def dfs(nod, visited, computers):
    if nod in visited:
        return 0
    
    visited.append(nod)
    for computer, status in enumerate(computers[nod]):
        if computer != nod and status == 1:
            dfs(computer, visited, computers)
    

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