Editorial for Matrix Reachability
Remember to use this editorial only when stuck, and not to copy-paste code from it. Please be respectful to the problem author and editorialist.
Submitting an official solution before solving the problem yourself is a bannable offence.
Submitting an official solution before solving the problem yourself is a bannable offence.
Approach
Use a breadth-first search starting from . Because the graph is given as an adjacency matrix,
the neighbors of a vertex are found by scanning that vertex's entire row of the matrix.
Whenever row entry is
1 and vertex has not been visited yet, mark it visited and add it
to the BFS queue. At the end, print all visited vertices in increasing order.
The BFS processes each reachable vertex once, and scanning one matrix row costs . The total
time complexity is
, and the memory complexity is
for the matrix.
Solution (Python)
import sys
from collections import deque
def main() -> None:
data = sys.stdin.buffer.readline().split()
n = int(data[0])
start = int(data[1]) - 1
matrix = [list(map(int, sys.stdin.buffer.readline().split())) for _ in range(n)]
visited = [False] * n
visited[start] = True
queue = deque([start])
while queue:
node = queue.popleft()
row = matrix[node]
for nxt, has_edge in enumerate(row):
if has_edge and not visited[nxt]:
visited[nxt] = True
queue.append(nxt)
answer = [str(i + 1) for i, reachable in enumerate(visited) if reachable]
print(" ".join(answer))
if __name__ == "__main__":
main()
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