Programming competitions and contests, programming community. use this archive https://web.archive.org/web/20181030143808/http://wcipeg.com/wiki/Convex_hull_trick. → Pay attention Before contest Codeforces Round #695 (Div. 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. He asked if N or K will be removed from the complexity, I answered. Atcoder: ARC 111 post-contest discussion Programming competitions and contests, programming community. 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. In combinatorics, C(n.m) = C(n-1,m) + C(n-1,m-1). i dont know how to use knuth optimizaation?? For two indices a < b, if dp[a] > dp[b], then f(a, i) >= f(b, i) for any i > b. The following table summarizes methods known to me. 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. It also helps you to manage and track your programming comepetions training for you and your friends. We define dp[i] to be the minimum number of Guernsey majority or tied districts with the first i pastures. 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. Codeforces. We'll define bad(a, b) = dp[a] != dp[b] ? 2) 2 days Can anyone explain how to bring the DP of NKLEAVES to the form of Convex Hull Optimization 2 from above? Alien's optimization? 1.Knuth Optimization. [Beta] Harwest — Git wrap your submissions this Christmas! 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. Codeforces. Applying that to NKLEAVES we get: where $$C = sum[j] \cdot j - wsum[j]$$$is a constant when we fix $$j$$$, $$F(k) = wsum(k) + dp[i - 1][k]$$$, $$b(k) = -sum(k)$$$, $$a(j) = j$$$. But wasn't able to formally prove it. Here is a list I gathered a few weeks ago: Arabic (Youtube Videos and Playlists): 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. We then use CHT to get values for all $$j$$$ on $$i$$$-th row in $$O(n)$$$ time. ... Codeforces Global Round 1 - coding contest stream 1/2 - … Notes: A[i][j] — the smallest k that gives optimal answer, for example in dp[i][j] = dp[i - 1][k] + C[k][j] C[i][j] — some given cost function; We can generalize a bit in the Programming competitions and contests, programming community. In dynamic Programming all the subproblems are solved even those which are not needed, but in recursion only required subproblem are solved. The same is also true if dp[a] == dp[b] && suffix[a] >= suffix[b]. 1 1 1 We'll define bad(a, b) = dp[a] != dp[b] ? Dynamic programming overkill. 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. Cp-algorithms has added official translation of Divide and Conquer Optimization, thought it might be useful to add. 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