PAST3N

• Third Algorithm Practical Skills Test PAST3

N - Replace and Reorder

• Thoughts.

  • A naive implementation of a query to sort a range is O(NlogN)
    • Obviously, they won't make it in time.
    • It has to be about O(logN).
    • How?
    • Swap from already sorted to swap at a high Q times, so the columns before sorting are pretty close to already sorted
    • Let's treat the sorted columns as a single chunk.
    • Represent $[x, y)$ as a tuple (x, y)?
    • Swap at z is $[x, y) → [x, z), [z+1, z+2), [z, z+1), [z+2, y)$.
    • Segments do not overlap, so you only need to look at the top and sort.
      • Then stick them together if they are consecutive?
    • The computational complexity of the sort is O(MlogM) using the number of times M swapped up to that point.
      • It consumes Q to increase this, so it can't be too heavy.
      • If you do the sort X times, M averages Q/X, so the overall computational complexity of the sort query is O(X Q/X log(Q/X)) = O(Q log(Q/X))
        • The average computational complexity of the entire query is less than O(logN)
        • We'll be there in time.
    • How to implement this.
      • No arrays because swapping causes object insertion.
      • But when it comes to dividing by z, I want to find the object to be divided quickly.
        • If you're going to do a binary search, it has to be an array.
        • If you make a linked list, the search is in linear order.
      • Something like this "no array, no linked list" case, often a phenic tree.
      • A set represented by a phenic tree with 1's can be dichotomized in logarithmic order like an array, but insertion and deletion are in constant order.
      • However, this is unordered, so we need to order it separately in the linked list.
      • I want to find the corresponding node in the linked list in constant order after finding it in a binary search.
      • This can be accomplished with [Scarce Linked List
        • Use mostly empty, having reserved an array of size N first.
    • So this will work.
      • 

• mounting

  • How do you do the sorting?
  • Oh, not with the previous diagram.
    • This has the "start value of the fragment" in the phenic tree, but the actual sort and swap query is on the "position in the array".
    • It has to be searchable.
  • I'd rather not need a linked list because the before and after values can be obtained from a bifurcated search of the phenic tree.
  • 
  • The difference between start and end shows where the "next fragment" is in constant order.
  • The implementation of the phenic tree was 1-origin, which was confusing, so the interface was changed to 0-origin.
  • I'm getting confused because it's so hard to distinguish between cases.
    • 

• pause

• If you look for explanations on the internet, it seems like a simpler way to do it.

• Swap increases number of events (e.g. accidents, crimes, meetings, housing starts, hits on a web page) by up to 1, swap with bubble sort decreases number of falls by 1

• Just keep track of where they've been swapped and only check there when sorting.

• I'm not sure if that's really true, but I'll try to implement that policy.

• Sample passed, judge 11 TLE.

• I didn't erase the flag once I set it.

• If you turn it off, the sample will not pass.

  • It was buggy and just happened to pass by not clearing the flag.

• A swap may result in a fall in the foreground

• Flag where it can fall over during a swap (including those done at sort time).

• Sorting is done while erasing flags.

• AC'd with this.

AC: python

def solve(N, Q, QS):
    values = list(range(1, N + 1))
    ft = FenwickTree(N)

    def swap(i):
        values[i], values[i + 1] = values[i + 1], values[i]
        if i > 0:
            ft.set(i - 1, 1)
        ft.set(i, 1)
        if i < N - 1:
            ft.set(i + 1, 1)

    for t, x, y in QS:
        if t == 1:
            x -= 1
            # swap query
            swap(x)
        else:
            x -= 1
            y -= 1
            # sort query
            s = ft.sum(x) + 1
            pos = ft.bisect(s) - 1
            while pos < y:
                ft.set(pos, 0)
                while pos >= x and values[pos] > values[pos + 1]:
                    swap(pos)
                    pos -= 1
                pos = ft.bisect(s) - 1

    return values

• You can only decrease the amount of increase.

This page is auto-translated from /nishio/PAST3N using DeepL. If you looks something interesting but the auto-translated English is not good enough to understand it, feel free to let me know at @nishio_en. I'm very happy to spread my thought to non-Japanese readers.