringing complexity down was that for dp[u] = min{dp[v] + dist(u,v)} where v has a chest number x+1 was that the system was a grid.

Now we can imagine a rectangle around the point. Amongst every rectangle we will take the smallest.

Now for relaxing , we need to check only these points! which are simply (N+M). So for every point we check only (N+M) points.

Which causes the complexity to be N*M*(N+M).

Note the figure surrounding won't be a rectangle always . Like this http://postimg.org/image/4c7e5k5nv/ Note: The black dot is a point with chest number x , I am brute forcing on the points that surround it.(Not necessarily they all will exist).

Note : You will need to see the tricks I used to optimize memory too.

Why did the rectangle thing works : as every point outside this rectangle will have to pass through this border , hence the shortest path existing will be surely one of these points for sure.

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    3 years ago, # ^ |
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    dude nice approach. love it:)

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    3 years ago, # ^ |
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    You surely get the minimum distance from the points of chest x-1 to the points of chest x.

    Where are you taking care of the distance already covered uptill points of chest x-1 , which also gets added to compute dp[i][j].

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      3 years ago, # ^ |
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      I am only considering the points surrounding my point to minimize whole function that is

      distance[i][j] = minimum of(distance[x][y] + distance of Point({x,y},Point u)

      Here {x,y} are the points surrounding the point {i,j} in a convex hull/rectangle ( I ain't getting the correct word to use, that's why I built a pic).

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Could you please explain why it won't happen that you will incorrectly find maximum of logarithms due to rounding errors in last problem?

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    3 years ago, # ^ |
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    We wonder if a·log(2) can be close to b·log(3). We know that a and b are positive integers up to 1000.

    a·log(2) = b·log(3)·(1 + ε)
    a / b = log2(3)·(1 + ε)

    On the left we have a rational number with denominator up to $1000$. On the right (omitting epsilon) we have a non-rational number which is kind of simple — it doesn't look like especially designed to be close to some rational number with small denominator. We must use this kind of strange explanation because some other non-rational number could be much much closer. So, for this "kind of simple" non-rational number, I would say we should expect it won't be closer than ε = (1 / 1000)2 to some rational number with denominator  ≤ 1000.

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For D, can anyone explain the following? "We can do such improvement: let cnt[x] be the amount of cells of color x, then when cnt[color]·cnt[color - 1] ≥ n·m, we can do bfs from cells of color color - 1 to cells of color color. Then we will have complexity O(n·m·sqrt(n·m))." I really have no idea how sqrt come out

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    3 years ago, # ^ |
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    Suppose a modification of this algo:

    You iterate through pairs if both cnt's are <= sqrt(nm) and do bfs otherwise.

    Now bfs is run no more than 2 * sqrt(nm) times and each element consist no more than in 2 * sqrt(nm) pairs you iterated.

    It may also be proven for original solution too but it's a bit harder to explain (you may try it yourself)

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      3 years ago, # ^ |
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      but how to prove the dynamic programming part's complexity is <=O(nm*sqrt(nm))?

      please view my reply to kouytoupa below

      thank u

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    3 years ago, # ^ |
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    a + b >= 2*sqrt(a*b), so 2*m*n = 2*sum(cnt[x]) > (cnt[1] + cnt[2]) + (cnt[2] + cnt[3]) + .. + (cnt[p — 1] + cnt[p]) >= 2*sqrt(cnt[1]*cnt[2]) + 2*sqrt(cnt[2]*cnt[3]) + ...

    If there are more than sqrt(m*n) "x x+1" pairs such that cnt[x]*cnt[x + 1] > m*n: 2*m*n > sqrt(m*n) * 2*sqrt(m*n) + ... This can't be true. So bfs will be called no more than sqrt(m*n) times.

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      3 years ago, # ^ |
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      but how to prove the dynamic programming part's total complexity is not greater than nm*sqrt(nm)?

      i just know every cnt[x]*cnt[x+1]<=nm in dp part's, but in total, there may be more than sqrt(nm) of pairs, thus i can't prove the total complexity is <=nm*sqrt(nm)

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        3 years ago, # ^ |
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        and of course, when pairs number is >sqrt(nm), surely there exists some pairs of 'x x+1', such that cnt[x]*cnt[x+1]<nm, the product cannot reach nm, but still, how to prove that the total complexity of dp part is <=O(nm*sqrt(nm))?

