Problem solution · Python

ABC310 D — Peaceful Teams

ABC310 D — Peaceful Teams: 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
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

Depth-first search

For ABC310 D — Peaceful Teams, 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

  • 62 lines of Python from the credited upstream file abc310_d.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 codeABC310 D — Peaceful Teams · PythonPython
Use this to learn the idea, then write your own version.
# -*- coding: utf-8 -*-  def main():    import sys     sys.setrecursionlimit(10**8)     input = sys.stdin.readline     n, t, m = map(int, input().split())     # チーム分けの方法を全探索(DFS)    # 相性の悪いペアを無向グラフとして扱う    ng = [[] for _ in range(n)]     for _ in range(m):        ai, bi = map(int, input().split())        ai -= 1        bi -= 1         ng[ai].append(bi)        ng[bi].append(ai)     def dfs(now, teams):        if now == n:            if len(teams) == t:                return 1            else:                return 0         count = 0         # Tチームのいずれかに配属できるか?        for team in teams:            ok = True             for player in team:                if player in ng[now]:                    ok = False                    break             if ok:                team.append(now)                count += dfs(now + 1, teams)                team.pop()         # Tチーム未満の場合、新しいチームに配属        if len(teams) < t:            teams.append([now])            count += dfs(now + 1, teams)            teams.pop()         return count     ans = dfs(0, [])    print(ans)  if __name__ == "__main__":    main() 

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