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Problem solution · Python

ABC396 E — Min of Restricted Sum

ABC396 E — Min of Restricted Sum: a Python solution using breadth-first search. Learn the idea, check the complexity, and read the full code, with credit to KATO-Hiro AtCoder Solutions.

Technique
Breadth-first search
Source
KATO-Hiro AtCoder Solutions
Length
77 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

Breadth-first search

For ABC396 E — Min of Restricted Sum, the implementation explores reachable states in layers, which is the standard shape for unweighted shortest paths and minimum-step transitions.

  1. Model each valid configuration as a state and each legal move as an edge.
  2. Seed the queue with the starting state and mark it immediately.
  3. Expand each state once, recording distance or reachability for unseen neighbours.

Code notes

  • 77 lines of Python from the credited upstream file abc396_e.py.
  • The implementation visibly relies on sequence storage, ordered lookup, work queue.
  • No explicit loop blocks detected.

Complexity

Verify that each state and transition is processed only a bounded number of times; that determines the traversal cost.

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 codeABC396 E — Min of Restricted Sum · PythonPython
Use this to learn the idea, then write your own version.
# -*- coding: utf-8 -*-  def main():    import sys    from collections import deque     input = sys.stdin.readline     n, m = map(int, input().split())    graph = [[] for _ in range(n)]     for _ in range(m):        ai, bi, zi = map(int, input().split())        ai -= 1        bi -= 1         graph[ai].append((bi, zi))        graph[bi].append((ai, zi))     pending = -1     def bfs(start):        q = deque([start])        colors[start] = 0        zero_or_one[0].append(start)         while q:            cur = q.popleft()            color = colors[cur]             for to, zi in graph[cur]:                expected_color = color ^ (zi >> digit & 1)                 if colors[to] != pending:                    if colors[to] != expected_color:                        return False                else:                    colors[to] = expected_color                    zero_or_one[expected_color].append(to)                    q.append(to)         return True     ans = [0] * n     # XORは、桁ごとで独立に扱える    for digit in range(30):        colors = [pending] * n         # 0 / 1 の割り当てを2色で塗り分ける問題と言い換え        for i in range(n):            if colors[i] != pending:                continue             # 連結成分単位で管理            zero_or_one = [[], []]             if not bfs(i):                print(-1)                exit()             # 各bitに対して、1の割り当てる個数を減らす            zero_count, one_count = len(zero_or_one[0]), len(zero_or_one[1])             if zero_count < one_count:                zero_or_one[0], zero_or_one[1] = zero_or_one[1], zero_or_one[0]             for one_id in zero_or_one[1]:                ans[one_id] |= 1 << digit     print(*ans)  if __name__ == "__main__":    main()

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