Problem:
Score : 500 points

Problem Statement

Given is a connected undirected graph with N vertices and M edges. The vertices are numbered 1 to N, and the edges are described by a grid of characters S.
If S_{i,j} is 1, there is an edge connecting Vertex i and j; otherwise, there is no such edge.

Determine whether it is possible to divide the vertices into non-empty sets V_1, \dots, V_k such that the following condition is satisfied. If the answer is yes, find the maximum possible number of sets, k, in such a division.

Every edge connects two vertices belonging to two "adjacent" sets. More formally, for every edge (i,j), there exists 1\leq t\leq k-1 such that i\in V_t,j\in V_{t+1} or i\in V_{t+1},j\in V_t holds.

Constraints

2 \leq N \leq 200

S_{i,j} is 0 or 1.

S_{i,i} is 0.

S_{i,j}=S_{j,i}

The given graph is connected.

N is an integer.

Input

Input is given from Standard Input in the following format:

N
S_{1,1}...S_{1,N}
:
S_{N,1}...S_{N,N}

Output

If it is impossible to divide the vertices into sets so that the condition is satisfied, print -1.
Otherwise, print the maximum possible number of sets, k, in a division that satisfies the condition.

Sample Input 1

2
01
10

Sample Output 1

2

We can put Vertex 1 in V_1 and Vertex 2 in V_2.

Sample Input 2

3
011
101
110

Sample Output 2

-1

Sample Input 3

6
010110
101001
010100
101000
100000
010000

Sample Output 3

4

Solution in Python (3.4.3) (judge verdict: Accepted)

from collections import deque
n = int(input())
ls = [list(input()) for i in range(n)]
graph = [[] for i in range(n+1)]
visited = [0 for i in range(n+1)]
parity = [-1 for i in range(n+1)]
dist = [0 for i in range(n+1)]
for i in range(n):
  for j in range(i,n):
    if ls[i][j] == "1":
      graph[i+1].append(j+1)
      graph[j+1].append(i+1)
que = deque([(1,0)])
parity[1] = 0
visited[1] = 1
while que:
  x,cnt = que.popleft()
  dist[x] = cnt
  for y in graph[x]:
    if visited[y] == 0:
      que.append((y,cnt+1))
      visited[y] = 1
      parity[y] = 1-parity[x]
# k = max(dist)
# farls = []
# for i in range(1,n+1):
#   if dist[i] == k:
#     farls.append(i)
ans = []
for far in range(1,n+1):
  que.append((far,0))
  visited = [0 for i in range(n+1)]
  visited[far] = 1
  dist = [0 for i in range(n+1)]
  while que:
    x,cnt = que.popleft()
    dist[x] = cnt
    for y in graph[x]:
      if visited[y] == 0:
        que.append((y,cnt+1))
        visited[y] = 1
      if parity[y] == parity[x]:
        print(-1)
        exit()
  ans.append(max(dist))
print(max(ans)+1)