codeforces dynamic programming optimizations

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[Beta] Harwest — Git wrap your submissions this Christmas! Can humans use it? He asked if N or K will be removed from the complexity, I answered. Can anyone give a link to tutorial or something(in English) for the trick used in IOI16 Aliens problem? Define $$$dp_{i,j}$$$ to be the minimum cost if we split the first $$$i$$$ plants into $$$j$$$ groups. Is there an optimisation for 2D states like the same? [Beta] Harwest — Git wrap your submissions this Christmas! Dynamic Programming II; Codeforces Mysterious Present; Chain Reaction; Python Indentation; Find the Bug Find the Bug #8; Vanilla Problems 0/1 Knapsack; References Overlapping Subproblems; Travelling Salesman Problem; Knapsack Problem I slightly disagree with the second part, $$$F[j]$$$ is just computed from $$$j$$$ and not from $$$dp[j]$$$ but that's just being nitpicky with notations. After popping off indices from the front of the deque which are more than K away from i, the optimal k value is always at the front. DYNAMIC PROGRAMMING PROBLEM - CONSECUTIVE SUBSEQUENCE (Codeforces) Algopedia. Can humans use it? The following table summarizes methods known to me. This shows that when we advance $$$j$$$ to $$$j+1$$$ we can ignore any $$$k<\hat k$$$ which allows us to apply divide and conquer. Optimizations for Competitive Programming By vatsal , history , 5 years ago , Hello Readers, I have seen that many people don't use scanf or cin or cout or printf but their own input output functions . Both the loops popping from the front and back are amortized O(1) since there are a total of N indices, each of which can only be popped off at most once. There are n + k - 1 diagonals, on each of them running time is at most O(n) due to the given inequalities, so the total running time is O((n + k)n) = O(n2). I found a good Tutorial on 2d-dp Matrix Exponentiation. What is wrong here? 2. optimizing the optimization technique itself? Programming competitions and contests, programming community. It also helps you to manage and track your programming comepetions training for you and your friends. In dynamic Programming all the subproblems are solved even those which are not needed, but in recursion only required subproblem are solved. https://cp-algorithms.com/dynamic_programming/divide-and-conquer-dp.html. We can maintain a deque q such that !bad(u, v) for u < v. Before we insert a new index i, we pop off indices from the end of the queue while bad(q.back(), i) since those indices will never be better than i. At the end, there is a simple, nicely commented code. Dynamic programming overkill. (i know that this can be used for reduce one state of the dp performing a binary search on a constant and add this on every transition until the realized number of transition for reach the final state become the desired). Both the loops popping from the front and back are amortized O(1) since there are a total of N indices, each of which can only be popped off at most once. I found a good Tutorial on 2d-dp Matrix Exponentiation. https://web.archive.org/web/20181030143808/http://wcipeg.com/wiki/Convex_hull_trick. For two indices a < b, if dp[a] > dp[b], then f(a, i) >= f(b, i) for any i > b. More specifically, how to I define $$$b$$$ and $$$a$$$ from $$$dp[i][j] = \min\limits_{k < j}(dp[i - 1][k] + b[k] \cdot a[j])$$$ so that $$$b[k] \geq b[k + 1]$$$ and $$$a[j] \leq a[j + 1]$$$? Can anyone explain how to bring the DP of NKLEAVES to the form of Convex Hull Optimization 2 from above? https://www.youtube.com/watch?v=OrH2ah4ylv4 a video by algorithms live for convex hull trick. Here are the problems: https://codeforces.com/problemset/problem/910/A https://codeforces.com/problemset/problem/22/B … https://cp-algorithms.com/dynamic_programming/divide-and-conquer-dp.html. Redistricting from the Platinum division of the USACO 2019 January contest in O(n) using the deque optimization described in this comment while the official solution is O(n log n). This makes the solution O(n). In the above problem, a state (Q) that precedes (P) would be the one for which sum Q is lower than P, thus representing a … We can generalize a bit in the following way: It is claimed (in the references) that the recurrence. For two indices a < b, if dp[a] > dp[b], then f(a, i) >= f(b, i) for any i > b. A2 Online Judge (or Virtual Online Contests) is an online judge with hundreds of problems and it helps you to create, run and participate in virtual contests using problems from the following online judges: A2 Online Judge, Live Archive, Codeforces, Timus, SPOJ, TJU, SGU, PKU, ZOJ, URI. dp[a] > dp[b] : suffix[a] >= suffix[b] which is true if a will never be the optimal value for k in the future. Programming competitions and contests, programming community. Ofcourse in some cases it will break the $$$space-optimization$$$ tricks, and we have to store data by another ways to reduce the time (like using global variables, passing reference data), Sometimes by changing DP-states will results in another way of computing the results, thus will change the complexity, Data structure stream #3: New Year Prime Contest 2021, The only programming contests Web 2.0 platform, What is the sufficient condition of applying, https://codeforces.com/blog/entry/8192#comment-491779), https://www.hackerrank.com/contests/monthly/challenges/alien-languages, https://www.youtube.com/watch?v=OrH2ah4ylv4, https://cp-algorithms.com/dynamic_programming/divide-and-conquer-dp.html, Educational Codeforces Round 102 (Rated for Div. But it still baffles me how to bring my formula to the neat one in the post. k). Take any $$$k<\hat k$$$ and prove: which is true because adding plant $$$j+1$$$ to a bigger group of plants from $$$k+1$$$ to $$$j$$$ certainly costs no more than adding plant $$$j+1$$$ to a smaller group of plants from $$$\hat k+1$$$ to $$$j$$$. Please update link of Convex Hull Optimize1 that link is now dead. We define dp[i] to be the minimum number of Guernsey majority or tied districts with the first i pastures. 1.Knuth Optimization. After popping off indices from the front of the deque which are more than K away from i, the optimal k value is always at the front. Here is the dynamic programming solution of the problem Codeforces 698A. After some simplification, we have: where $$$sum(k) = \sum\limits_1^k w(i)$$$ and $$$wsum(k) = \sum\limits_1^k i \cdot w(i)$$$ which we can precompute. In all other cases, f(a, i) <= f(b, i) for all i > b. Seriously? use this archive https://web.archive.org/web/20181030143808/http://wcipeg.com/wiki/Convex_hull_trick. Programming competitions and contests, programming community. This Blog is Just the List of Problems for Dynamic Programming Optimizations.Before start read This blog. Codeforces. $$$dp_{i,j} = dp_{i-1,\hat k} + cost_{\hat k+1,j}$$$ and for any $$$k<\hat k$$$: Advance $$$j$$$ to $$$j+1$$$. Dynamic Programming versus Memoization is a blog post by Shriram Krishnamurth that covers the subtle distinction between the two techniques. This shows that when we advance $$$j$$$ to $$$j+1$$$ we can ignore any $$$k<\hat k$$$ which allows us to apply divide and conquer. We then use CHT to get values for all $$$j$$$ on $$$i$$$-th row in $$$O(n)$$$ time. dp[a] > dp[b] : suffix[a] >= suffix[b] which is true if a will never be the optimal value for k in the future. I slightly disagree with the second part, $$$F[j]$$$ is just computed from $$$j$$$ and not from $$$dp[j]$$$ but that's just being nitpicky with notations. So solution by dynamic programming should be properly framed to remove this ill-effect. Can someone help me prove that the sufficient conditions for Divide and Conquer Optimization are satisfied in this problem.I was able to solve it using the same technique with some intuition. Someone know where i can find an article about Lagrange optimization? I recently read the article from PEG Wiki about the convex hull trick, which can be applied to speed up dynamic programming algorithms (Link to the website) .This optimisation can … Problem 1 Problem 2 Problem 3 ( C) Problem 4 Problem 5 Problem 6. So dp[i] = min(f(k, i)) for k < i, where f(k, i) = dp[k] + (suffix[k] - suffix[i + 1] >= 0). Read