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Problem solution · Python

ABC299 E — Nearest Black Vertex

ABC299 E — Nearest Black Vertex: 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
102 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 ABC299 E — Nearest Black Vertex, 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

  • 102 lines of Python from the credited upstream file abc299_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.

Open problem ↗Original source ↗
Full codeABC299 E — Nearest Black Vertex · PythonPython
Use this to learn the idea, then write your own version.
# -*- coding: utf-8 -*-  def bfs(vertex_count: int, source: int, graph):    from collections import deque     d = deque()    d.append(source)    visited = [False] * vertex_count    inf = 10 ** 18    dist = [inf] * vertex_count    dist[source] = 0     while d:        cur = d.popleft()         if visited[cur]:            continue         visited[cur] = True         for to in graph[cur]:            if visited[to]:                continue             dist[to] = min(dist[to], dist[cur] + 1)            d.append(to)     return dist  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            # cost, vertex        graph[ai].append(bi)        graph[bi].append(ai)        # 頂点piからの距離がdi未満の頂点を白で塗る     # 上記以外を黒で塗る(全て白ならNo)    k = int(input())    dist = list()    pd = list()     none, white, black = -1, 0, 1    colors = [none] * n     for _ in range(k):        pi, di = map(int, input().split())        pi -= 1         # 前処理: 各頂点からの最短距離をBFSで求める        d = bfs(vertex_count=n, source=pi, graph=graph)        dist.append(d)        pd.append((pi, di))            for j, dij in enumerate(d):            if dij < di:                colors[j] = white        count = 0     for i, color in enumerate(colors):        if color == none:            colors[i] = black            count += 1        if count == 0:        print("No")        exit()        inf = 10 ** 18        # 頂点piと「黒で塗られた頂点のうち、頂点piとの距離の最小値」の距離がdiであるか判定    for i, (pi, di) in enumerate(pd):        d_min = inf         for j, (color, dij) in enumerate(zip(colors, dist[i])):            if color == black:                d_min = min(d_min, dij)         if d_min != di:            print("No")            exit()     print("Yes")    print(''.join(map(str, colors)))  if __name__ == "__main__":    main()

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