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Problem solution · Python

ABC219 D — Strange Lunchbox

ABC219 D — Strange Lunchbox: a Python solution using dynamic programming. Learn the idea, check the complexity, and read the full code, with credit to KATO-Hiro AtCoder Solutions.

Technique
Dynamic programming
Source
KATO-Hiro AtCoder Solutions
Length
33 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Dynamic programming

For ABC219 D — Strange Lunchbox, the implementation records answers for smaller states and reuses them to build the requested result without repeating work.

  1. Define precisely what one DP state represents.
  2. Establish the base cases before transitions are evaluated.
  3. Process states in dependency order and combine only already-known values.

Code notes

  • 33 lines of Python from the credited upstream file abc219_d.py.
  • The implementation visibly relies on ordered lookup, cached states.
  • No explicit loop blocks detected.

Complexity

Multiply the number of reachable states by the work performed for each transition, then include the stored state table in memory usage.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from KATO-Hiro AtCoder Solutions by KATO-Hiro and is used under the CC0-1.0 licence.

Open problem ↗Original source ↗
Full codeABC219 D — Strange Lunchbox · PythonPython
Use this to learn the idea, then write your own version.
# -*- coding: utf-8 -*-  def main():    import sys     input = sys.stdin.readline     n = int(input())    x, y = map(int, input().split())    inf = n + 1    dp = [[inf for _ in range(y + 1)] for __ in range(x + 1)]    dp[0][0] = 0     for _ in range(n):        ai, bi = map(int, input().split())         for j in range(x, -1, -1):            for k in range(y, -1, -1):                xi = min(j + ai, x)                yi = min(k + bi, y)                dp[xi][yi] = min(dp[xi][yi], dp[j][k] + 1)     ans = dp[x][y]     if ans == inf:        print(-1)    else:        print(ans)  if __name__ == "__main__":    main()

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