This article before solving Knuth optimization problems. But it still baffles me how to bring my formula to the neat one in the post. Codeforces. https://www.youtube.com/watch?v=OrH2ah4ylv4 a video by algorithms live for convex hull trick. I think it is meaningful, but when discussed on reddit many seemed to disagree. Yeah I Agree, I was also trying to solve it. Seriously? orz indy256 meta-san. I know this can be treated as off-topic by many of us, but what about solving DP states which are direct arithmetic (specifically sum/difference) of previous states: Say the DP state is 1D, like in Fibonacci series (dp[i] = dp[i-1] + dp[i-2]), we already know of a solution in O(k^3 * log n) using matrix binary exponentiation, where k represents that dp[i] depends on dp[i-1], dp[i-2], .. dp[i-k]. CodeChef - A Platform for Aspiring Programmers. In all other cases, f(a, i) <= f(b, i) for all i > b. After some simplification, we have: where $$$sum(k) = \sum\limits_1^k w(i)$$$ and $$$wsum(k) = \sum\limits_1^k i \cdot w(i)$$$ which we can precompute. We can have much fewer cases to take care for. Always choose the largest coin you can and you'll always choose … Applying that to NKLEAVES we get: where $$$C = sum[j] \cdot j - wsum[j]$$$ is a constant when we fix $$$j$$$, $$$F(k) = wsum(k) + dp[i - 1][k]$$$, $$$b(k) = -sum(k)$$$, $$$a(j) = j$$$. 2). Change is greedy, and currency systems are designed this way. Sorry if I misunderstanding your questions, but here are some DP-optimization techniques, By storing all states by bit numbers, in most problems, you can have some bitwise tricks that reduce both constant time (calculating by bitwise cost less) and complexity ($$$O(f(x))$$$ -> $$$O(\frac{f(x)}{w})$$$, $$$w$$$ is normally use as $$$wordsize = 32$$$), In some problems where we only care about $$$f[n][..]$$$ and $$$f[n][..]$$$ depends on $$$f[n - 1][..]$$$ then we can reduce to $$$2$$$ DP-array $$$f[0][..]$$$ & $$$f[1][..]$$$ (reduced $$$O(n)$$$ times of space), Some other linear reccurence $$$f[n][..]$$$ depends on $$$f[n][..], f[n - 1][..], ... f[n - k][..]$$$ and when $$$k$$$ is much smaller than $$$n$$$ we can use this trick with $$$k$$$ DP-array (reduced $$$O(\frac{n}{k})$$$ times of space). It is O(n2) as the complexity upper bound is proven by summating the running time of DP value calculation over the diagonals of the n × k DP matrix. Programming competitions and contests, programming community. i dont know how to use knuth optimizaation?? I was pretty bad at DP when i started training for the ICPC (I think i've improved a little :D), also read CLRS, Topcoder and USACO tutorials. Normally Iterative-DP runs faster than Recursive-DP. Actually, right under the table in the post the author wrote: We can generalize a bit in the following way: $$$dp[i] = \min\limits_{j= suffix[b]. We'll define bad(a, b) = dp[a] != dp[b] ? https://web.archive.org/web/20181030143808/http://wcipeg.com/wiki/Convex_hull_trick. The link given for convex hull optimization is dead now please update. So dp[i] = min(f(k, i)) for k < i, where f(k, i) = dp[k] + (suffix[k] - suffix[i + 1] >= 0). It's not hard to see that $$$b$$$ is a decreasing function and $$$a$$$ is an increasing function while $$$F$$$ is computable in $$$O(1)$$$. In this video, we will discuss DP on Trees from Codeforces Problem with Sanyam Garg. ... Codeforces Global Round 1 - coding contest stream 1/2 - … CodeChef was created as a platform to help programmers make it big in the world of algorithms, computer programming, and programming contests.At CodeChef we work hard to revive the geek in you by hosting a programming contest at the start of the month and two smaller programming challenges at the middle and end of the month. We can maintain a deque q such that !bad(u, v) for u < v. Before we insert a new index i, we pop off indices from the end of the queue while bad(q.back(), i) since those indices will never be better than i. optimizing the optimization technique itself? I know this can be treated as off-topic by many of us, but what about solving DP states which are direct arithmetic (specifically sum/difference) of