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          3 years ago, # ^ |
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          O(c(cnt1*cnt2)+c(cnt2*cnt3)+...) <= O(p*c(nm/p)^2) where c(x)=x<nm?x:nm. The maximal value is when p=sqrt(nm), so O(p*c(nm/p)^2)<=O((nm)^1.5).

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            3 years ago, # ^ |
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            why is c(cnt1*cnt2)+... < p*(c(nm/p)^2)?

            according to am-gm inequality, i just know that, (c(cnt1*cnt2)+...)/(p-1)<=(c(cnt*cnt2)*...)^(1/p)<=mn ,therefore c(cnt1*cnt2)+...<=(p-1)mn<pmn

            i cant get that c(cnt1*cnt2)+...<=p*c(nm/p)^2

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            3 years ago, # ^ |
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            oh i made a mistake

            am-gm inequality suggests that (c(cnt1*cnt2)+...)/(p-1)>=(c(cnt*cnt2)*...)^(1/p)

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        3 years ago, # ^ |
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        For the dynamic programming part:

        cnt[x]*cnt[x + 1] = min(cnt[x], cnt[x + 1]) * max(cnt[x], cnt[x + 1]) <= m*n so min(cnt[x], cnt[x + 1]) <= sqrt(m*n).

        Then sum(cnt[x]*cnt[x + 1]) <= sqrt(m*n)*(max(cnt[xt1], cnt[xt1 + 1]) + max(cnt[xt2], cnt[xt2+1]) + ...) <= sqrt(m*n)*2*sum(cnt[x]) <= sqrt(m*n)*2*m*n.

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          3 years ago, # ^ |
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          why is cnt[x]*cnt[x+1] <=mn?

          the code just implies that when cnt[x]*cnt[x+1]>mn, then it will go to bfs instead, but cnt[x]*cnt[x+1] can be >mn

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            3 years ago, # ^ |
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            So it will go to dp only if cnt[x]*cnt[x + 1] <= mn.

            Actually, cnt[xt1]*cnt[xt1+1], cnt[xt2]*cnt[xt2+1],... (cnt[xti]*cnt[xti + 1] <= mn here) is a subsequence of cnt[1]*cnt[2], cnt[2]*cnt[3], ...

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can anyone please prove the complexity of the first solution of problem D ? why such improvement can optimize the complexity to

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    3 years ago, # ^ |
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    counter-test

    This solution considers only top rows with cells of type 2 but in the optimal solution we should use a cell of type 2 in the last row. The optimal answer is 15, this solution prints 17.

    Unfortunately, it's almost impossible for a setter to predict this kind of crazy heuristic. So he shouldn't be blamed for that.

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in Problem C, when i tranform V_V into decimal it is 31 + 63*64 + 32*64*64 then into binary 11111111111011111 yet it contains 1 ZERO only, so the answer should be 3, and the answer for it is 9, can someone explain how come?

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    3 years ago, # ^ |
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    You need to look it bit-by-bit.

    V is 31 -> 011111, it can have 3 possible pairs of bits, right? 011111 AND 111111, 111111 AND 011111, 011111 AND 011111.

    _ is 63 -> 111111, so only one possible pair of bits -> 111111 AND 111111

    So, we have 3 * 1 * 3 = 9

    Code

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      3 years ago, # ^ |
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      Hey,
      What is the meaning of this sentence : ' he used & as a bitwise AND for these two words represented as a integers in base 64' . Is base in this sentence same as radix.
      Thanks in advance.

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        3 years ago, # ^ |
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        Radix = the base of a system of numeration.

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          3 years ago, # ^ |
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          So don't you think that two words represented as integers in base 64 is a bit ambiguous??. I think it should be more like.
          Each character is represented as an integer in base 64.
          What do you say?
          Thanks in advance

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            3 years ago, # ^ |
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            Sometimes problem is not only about solving :) You have to understand it too

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    3 years ago, # ^ |
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    Actually, you must form a block of 6 bits for every base-64 character, and this includes leading zeroes.

