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Problem solution · Python

ABC160 D — Line++

ABC160 D — Line++: 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
65 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 ABC160 D — Line++, 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

  • 65 lines of Python from the credited upstream file abc160_d.py.
  • The implementation visibly relies on 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 codeABC160 D — Line++ · PythonPython
Use this to learn the idea, then write your own version.
# -*- coding: utf-8 -*- def main():    from collections import deque    import sys    input = sys.stdin.readline     n, x, y = map(int, input().split())    x -= 1    y -= 1    inf = 10 ** 7    ans = [0 for _ in range(n)]     # See:    # https://www.youtube.com/watch?v=zG1L4vYuGrg&feature=youtu.be    # KeyInsight:    # △: 全探索 + 深さ優先探索    # ×: どのように実装すればいいか、見当がつかなかった    for i in range(n):        # グラフを陽に持たない場合        # 始点からの距離を管理        dist = [inf for _ in range(n)]  # 距離無限大で初期化        dist[i] = 0         # キューを用意して、頂点を管理        d = deque()        d.append(i)         # BFS        while d:            v = d.popleft()  # 最も左側の内容(=最初に突っ込んだ値)を取り出す            dj = dist[v]             # 場合分け            # 範囲外参照を回避しながら、次の頂点に移動したときの距離を計算            if v - 1 >= 0 and dist[v - 1] == inf:                dist[v - 1] = dj + 1                # 次の頂点をキューに突っ込む                d.append(v - 1)            if v + 1 < n and dist[v + 1] == inf:                dist[v + 1] = dj + 1                d.append(v + 1)            if v == x and dist[y] == inf:                dist[y] = dj + 1                d.append(y)            if v == y and dist[x] == inf:                dist[x] = dj + 1                d.append(x)         # 距離ごとの個数を計算        for j in range(n):            ans[dist[j]] += 1     # 始点と終点を区別    for k in range(n):        ans[k] //= 2     # 答えを出力    for ii in range(1, n):        print(ans[ii])  if __name__ == '__main__':    main()

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