PAST2K

from Second Algorithm Practical Skills Test PAST2K

• Thoughts.
  • The issue of changing and deleting to make corresponding parentheses.
  • Conversion and deletion costs are fixed for each character.
  • N=3000
  • There is no advantage in converting and then deleting, so there are three options: do nothing, convert, or delete.
  • If we set the parentheses as plamy 1 and the deletion as 0, we could make a DP with Y as the “sum up to the Xth”.
    • Corresponding parentheses are non-negative in the middle and zero at the end
    • Values range from 0 to N
    • N^2, so it’s not too late.
• AC python

def solve(N, S, CS, DS):
    INF = 9223372036854775807
    table = [INF] * (N + 1)  # table[N] is sentinel
    table[0] = 0
    for i in range(N):
        new_table = [INF] * (N + 1)
        if S[i] == "(":
            d = 1
        else:
            d = -1

        for j in range(N):
            new_table[j] = min(
                table[j - d],  # no change
                table[j + d] + CS[i],  # change
                table[j] + DS[i],  # delete
            )
        table = new_table
    return table[0]

- [[Bracketed rows are up and down]]
- [[dynamic programming]]
- [[Watchdogs up and down the range]]

---

This page is auto-translated from /nishio/PAST2K 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.