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AminAnvari's blog
By AminAnvari, 4 years ago, In English,
Hi :)
Some problem about powerful data structure and algorithm SQRT decomposition.
If you want to learn this algorithm click here.
And a big thank to mr_agha_seyed for helping.
220B - Little Elephant and Array
86D - Powerful array
13E - Holes
455D - Serega and Fun
375D - Tree and Queries
446C - DZY Loves Fibonacci Numbers
487D - Conveyor Belts
506D - Mr. Kitayuta's Colorful Graph
348C - Subset Sums
617E - XOR and Favorite Number
444C - DZY Loves Colors
398D - Instant Messanger
342E - Xenia and Tree
Maybe some of them can solve by other data structures like segment tree.
• Vote: I like it
• +168
• Vote: I do not like it
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4 years ago, # |
  Vote: I like it +5 Vote: I do not like it
cheers! :) :D
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4 years ago, # |
  Vote: I like it +2 Vote: I do not like it
Would be nice to have SQRT decomposition tag for those problems. Now, I guess, all of them in data structures
http://codeforces.com/problemset/tags/data%20structures
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4 years ago, # |
  Vote: I like it +6 Vote: I do not like it
timus 1846 also nice sqrt-decomp problem.
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4 years ago, # ^ |
  Vote: I like it 0 Vote: I do not like it
Can you, please, elaborate a bit?
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4 years ago, # ^ |
Rev. 5   Vote: I like it +3 Vote: I do not like it
I'm thinking of splitting the queries in chunks of sqrt(N) and processing them similar to the disjoint-set-union split as described on e-maxx.
First of all, we will call a set a DS which can handle the insert and erase operations in O(1).
Let active set and temp. set be 2 empty sets
For every chunk of sqrt(N) queries:
1. Go through all the remove operations (query type '-') and, if possible, remove those numbers from the active set of numbers, and add them in a temp. set. This loop will take O(sqrt(N)).
2. Let currGcd be the gcd of all the keys in our active set. There are at most O(N) of them.
3. Process the chunk again and take 2 cases:
• The query asks to remove X. We will erase it from our temp. set.
• The query asks to insert X. We will insert it in the temp. set.
The answer for every query will be the gcd of all the keys in the temp. set and our currGcd. There are at most O(sqrt(N)) items in the temp. set, so this loop will take O(N).
At the end of a chunk, move all the items from the temp set to the active set (and clear the temp set).
The overall complexity is O(N * sqrt(N))
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4 years ago, # ^ |
Rev. 2   Vote: I like it 0 Vote: I do not like it
можно хранить наши элементы по блокам (gcd всех элементов в нем) + хранить мапу, где запоминаем, в каком блоке лежит число, для ответа пройдемся по блокам и пересчитаем gcd, для добавления/удаления пересчитаем gcd только в одном блоке код
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3 years ago, # ^ |
  Vote: I like it 0 Vote: I do not like it
It easily solves with segment tree, each leaf node stores x, non-leaf nodes gcd of its children.
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4 years ago, # |
  Vote: I like it +11 Vote: I do not like it
Are Mo Algorithm and SQRT decomposition considered the same?
In SQRT decomposition we divide the array in buckets, solve the task in these buckets and answer each query in O() time (worst case). However, using Mo Algorithm the queries are sorted according to the bucket the left end of the queries falls into ...
I think both methods have the same time complexity, but are they considered the same?
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4 years ago, # ^ |
  Vote: I like it +23 Vote: I do not like it
SQRT decomposition combines several approaches and algorithms. Mo Algorithm is one of them.