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Problem solution · Python

ABC282 D — Make Bipartite 2

ABC282 D — Make Bipartite 2: a Python solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to KATO-Hiro AtCoder Solutions.

Technique
Direct simulation
Source
KATO-Hiro AtCoder Solutions
Length
75 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

Direct simulation

For ABC282 D — Make Bipartite 2, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 75 lines of Python from the credited upstream file abc282_d.py.
  • The implementation visibly relies on sequence storage, ordered lookup.
  • No explicit loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

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.

Open problem ↗Original source ↗
Full codeABC282 D — Make Bipartite 2 · PythonPython
Use this to learn the idea, then write your own version.
# -*- coding: utf-8 -*-  def main():    import sys     input = sys.stdin.readline     n, m = 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)     # See:    # https://prd-xxx.hateblo.jp/entry/2017/10/13/004256    no_color, black, white = 0, 1, -1    colors = [no_color] * n # 0: 未着色、1: 黒、-1: 白     def is_bipartite(i, black_count = 0, white_count = 0):        stack = [(i, black)] # 頂点、色         while stack:            vertex, color = stack.pop()             if colors[vertex] != no_color:                continue             colors[vertex] = color             if color == black:                black_count += 1            else:                white_count += 1                    for to in graph[vertex]:                if colors[to] == color:                    return False, 0, 0                                if colors[to] == no_color:                    stack.append((to, -color)) # 色を反転         return True, black_count, white_count         def nC2(n):        return n * (n - 1) // 2        ans = nC2(n) - m        # 各頂点を始点として、二部グラフか判定(連結成分が複数であることへの対応)    for i in range(n):        if colors[i] != no_color:            continue         flag, black_count, white_count = is_bipartite(i)         if flag:            ans -= nC2(black_count)            ans -= nC2(white_count)        else:            print(0)            exit()     print(ans)     if __name__ == "__main__":    main()

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