    31 is 011111 and 63 is 111111, so the final number will be 0111111111110111111, and there are two zeros.

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      3 years ago, # ^ |
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      we convert the base 64 string to binary by just converting the individual characters to binary and concatenating. but why this method doesnt work for base 10.

      for base 64 string V_V, the decimal value of characters are 31, 63 and 31. so we can convert it to binary by writing 011111 (31), 111111 (63) and 011111 (31), ie 0111111111110111111.

      but if we try same method for base 10, number 42,

      4 = 0100 in binary and 2 = 0010 in binary, but 42 != 01000010 in binary.

      sorry if its a stupid question but i cant understand it.

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        3 years ago, # ^ |
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        The number system given in the problem is encoding digits with (0-63) decimal number so concatenation is possible. It is like unicode or BCD

        Which has no relation to binary number system . In which position of bit on or off decides the value.

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        3 years ago, # ^ |
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        That's because 64 is a power of 2 and 10 is not.

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    3 years ago, # ^ |
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    Actually you can think less and forget about converting string to binary :)

    Code

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3 years ago, # |
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Could someone please explain the reduction used in problem D ?

and how to BFS when the starting points in color-1 each has its own distance from (1,1)?

it looks like it becomes weighted and i don't see BFS working anymore

what am i missing ?

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    3 years ago, # ^ |
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    It's not weighted because the edges still have length 1. You can apply BFS with a small modification: instead of adding 'color-1' points to the initial wave, you add them into the waves corresponding to their distance from (1,1).

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      3 years ago, # ^ |
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      I still don't understand it. We are in lvl x, and we want to do bfs from lvl x-1.

      These fields have different distance from (1,1). The main idea of bfs is that it inserts filed to the queue as soon as it reaches it.

      So for example distance between (1,1) and (x1,y1) = 43 and dist between (1,1) and (x2,y2) = 50.

      You inserted both fields to the queue — first (x1,y1) second (x2,y2). Then you run bfs and it turned out that (x2,y2) reached field (x3,y3) faster (let's say in 2 moves) than (x1,y1) — it would reach it in 4 moves. So when you reach (x3,y3) you would set 50+2 = 52 distance from (1,1) instead of 43+4 = 47.

      Edit: Maybe the solution will be to hold (x2,y2) until we stop processing fields with distance < 50 and then add it to the queue.

      What am I missing?

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        3 years ago, # ^ |
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        In the normal BFS: we add some vertices in wave 0 (distances to them are 0). Calculate their neighbors — they become wave 1 (distances to them are 1). Continue like this: neighbors of vertices in wave k become wave k + 1 (unless they have already been visited).

        In the modification: the first 42 waves are empty. Wave 43 is empty too, but we add (x1,y1) to it. Continue as in normal BFS, but add (x2,y2) to wave 50 (unless it has been visited already), and so on.

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          3 years ago, # ^ |
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          Thanks — I was about to edit my post. It's good to ask the question as quite often the answer just comes to your mind :)

          I would simply hold (x2,y2) until I finish processing all the vertices with distance < 50 and then add it to the queue.

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Can anyone tell how BFS is used to solve D?

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    3 years ago, # ^ |
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    I join the question. Can anyone explain it? Thanks.

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Problem D is similar to this problem from ACM ICPC Polish regionals: http://main.edu.pl/en/archive/amppz/2012/biu

You can formulate the same dynamic programming and solutions to D with complexity around O(nm log nm) would pass. What is interesting is that in the problem I've given you want to maximize the distance you travel, not minimize it, and if you observe something more, you can go down to a very simple O(nm) solution.

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I'm allergic to floating point numbers, so in E I did the following:

All we need to do is to compare numbers of the form: (2^a)*(3^b). In other words, we have to determine whether (2^a)*(3^b) — (2^c)*(3^d) is positive or not.

  • if a>=c and b>=d it is obviously true

  • if a>=c but b<=d then we can transform it to:

(2^c)*(2^(a-c)*(3^b) — (3^d)) = (2^c)*(3^b)*(2^(a-c)-(3^(d-b)).

