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Problem solution · Python

ABC391 E — Hierarchical Majority Vote

ABC391 E — Hierarchical Majority Vote: 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
45 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 ABC391 E — Hierarchical Majority Vote, 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

  • 45 lines of Python from the credited upstream file abc391_e.py.
  • The implementation visibly relies on 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 codeABC391 E — Hierarchical Majority Vote · 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())    s = input().rstrip()    m = len(s)    # # 木dp: dp[v][i]: 頂点vにおいて、0もしくは1にするときのコストの最小値    dp = [[1 for _ in range(2)] for _ in range(m)]     # 元の01列と同じ場合のコストは0    for i, si in enumerate(s):        dp[i][int(si)] = 0     inf = 10**18     while len(dp) > 1:        size = len(dp)        ndp = [[inf for _ in range(2)] for _ in range(size // 3)]         # 部分木を3つずつ見る        for left in range(0, size, 3):            for pattern in range(1 << 3):                cost = 0                 for delta in range(3):                    cost += dp[left + delta][pattern >> delta & 1]                 # 多数決をpopcountで判定                majority = 1 if pattern.bit_count() >= 2 else 0                ndp[left // 3][majority] = min(ndp[left // 3][majority], cost)         dp = ndp     ans = max(dp[0])    print(ans)  if __name__ == "__main__":    main()

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