Problem solution · Python

ABC317 E — Avoid Eye Contact

ABC317 E — Avoid Eye Contact: 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
100 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 ABC317 E — Avoid Eye Contact, 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

  • 100 lines of Python from the credited upstream file abc317_e.py.
  • The implementation visibly relies on sequence storage, ordered lookup, work queue.
  • 1 loop block 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 codeABC317 E — Avoid Eye Contact · PythonPython
Use this to learn the idea, then write your own version.
# -*- coding: utf-8 -*-  def main():    import sys    from collections import deque    from typing import Any, List, Tuple     input = sys.stdin.readline     h, w = map(int, input().split())    a = [list(input().rstrip()) for _ in range(h)]     # 人の視線に入る部分をチェック    # 共通部分をまとめる    dxy = [(-1, 0), (1, 0), (0, -1), (0, 1), (-1, -1), (1, -1), (-1, 1), (1, 1)]    dxy = dxy[:4]    dirs = ["<", ">", "^", "v"]    eye = "!"     for (dx, dy), dir in zip(dxy, dirs):        for i in range(h):            for j in range(w):                if a[i][j] != dir:                    continue                 ny, nx = i, j                 while True:                    ny += dy                    nx += dx                     if nx < 0 or nx >= w or ny < 0 or ny >= h:                        break                    if not (a[ny][nx] == "." or a[ny][nx] == eye):                        break                     a[ny][nx] = eye     # 始点・終点の位置を調べる    sy, sx, gy, gx = -1, -1, -1, -1     for i in range(h):        for j in range(w):            if a[i][j] == "S":                sy, sx = i, j            elif a[i][j] == "G":                gy, gx = i, j     # BFSで始点から終点まで移動    blocked = [">", "v", "<", "^", "#", "!"]     def bfs_for_grid(        grid: List[List[Any]], h: int, w: int, sy: int = 0, sx: int = 0    ) -> Tuple[List[List[bool]], List[List[int]]]:        d = deque()        d.append((sy, sx))        visited = [[False] * w for _ in range(h)]        pending = -1        dist = [[pending] * w for _ in range(h)]        dist[sy][sx] = 0  # Initialize        dxy = [(-1, 0), (1, 0), (0, -1), (0, 1)]         while d:            y, x = d.popleft()             if dist[y][x] == pending:                continue             if visited[y][x]:                continue             visited[y][x] = True             for dx, dy in dxy:                nx = x + dx                ny = y + dy                 if nx < 0 or nx >= w or ny < 0 or ny >= h:                    continue                if visited[ny][nx]:                    continue                if grid[ny][nx] in blocked:                    continue                if dist[ny][nx] != pending and dist[ny][nx] <= dist[y][x]:                    continue                 dist[ny][nx] = dist[y][x] + 1  # Update ans                d.append((ny, nx))         return visited, dist     visited, dist = bfs_for_grid(grid=a, h=h, w=w, sy=sy, sx=sx)     print(dist[gy][gx])  if __name__ == "__main__":    main() 

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