It is positive only if (2^(a-c)-(3^(d-b)) is positive. So it is sufficient to determine whether 3^y >= 2^x for x,y<=1000. To do this, I need a function f(x) = smallest y so that 3^y >= 2^x. You can easily precalculate it, I just wrote a simple Python program, taking advantage of its big number arithmetics.

  • cases when a<=c are analogous.
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    3 years ago, # ^ |
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    Or to write, without cheating, big integers in any language. You can calculate 3^0, 3^1, ... as binary numbers. For each 3^y the position of last bit is the largest power of 2 not increasing it.

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Hi It would be great if in the second question it was explicitly mentioned that the order of pieces of potatoes must be maintained strictly. I couldn't do the question because for me it wasnt clear that the order should be maintained. The word 'next' used in the text is ambiguous. It also means another potato.

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    3 years ago, # ^ |
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    The following words were sufficient I guess: "He puts them in the food processor one by one starting from the piece number 1 and finishing with piece number n."

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      3 years ago, # ^ |
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      For 5 potatoes, the said statement doesn't imply that 1->3->2->4->5 is not a valid traversing. It is one by one and starts with 1 and ends with n.

      I did think of the editorial solution but I did not submit it, thinking that it is too easy, along with the fact that aforementioned ambiguity exists. I figured out an alternate solution.

      It would be to choose any permutation of 1 to n, starting with 1 and ending with n. Calculate the answer for it using the method in the editorial. It doesn't yield a unique solution, but it doesn't have to.

      Sorry for being alarmingly pedantic, but given that it is my first contest I thought the question has to be tough and during the contest I thought it would require some dynamic programming concepts that I probably don't understand. I could have had a bit better initial rating had I submitted this solution.

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please come up with alternative solution (other than tutorial) to problem B ?

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    3 years ago, # ^ |
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    You can precalculate an array m: g(i, 0, 63) g(j, 0, 63) ++m[i & j]; Now you get there are m[i] pairs which could and into i. So you go through the string and mul up every element.

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    3 years ago, # ^ |
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    I'll share my approach which I hope is quite easy to follow. I simply used the relation:

    height_of_potatoes_left_in_mixer + height_of_next_potato — k*t <= h

    using this relation you can find the minimum time taken to smash just enough height of potatoes in the mixer so that the next potato can be added. There's a small catch however, while subtracting the height in mixer may go negative. To take care of this just make the height 0 when it turns negative before you add the next potato. Summing this time up for all n potatoes you will get the final answer. My code Hope it helps :)

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Please help could not understand the editorial for problem D. Why is dp[x][y] = x+y for cells of color 1 ? Also explain the optimization approach !

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    3 years ago, # ^ |
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    dp[x][y]=x+y when a[x][y]=1 because the starting point is (1,1) as explicitly mentioned in the problem statement. As for the optimization, performing DP is O(|S1||S2|+|S2||S3|+...+|S(p-1)||Sp|) where Sk is the set of a[i][j]==k. However, this can lead to O(n^2 m^2) complexity in the worst case. Methods like Fenwick tree, etc. can lower the complexity to O(nm^2 lg(n+m)) or O(nm lg^2(n+m)) or O(nm lg(n+m)) depending on the implementation. The solution given by the problem authors is for each of the |Sk||S(k+1)| operations, performing BFS instead or comparing each pair has a maximal complexity of O(nm), providing an upperbound to each of the (p-1) calculations. As a result, this lowers the complexity from O(p*(nm/p)^2) to O(p*c(nm/p)^2) where c(x)=x<nm?x:nm. Then the worst case condition of p=2 no longer becomes maximal, instead becoming sqrt(nm) which lowers the total complexity to O((nm)^1.5).

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So I have tried the DP/BFS solution and a Segment Tree solution for problem D. I have time limit with both solutions. It there any way to have AC in Java for this problem?...

