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Problem solution · Python

ABC222 E — Red and Blue Tree

ABC222 E — Red and Blue Tree: 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
86 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 ABC222 E — Red and Blue Tree, 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

  • 86 lines of Python from the credited upstream file abc222_e.py.
  • The implementation visibly relies on sequence storage, ordered lookup, work queue, cached states.
  • 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 codeABC222 E — Red and Blue Tree · 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, k = map(int, input().split())    a = list(map(int, input().split()))    graph = [[] for _ in range(n)]     for i in range(n - 1):        ai, bi = map(int, input().split())        ai -= 1        bi -= 1         # 頂点番号 + 辺のidもセットで保持        graph[ai].append((bi, i))        graph[bi].append((ai, i))     # 前計算: 与えられた経路でそれぞれ辺を何回通るか    count = [0] * (n - 1)     # See:    # https://atcoder.jp/contests/abc222/submissions/26452884    def bfs(start):        inf = -1        dist = [inf] * n        dist[start] = 0        q = deque([start])         while q:            qi = q.popleft()             for to, _ in graph[qi]:                if dist[to] != inf:                    continue                 dist[to] = dist[qi] + 1                q.append(to)         return dist     for start, goal in zip(a, a[1:]):        start -= 1        goal -= 1         dist = bfs(start)        cur = goal         # 終点から始点までの経路をたどり、通過する辺を数える        while cur != start:            for to, edge_id in graph[cur]:                if dist[to] < dist[cur]:                    cur = to                    count[edge_id] += 1                     break     # 合計S回のうち、赤い辺を何回通るか?    # R - B = K、R + B = Sから、変数Bを削除    r = k + sum(count)     if r < 0 or r % 2 == 1:        print(0)        exit()     # 赤い辺をいくつか選んで、合計r回にできるか?    r //= 2    dp = [0] * (r + 1)    dp[0] = 1    mod = 998244353     for ci in count:        for i in range(r - ci, -1, -1):            dp[i + ci] += dp[i]            dp[i + ci] %= mod     print(dp[r])  if __name__ == "__main__":    main()

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