Editorial for Connected Components

Use this editorial only when stuck, and do not copy-paste code from it.

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Approach

Build an adjacency list for the undirected graph.

Scan vertices from $1$ to $n$ . If the current vertex has not been labelled yet, it starts a new component. Run BFS from that vertex and assign the same component number to every reachable vertex. Because BFS visits exactly the vertices connected to the start vertex, each connected component gets one label. Scanning vertices in increasing order gives the required component numbering.

Each vertex is enqueued at most once and each undirected edge is inspected twice, so the time complexity is $O(n + m)$ . The memory complexity is $O(n + m)$ .

Solution (Python)

Code 1

import sys
from collections import deque


def main() -> None:
    data = list(map(int, sys.stdin.buffer.read().split()))
    if not data:
        return

    n, m = data[0], data[1]
    graph = [[] for _ in range(n)]
    pos = 2
    for _ in range(m):
        a = data[pos] - 1
        b = data[pos + 1] - 1
        pos += 2
        graph[a].append(b)
        graph[b].append(a)

    labels = [0] * n
    component = 0
    queue: deque[int] = deque()

    for start in range(n):
        if labels[start] != 0:
            continue

        component += 1
        labels[start] = component
        queue.append(start)

        while queue:
            v = queue.popleft()
            for to in graph[v]:
                if labels[to] == 0:
                    labels[to] = component
                    queue.append(to)

    sys.stdout.write(str(component) + "\n")
    sys.stdout.write(" ".join(map(str, labels)) + "\n")


if __name__ == "__main__":
    main()

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Editorial for Connected Components

Approach

Solution (Python)

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