Problem solution · Python

ABC139 E — League

ABC139 E — League: 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
113 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 ABC139 E — League, 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

  • 113 lines of Python from the credited upstream file abc139_e.py.
  • The implementation visibly relies on sequence storage, hash lookup, 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.

Full codeABC139 E — League · PythonPython
Use this to learn the idea, then write your own version.
# -*- coding: utf-8 -*-  from collections import dequefrom typing import List, Tuple  class TopologicalSorting:    """    See:    https://atcoder.jp/contests/abc291/submissions/39241055    """     def __init__(self, vertex_count: int) -> None:        self.vertex_count = vertex_count        self.graph = [[] for _ in range(vertex_count)]        self.indegrees = [0] * vertex_count     def add_edge(self, frm: int, to: int) -> None:        """        Args:            frm(from) -> to: Vertex number (0-indexed).        """        assert 0 <= frm < self.vertex_count        assert 0 <= to < self.vertex_count         self.graph[frm].append(to)        self.indegrees[to] += 1     def sort(self) -> Tuple[bool, List[int]]:        """        Returns:            is_DAG: Is it DAG (Directed Acyclic Graph) ?            orders: Order of vertices (0-indexed).        """        que = deque([i for i in range(self.vertex_count) if self.indegrees[i] == 0])        results = list()        cost = [1] * self.vertex_count         if len(que) == 0:            return False, []         while que:            # 起点となる頂点が複数存在してもok            # if len(que) >= 2:            #     return False, []             vertex = que.popleft()            results.append(vertex)             for to in self.graph[vertex]:                self.indegrees[to] -= 1                # トポロジカルソートをしながらコストを更新                cost[to] = max(cost[to], cost[vertex] + 1)                 if self.indegrees[to] == 0:                    que.append(to)         if len(results) == self.vertex_count:            return True, cost        else:            return False, []  def main():    import sys    from collections import defaultdict     input = sys.stdin.readline     n = int(input())    a = [list(map(lambda x: int(x) - 1, input().split())) for _ in range(n)]     # 試合を頂点、各選手が行う試合の順番を有向グラフと捉える    # DAG + 最長経路問題    # 選手と試合のidの対応づけ    ids = defaultdict(int)    id = 0     for i in range(n):        for j in range(i + 1, n):            ids[(i, j)] = id            id += 1     def to_id(i, j):        if i > j:            i, j = j, i         return ids[(i, j)]     vertex_count = n * (n - 1) // 2    ts = TopologicalSorting(vertex_count)     for i in range(n):        # 選手番号から試合のidに変換        for j in range(n - 1):            a[i][j] = to_id(i, a[i][j])         # 各選手に対して、指定された日程の順番に辺を張る        for ui, vi in zip(a[i], a[i][1:]):            ts.add_edge(vi, ui)     is_DAG, dist = ts.sort()     if is_DAG:        print(max(dist))    else:        print(-1)  if __name__ == "__main__":    main() 

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