Problem solution · Python

ABC461 C — Variety

ABC461 C — Variety: a Python solution using sorting and greedy selection. Learn the idea, check the complexity, and read the full code, with credit to KATO-Hiro AtCoder Solutions.

Technique
Sorting and greedy selection
Source
KATO-Hiro AtCoder Solutions
Length
55 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

Sorting and greedy selection

For ABC461 C — Variety, the implementation first exposes a useful order, then scans that order while making locally justified choices.

  1. Choose the key that reveals the greedy or grouping structure.
  2. Sort the relevant records by that key.
  3. Scan in order, maintaining the invariant that makes each local choice safe.

Code notes

  • 55 lines of Python from the credited upstream file abc461_c.py.
  • The implementation visibly relies on sequence storage, hash lookup, ordered lookup.
  • No explicit loop blocks detected.

Complexity

Sorting is typically the dominant term unless the subsequent scan uses a more expensive nested operation.

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.

Full codeABC461 C — Variety · PythonPython
Use this to learn the idea, then write your own version.
# -*- coding: utf-8 -*-  def main():    import sys    from collections import defaultdict     input = sys.stdin.readline     n, k, m = map(int, input().split())    pending = -1    c_max = [pending] * n    d = defaultdict(list)     for _ in range(n):        ci, vi = map(int, input().split())        ci -= 1         d[ci].append(vi)        c_max[ci] = max(c_max[ci], vi)     candidates = []     for i, c_max_i in enumerate(c_max):        if c_max_i == pending:            continue         candidates.append(c_max_i)     candidates.sort(reverse=True)     remains = []     while len(candidates) > m:        remains.append(candidates.pop())     for key, values in d.items():        values = sorted(values)        values.pop()         remains += values     remains.sort()     while len(candidates) < k:        value = remains.pop()        candidates.append(value)     ans = sum(candidates)    print(ans)  if __name__ == "__main__":    main() 

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