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Problem solution · Python

ABC375 E — 3 Team Division

ABC375 E — 3 Team Division: 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
57 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 ABC375 E — 3 Team Division, 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

  • 57 lines of Python from the credited upstream file abc375_e.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 codeABC375 E — 3 Team Division · 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())    ab = [tuple(map(int, input().split())) for _ in range(n)]    x = 0     for _, bi in ab:        x += bi     if x % 3 != 0:        print(-1)        exit()     x //= 3     # 部分和問題 = DP    inf = 10**12    dp = [[inf for _ in range(x + 1)] for _ in range(x + 1)]    dp[0][0] = 0     for ai, bi in ab:        ndp = [[inf for _ in range(x + 1)] for _ in range(x + 1)]         # 元のチームと異なる場合にコストを加算        cost1 = 0 if ai == 1 else 1        cost2 = 0 if ai == 2 else 1        cost3 = 0 if ai == 3 else 1         # 3チーム目は残りの2チームで決まるため、状態を持たなくてもよい        for j1 in range(x + 1):            for j2 in range(x + 1):                nj1, nj2 = j1 + bi, j2 + bi                 if nj1 <= x:                    ndp[nj1][j2] = min(ndp[nj1][j2], dp[j1][j2] + cost1)                if nj2 <= x:                    ndp[j1][nj2] = min(ndp[j1][nj2], dp[j1][j2] + cost2)                 ndp[j1][j2] = min(ndp[j1][j2], dp[j1][j2] + cost3)         dp = ndp[:]     if dp[x][x] == inf:        print(-1)    else:        print(dp[x][x])  if __name__ == "__main__":    main()

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