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Problem solution · Python

ABC291 F — Teleporter and Closed off

ABC291 F — Teleporter and Closed off: a Python solution using sliding window or two pointers. Learn the idea, check the complexity, and read the full code, with credit to KATO-Hiro AtCoder Solutions.

Technique
Sliding window or two pointers
Source
KATO-Hiro AtCoder Solutions
Length
62 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

Sliding window or two pointers

For ABC291 F — Teleporter and Closed off, the implementation maintains a moving interval and updates only the information that enters or leaves the window.

  1. Choose the invariant that makes a window valid or useful.
  2. Advance the right boundary and add the new element.
  3. Move the left boundary only as needed while maintaining the invariant and updating the answer.

Code notes

  • 62 lines of Python from the credited upstream file abc291_f.py.
  • The implementation visibly relies on sequence storage, ordered lookup.
  • No explicit loop blocks detected.

Complexity

Confirm that neither pointer moves backwards; if so, the scan is usually linear apart from the window’s data-structure operations.

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 codeABC291 F — Teleporter and Closed off · PythonPython
Use this to learn the idea, then write your own version.
# -*- coding: utf-8 -*-  def main():    import sys     input = sys.stdin.readline     n, m = map(int, input().split())    s = [input().rstrip() for _ in range(n)]     # 頂点1→頂点v、頂点v→頂点nを前計算    inf = 10**12    dist_1v = [inf] * n    dist_vn = [inf] * n    dist_1v[0], dist_vn[n - 1] = 0, 0     for i in range(n):        for j in range(m):            if s[i][j] == "0":                continue             dist_1v[i + j + 1] = min(dist_1v[i + j + 1], dist_1v[i] + 1)     for i in range(n - 1, -1, -1):        for j in range(m):            if s[i][j] == "0":                continue             dist_vn[i] = min(dist_vn[i], dist_vn[i + j + 1] + 1)     ans = list()     # 頂点kをスキップするような頂点を探索    # 左側: [k - m, k)、右側[k + 1, k + 1 + m)    for k in range(1, n - 1):        candidate = inf         for left in range(k - m, k):            if left < 0:                continue             for right in range(k + 1, k + 1 + m):                if right >= n or (right - left) > m:                    break                 if s[left][right - left - 1] == "0":                    continue                 candidate = min(candidate, dist_1v[left] + dist_vn[right] + 1)         if candidate == inf:            candidate = -1         ans.append(candidate)     print(*ans)  if __name__ == "__main__":    main()

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