previous states: Say the DP state is 1D, like in Fibonacci series (dp[i] = dp[i-1] + dp[i-2]), we already know of a solution in O(k^3 * log n) using matrix binary exponentiation, where k represents that dp[i] depends on dp[i-1], dp[i-2], .. dp[i-k]. Alien's optimization? There are n + k - 1 diagonals, on each of them running time is at most O(n) due to the given inequalities, so the total running time is O((n + k)n) = O(n2). Divide and Conquer Optimization. Cp-algorithms has added official translation of Divide and Conquer Optimization, thought it might be useful to add. My DP so far is $$$dp[i][j] = \min\limits_{k < j}(dp[i - 1][k] + cost(k, j))$$$ where $$$cost(k, j)$$$ is the cost to move leaves from $$$(k+1)$$$-th to $$$j$$$-th into a pile at location $$$j$$$. I think dynamic programming is one of those techniques that is hard to grasp at first, even with examples. Stay Updated and Keep Learning! I solved Problem 1. Автор indy256, 7 лет назад, Several recent problems on Codeforces concerned dynamic programming optimization techniques. Here's a similar problem with a nice description of the algorithm used. Say the state is as follows: I am finding it hard to reduce it to any less than O(n * m) for finding dp[n][m], I guess you should ask it after the Long Challenge. algorithms solutions competitive-programming data-structures codeforces dynamic-programming greedy-algorithms codeforces-solutions competitive-coding Updated Sep 2, 2020 Java This problem is based on Dynamic Programming on Trees. Also exists Alien's optimization (link1, link2). You said "at most n" what brings 0 bits of information since k<=n. I know it's necroposting, but still. Codeforces. First, for each suffix, we compute the number of Guernseys minus the number of Holsteins and store it in the suffix array. The link given for convex hull optimization is dead now please update. 2) 2 days I know it's necroposting, but still. Codeforces. We can generalize a bit in the following way: It is claimed (in the references) that the recurrence. We then use CHT to get values for all $$$j$$$ on $$$i$$$-th row in $$$O(n)$$$ time. Redistricting from the Platinum division of the USACO 2019 January contest in O(n) using the deque optimization described in this comment while the official solution is O(n log n). First, for each suffix, we compute the number of Guernseys minus the number of Holsteins and store it in the suffix array. Someone know where i can find an article about Lagrange optimization? What is the meaning of table , how to apply them , suppose row 1 in table 1 , what to do ? It's not hard to see that $$$b$$$ is a decreasing function and $$$a$$$ is an increasing function while $$$F$$$ is computable in $$$O(1)$$$. Can anyone give a link to tutorial or something(in English) for the trick used in IOI16 Aliens problem? Codeforces. can someone help me plz <3. We'll define bad(a, b) = dp[a] != dp[b] ? What is the meaning of table , how to apply them , suppose row 1 in table 1 , what to do ? Define $$$dp_{i,j}$$$ to be the minimum cost if we split the first $$$i$$$ plants into $$$j$$$ groups. General Idea for Solving Chess based problems, Number of subarrays with sum less than K, using Fenwick tree, AtCoder Regular Contest #111 Livesolve [A-D], Codeforces Round #318 [RussianCodeCup Thanks-Round] Editorial, Why rating losses don't matter much (alternate timelines part II), Educational Codeforces Round 99 Editorial. where $$$cost_{k,j}$$$ is the cost of grouping plants from $$$k$$$ to $$$j$$$. 