Problem solution · Python

ABC417 E — A Path in A Dictionary

ABC417 E — A Path in A Dictionary: a Python solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to KATO-Hiro AtCoder Solutions.

Technique
Depth-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

Depth-first search

For ABC417 E — A Path in A Dictionary, the implementation follows one branch at a time, making it suitable for components, trees, backtracking, or dependency exploration.

  1. Define the state carried into one recursive or stack frame.
  2. Mark or choose the current state before exploring children.
  3. Combine child results or undo the choice when the branch finishes.

Code notes

  • 65 lines of Python from the credited upstream file abc417_e.py.
  • The implementation visibly relies on sequence storage, ordered lookup.
  • No explicit loop blocks detected, together with recursive traversal.

Complexity

Count unique states for graph traversal; for backtracking, count the branching factor and maximum depth.

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 codeABC417 E — A Path in A Dictionary · PythonPython
Use this to learn the idea, then write your own version.
# -*- coding: utf-8 -*-  def solve():    n, m, x, y = map(int, input().split())    x -= 1    y -= 1     graph = [[] for _ in range(n)]     for _ in range(m):        ai, bi = map(int, input().split())        ai -= 1        bi -= 1         graph[ai].append(bi)        graph[bi].append(ai)     for g in graph:        g.sort()     visited = [False] * n    ans = []     def dfs(cur, parent=-1):        ans.append(cur + 1)         if cur == y:            return True         visited[cur] = True         for to in graph[cur]:            if to == parent:                continue            if visited[to]:                continue             if dfs(to, cur):                return True         ans.pop()         return False     dfs(x)    print(*ans)  def main():    import sys     sys.setrecursionlimit(10**6)     input = sys.stdin.readline     t = int(input())     for _ in range(t):        solve()  if __name__ == "__main__":    main() 

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