프로그래머스 | dfs/bfs | 네트워크

https://school.programmers.co.kr/learn/courses/30/lessons/43162?language=python3

 

def dfs(n, v, computers, visited):
    visited[v] = True
    for near_v in range(n):
        if v != near_v and computers[v][near_v] == 1:
            if not visited[near_v]:
                dfs(n, near_v, computers, visited)

def solution(n, computers):
    answer = 0
    visited = [False] * n
    
    for i in range(n):
        if not visited[i]:
            dfs(n, i, computers, visited)
            answer += 1
    
    return answer