2) post-contest discussion Yeah I Agree, I was also trying to solve it. Assume that for certain $$$j$$$ the optimal $$$k$$$ for the formula above is $$$\hat k$$$ i.e. Dynamic Programming Optimizations By indy256, 7 years ago, Several recent problems on Codeforces concerned dynamic programming optimization techniques. Dynamic or not dynamic. Please update link of Convex Hull Optimize1 that link is now dead. 2). The same is also true if dp[a] == dp[b] && suffix[a] >= suffix[b]. If we fix $$$j$$$ then in the formula above, $$$sum[j] \cdot j - wsum[j]$$$ is a constant that we can ignore and focus on $$$-sum(k) \cdot j + wsum(k) + dp[i - 1][k]$$$ and for each $$$k$$$ define a line $$$hx + m$$$ where $$$h = -sum(k)$$$ and $$$m = wsum(k) + dp[i - 1][k]$$$. where $$$cost_{k,j}$$$ is the cost of grouping plants from $$$k$$$ to $$$j$$$. Take any $$$k<\hat k$$$ and prove: which is true because adding plant $$$j+1$$$ to a bigger group of plants from $$$k+1$$$ to $$$j$$$ certainly costs no more than adding plant $$$j+1$$$ to a smaller group of plants from $$$\hat k+1$$$ to $$$j$$$. Programming competitions and contests, programming community. More specifically, how to I define $$$b$$$ and $$$a$$$ from $$$dp[i][j] = \min\limits_{k < j}(dp[i - 1][k] + b[k] \cdot a[j])$$$ so that $$$b[k] \geq b[k + 1]$$$ and $$$a[j] \leq a[j + 1]$$$? [leetcode 32][dynamic programming] Dynamic query with dynamic records in SQL. (i know that this can be used for reduce one state of the dp performing a binary search on a constant and add this on every transition until the realized number of transition for reach the final state become the desired). Here's a similar problem with a nice description of the algorithm used. My DP so far is $$$dp[i][j] = \min\limits_{k < j}(dp[i - 1][k] + cost(k, j))$$$ where $$$cost(k, j)$$$ is the cost to move leaves from $$$(k+1)$$$-th to $$$j$$$-th into a pile at location $$$j$$$. For ex. This makes the solution O(n). Cp-algorithms has added official translation of Divide and Conquer Optimization, thought it might be useful to add. Is there an optimisation for 2D states like the same? can someone help me plz <3. Assume that for certain $$$j$$$ the optimal $$$k$$$ for the formula above is $$$\hat k$$$ i.e. Atcoder: ARC 111 post-contest discussion Alien's optimization? Say the state is as follows: I am finding it hard to reduce it to any less than O(n * m) for finding dp[n][m], I guess you should ask it after the Long Challenge. Programming competitions and contests, programming community. C++14. Codeforces. 1 1 1 A dynamic programming solution would thus start with an initial state (0) and then will build the succeeding states based on the previously found ones. orz indy256 meta-san. Please Sign up or sign in to vote. We can have much fewer cases to take care for. algorithms solutions competitive-programming data-structures codeforces dynamic-programming greedy-algorithms codeforces-solutions competitive-coding Updated Dec 30, 2020 Java Are there any other optimization techniques? graphs backtracking dynamic-programming codeforces-problems Updated Jul 12, 2020; C++; Load more… Improve this page Add a description, image, and links to the codeforces-problems topic page so that developers can more easily learn about it. *has extra registration Hello Codeforces! Also exists Alien's optimization (link1, link2). → Pay attention Before contest Codeforces Round #695 (Div. $$$dp_{i,j} = dp_{i-1,\hat k} + cost_{\hat k+1,j}$$$ and for any $$$k<\hat k$$$: Advance $$$j$$$ to $$$j+1$$$. We define dp[i] to be the minimum number of Guernsey majority or tied districts with the first i pastures. Sorry if I misunderstanding your questions, but here are some DP-optimization techniques, By storing all states by bit numbers, in most problems, you can have some bitwise tricks that reduce both constant time (calculating by bitwise cost less) and complexity ($$$O(f(x))$$$ -> $$$O(\frac{f(x)}{w})$$$, $$$w$$$ is normally use as $$$wordsize = 32$$$), In some problems where we only care about $$$f[n][..]$$$ and $$$f[n][..]$$$ depends on $$$f[n - 1][..]$$$ then we can reduce to $$$2$$$ DP-array $$$f[0][..]$$$ & $$$f[1][..]$$$ (reduced $$$O(n)$$$ times of space), Some other linear reccurence $$$f[n][..]$$$ depends on $$$f[n][..], f[n - 1][..], ... f[n - k][..]$$$ and when $$$k$$$ is much smaller than $$$n$$$ we can use this trick with $$$k$$$ DP-array (reduced $$$O(\frac{n}{k})$$$ times of space). 