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Problem solution · Python

ABC060 D — Simple Knapsack

ABC060 D — Simple Knapsack: 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
41 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 ABC060 D — Simple Knapsack, 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

  • 41 lines of Python from the credited upstream file arc073_b.py.
  • The implementation visibly relies on hash lookup, 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 codeABC060 D — Simple Knapsack · PythonPython
Use this to learn the idea, then write your own version.
# -*- coding: utf-8 -*-  def main():    from collections import defaultdict    import sys     input = sys.stdin.readline     n, large_w = map(int, input().split())    w = [0] * (n)    v = [0] * (n)     for i in range(n):        w[i], v[i] = map(int, input().split())        # See:    # https://atcoder.jp/contests/abc060/submissions/1253016    # 物の重さをリスト・配列ではなく、defaultdictで管理    dp = defaultdict(int)    dp[0] = 0     for i in range(n):        ndp = defaultdict(int)         for j in dp.keys():            # i番目の物を選ぶ            if (j + w[i]) <= large_w:                ndp[j + w[i]] = max(ndp[j + w[i]], dp[j] + v[i])             # i番目の物を選ばない            ndp[j] = max(ndp[j], dp[j])         dp = ndp        print(max(dp.values()))  if __name__ == "__main__":    main()

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