Problem solution · Python

ABC213 E — Stronger Takahashi

ABC213 E — Stronger Takahashi: 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
69 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 ABC213 E — Stronger Takahashi, 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

  • 69 lines of Python from the credited upstream file abc213_e.py.
  • The implementation visibly relies on sequence storage, 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 codeABC213 E — Stronger Takahashi · PythonPython
Use this to learn the idea, then write your own version.
# -*- coding: utf-8 -*-  def main():    from collections import deque     h, w = map(int, input().split())    s = [list(input()) for _ in range(h)]    inf = 10 ** 10    costs = [[inf for _ in range(w)] for _ in range(h)]    costs[0][0] = 0    visited = [[False for _ in range(w)] for _ in range(h)]    dxy = [(0, 1), (1, 0), (0, -1), (-1, 0)]    d = deque()    d.append((0, 0))     while d:        cur_y, cur_x = d.popleft()                if visited[cur_y][cur_x]:            continue         visited[cur_y][cur_x] = True        cost = costs[cur_y][cur_x]         # 壁を破壊せずに移動        for dx, dy in dxy:            nx = cur_x + dx            ny = cur_y + dy             if nx < 0 or nx >= w:                continue            if ny < 0 or ny >= h:                continue            if s[ny][nx] == '#':                continue                               if costs[ny][nx] <= cost:                continue             costs[ny][nx] = cost            d.appendleft((ny, nx))                # 壁を破壊して移動        for dx2 in range(-2, 3):            for dy2 in range(-2, 3):                # マンハッタン距離が3より大きい                if abs(dy2) + abs(dx2) > 3:                    continue                                nx2 = cur_x + dx2                ny2 = cur_y + dy2                 if nx2 < 0 or nx2 >= w:                    continue                if ny2 < 0 or ny2 >= h:                    continue                if costs[ny2][nx2] <= cost + 1:                    continue                 costs[ny2][nx2] = cost + 1                d.append((ny2, nx2))      print(costs[h - 1][w - 1])  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 ↗