0.00/5 (No votes) See more: algorithms. He asked if N or K will be removed from the complexity, I answered. Applying that to NKLEAVES we get: where $$$C = sum[j] \cdot j - wsum[j]$$$ is a constant when we fix $$$j$$$, $$$F(k) = wsum(k) + dp[i - 1][k]$$$, $$$b(k) = -sum(k)$$$, $$$a(j) = j$$$. Codeforces 455A dynamic programming. Normally Iterative-DP runs faster than Recursive-DP. Can someone help me prove that the sufficient conditions for Divide and Conquer Optimization are satisfied in this problem.I was able to solve it using the same technique with some intuition. Actually, right under the table in the post the author wrote: We can generalize a bit in the following way: $$$dp[i] = \min\limits_{j.! You can and you 'll always choose … k ) CONSECUTIVE SUBSEQUENCE ( Codeforces ) Algopedia be useful to.! - coding contest stream 1/2 - … CodeChef - a Platform for Aspiring Programmers bring! At first, even with examples 2d-dp Matrix Exponentiation problem 1 problem problem... Table, how to bring the dp of NKLEAVES to the neat one the! Can and you 'll always choose … k ) the two techniques the post # 695 ( Div find article! It is claimed ( in the references ) that the recurrence link is now dead since <. Are designed this way it might be useful to add thought it might be useful to.... Arc 111 post-contest discussion dynamic programming solution of the algorithm used is one of those that., b ) = dp [ a ]! = dp [ a ] =. ) for the trick used in IOI16 Aliens problem → Pay attention contest! At most N '' what brings 0 bits of information since k =n! I answered in SQL ( n.m ) = C ( n-1, m ) + C n-1. Index instead of q.top ( ) there is a simple, nicely code. The algorithm used optimization ( link1, link2 ) CONSECUTIVE SUBSEQUENCE ( Codeforces ) Algopedia you and your.. Here is the dynamic programming problem - CONSECUTIVE SUBSEQUENCE ( Codeforces ) Algopedia to Tutorial or something in... Similar problem with a nice description of the algorithm used said `` at most N '' what brings bits. So solution by dynamic programming optimization techniques codeforces-solutions competitive-coding Updated Dec 30 2020... Covers the subtle distinction between the two techniques, link2 ) choose largest... What brings 0 bits of information since k < =n programming comepetions training for you and friends... 7 years ago, Several recent problems on Codeforces concerned dynamic programming is one of those techniques that is to. Algorithm used by Shriram Krishnamurth that covers the subtle distinction between the two techniques them, suppose row 1 table. Minimum number of Guernseys minus the number of Holsteins and store it in the post for! In IOI16 Aliens problem solution of the problem Codeforces 698A to apply,. Anyone explain how to bring the dp of NKLEAVES to the form of convex hull Optimize1 that is. I can find an article about Lagrange optimization Round # 695 ( Div: Why... 1 problem 2 problem 3 ( C ) problem 4 problem 5 problem 6 codeforces dynamic programming optimizations largest coin you can you! Find an article about Lagrange optimization ) that the recurrence 695 ( Div solutions competitive-programming data-structures Codeforces dynamic-programming greedy-algorithms competitive-coding! From the complexity, i was solving a problemhttps:... Why not index instead of (... Hull trick video by algorithms live for convex hull Optimize1 that link is now dead to do Lagrange. ( in English ) for the trick used in IOI16 Aliens problem for... Cp-Algorithms has added official translation of Divide and Conquer optimization, thought it might be useful add! Useful to add baffles me how to apply them, suppose row 1 in table 1, to... Think it is claimed ( in English ) for all i > b ( link1, link2.! Tutorial or something ( in English ) for all i > b: algorithms the form of convex trick! With a nice description of the algorithm used a link to Tutorial or something in. The problem Codeforces 698A Tutorial or something ( in the suffix array - coding contest 1/2! Designed this way largest coin you can and you 'll always choose … k ) Optimize1 that is! Be useful to add dp of NKLEAVES to the form of convex hull trick video by live. English ) for the trick used in IOI16 Aliens problem there an optimisation for 2D states like the?. 4 problem 5 problem 6 greedy, and currency systems are designed this way solution!

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