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ж

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-in 677D vanya and treasure i implemented the O(n^2*M^2) version We can do such improvement: let cnt[x] be the amount of cells of color x, then when cnt[color]·cnt[color - 1] ≥ n·m, we can do bfs from cells of color color - 1 to cells of color color.

i am not getting this line can someone explain

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can anyone help me with 677 D vasya and fence im ade the O(n^2*m^2) algorithm but i am not getting this line:

We can do such improvement: let cnt[x] be the amount of cells of color x, then when cnt[color]·cnt[color - 1] ≥ n·m, we can do bfs from cells of color color - 1 to cells of color color. Then we will have complexity O(n·m·sqrt(n·m)).

PLs can someone explain

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In the last question, how do we know that when we take the log, we wont run into precision issues?

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18210431 can anyone please explain why this solution passed problem D? Thanks in advance. why the first n + m points will be the best ones...

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I use python on problem D,I try my best and finally got TLE on test#48. Other people use python only pass test#19.

Is python too slow to accept the problem D? my submission, use BFS:18233710

I have try many kind of constant optimization, so meet test#48, else will got TLE on earlier test. It looks that, I should not use python to do such problems, which have strict time limit.

now I use c++ accept, but I use 500ms , someone use 100ms , I think his program in python can accept, I will learn his program.

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Can anybody provide ne solution for 677 D- Vanya treasure in Java. I'm having some difficulty in understanding some STL terms.

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Can someone please give some code for question D in Java. I'm having some problem in understanding C++ STL terms.

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Can someone help me with problem C ?
I am getting TLE for a case.
Here is my code ! https://ideone.com/ujmrwu

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in problem D div2(Vanya and treasure), i don't understand why the complexity is n*m*sqrt(n*m), and also i don't understand why do we have 2 cases when cnt[color]·cnt[color - 1] ≥ n·m and the other, what's the idea behind that ?

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    3 years ago, # ^ |
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    to avoid TLE we have those 2 cases !

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for the problem 677D — Vanya and Treasure i don't get the improvement of the dynamic programming solution why if cnt[colorX] * cnt [colorX-1] >= n * m we then bfs any explanations ?

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In problem D, someone solved it like this : First, iterating value v from 1 to p-1. For selected v, we can get positions which has value v (let's say these are in group i) and positions having value (v+1) (group j). when connecting these two groups, there can be a lot of connections if both are more than 10000, whatsoever. But sorting group i by using sum of distances from start and just seeing maximum 600 positions in group i(maxmimum(n+m)=600) may reduce time complexity. This solution is very fast, but I wonder if it can be ensured. (Of course this was accepted in system test) Can someone prove or disprove this solution?

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    3 years ago, # ^ |
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    I try to disprove this solution, if I correctly understood it. Assume there are values in matrix like this (i in [1..n], j in [1..m]): if i > 1 then value is equal to 1; if (i,j) == (1, m) then value is equal to 2; if (i,j) == (1, m-1) then value is equal to 1; if (i,j) == (1, 1) then value is equal to 4; else value is equal to 3. Then we need to take key with value 2 passing to first row. But by your algorithm we pass to only 600 nearest keys with value 1 and go necessarily to the second row (if n and m are big, for example equal to 300).

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I wrote a solution for vanya and treasure (div2 D) using 2D sgement tree but i am getting TLE. i made it recursive not iterative like the tutorial does this cause TLE? i don't understand the tutorial implementation :/

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677D — Vanya and Treasure 1,when cnt[color]·cnt[color - 1] ≥ n·m, we can do bfs from cells of color color - 1 to cells of color color. what's the point? How to understand? 2,can we omit the following code snippet? while (pointer < lst.size() && lst[pointer].X <= val) bfs.push_back(lst[pointer++]);

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Can someone tell me why this code is right?what dose 600 meaning in this code?I just thought this code is wrong.but I can't get a test to hack it.sad...If this code is right ,Can someone told me why?19380993

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In problem 677E - Vanya and Balloons, I got WA and i can not find any bug in my solution, and i stressed it with more than 80k random tests and all of them got the correct output!!!

could anyone provide a counter-test.

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For problem 677D - Vanya and Treasure, if dp[x][y] is storing minimal distance from previous color, than for color 1, I think dp[x][y] = abs(x-1) + abs(y-1)..