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apnakaamkar's blog

By apnakaamkar, history, 4 years ago, In English

Could anyone please exlain me how to solve this dp problem? Link:https://atcoder.jp/contests/dp/tasks/dp_n Help?

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4 years ago, # |
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Check this out... https://www.codechef.com/JULY19B/problems/CIRMERGE/

It also has a nice editorial.

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4 years ago, # |
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Its easier to understand the recurrence relation from a solution done made with recursion (and memoisation), so here you go — https://atcoder.jp/contests/dp/submissions/15834404

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    5 months ago, # ^ |
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    please can you tell me what's the problem with greedy solution in this problem? when i use this solution it give me WA

    the greedy solution based on at each move we select 2 adjacent elements with the lowest sum, and merge them until one element remaining.

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      4 months ago, # ^ |
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      In greedy approach, we are making sure that cost incurred in each of the progressive step is minimum but this may not result in minimum total cost incurred.

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4 years ago, # |
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this video by Errichto helped me a lot while solving the dp atcoder contest : https://www.youtube.com/watch?v=FAQxdm0bTaw&t=8731s

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    4 months ago, # ^ |
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    hint for someone who wants to try: dp[i][j] state refers to the minimum total cost of combining interval [i,j] into one vertex. answer will be dp[0][n-1]

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4 years ago, # |
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Is there anything I am missing? If someone can help ?

Link : My solution

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    3 years ago, # ^ |
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    Update to cost=1e18; and it works.

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3 years ago, # |
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Am i missing something in this solution: https://atcoder.jp/contests/dp/submissions/22675988

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    3 years ago, # ^ |
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    lli i,a, b = sum[endi]-sum[start-1],c=INT_MAX; c can be larger than INT_MAX

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3 years ago, # |
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If you want to learn the concept behind this problem then go through Matrix chain multiplication DP, it is a very famous DP problem and has clear explanations(you can check Cormen or any online articles). This N-slimes problem is very similar to that but instead of multiplication, we do addition.

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5 months ago, # |
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can someone tell me what's the problem with greedy solution in this problem? when i use this solution it give me WA

the greedy solution based on at each move we select 2 adjacent elements with the lowest sum, and merge them until one element remaining.