Problem:
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Breadth First Search

Write a program which reads an directed graph $G = (V, E)$, and finds the shortest distance from vertex $1$ to each vertex (the number of edges in the shortest path). Vertices are identified by IDs $1, 2, ... n$.

Input

In the first line, an integer $n$ denoting the number of vertices, is given. In the next $n$ lines, adjacent lists of vertex $u$ are given in the following format:

$u$ $k$ $v_1$ $v_2$ ... $v_k$

$u$ is ID of the vertex and $k$ denotes its degree.$v_i$ are IDs of vertices adjacent to $u$.

Constraints

$1 \leq n \leq 100$

Output

For each vertex $u$, print $id$ and $d$ in a line. $id$ is ID of vertex $u$ and $d$ is the distance from vertex $1$ to vertex $u$. If there are no path from vertex $1$ to vertex $u$, print -1 as the shortest distance. Print in order of IDs.

Sample Input 1

4
1 2 2 4
2 1 4
3 0
4 1 3

Sample Output 1

1 0
2 1
3 2
4 1

Reference

Introduction to Algorithms, Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein. The MIT Press.

Solution in Python3 (judge verdict: Accepted)

cin = open(0).read().strip().split('\n')
n = int(cin[0])
v = [list(map(lambda x: int(x)-1, a.split(' ')[2:])) for a in cin[1:]]

import queue
q = queue.Queue()
q.put(0)

dist = [-1]*n
dist[0] = 0
seen = [False]*n
while not q.empty():
    num = q.get()
    seen[num] = True
    for a in v[num]:
        if dist[a] != -1: continue
        dist[a] = dist[num] + 1
        q.put(a)

for idx, num in enumerate(dist):
    print(idx+1, num)