Problem solution · Python

ABC289 E — Swap Places

ABC289 E — Swap Places: 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
67 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 ABC289 E — Swap Places, 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

  • 67 lines of Python from the credited upstream file abc289_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 codeABC289 E — Swap Places · PythonPython
Use this to learn the idea, then write your own version.
# -*- coding: utf-8 -*-  from collections import deque  def solve():    n, m = map(int, input().split())    colors = list(map(int, input().split()))    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)        # (高橋君のいる頂点ti、青木君のいる頂点ai)を状態とする最短経路問題に言い換え    # 状態数O(n ** 2)、遷移: O(m ** 2)    q = deque()    inf = 10 ** 12    dist = [[inf for _ in range(n)] for _ in range(n)]     def push(i, j, d):        if dist[i][j] != inf:            return                dist[i][j] = d        q.append((i, j))        push(0, n - 1, 0)     while q:        a, b = q.popleft()        di = dist[a][b]         for na in graph[a]:            for nb in graph[b]:                if colors[na] == colors[nb]:                    continue                 push(na, nb, di + 1)        ans = dist[n - 1][0]     if ans == inf:        ans = -1        print(ans)  def main():    import sys     input = sys.stdin.readline     t = int(input())     for _ in range(t):        solve()  if __name__ == "__main__":    main() 

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