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 PyPy3 (2.4.0) (judge verdict: Accepted)

from collections import deque
n = int(input())
mat=[[10**18 for i in range(n+1)] for j in range(n+1)]
ab = [[] for i in range(n)]

for i in range(n):
    si = input()
    for j in range(n):
        if int(si[j]) == 1:
            #print(i,j)
            mat[i+1][j+1] = 1
            ab[i].append(j)

colors = [0 for i in range(n)] 

#頂点vをcolor(1 or -1)で塗り、再帰的に矛盾がないか調べる。矛盾があればFalse
def dfs(v,color):
    #今いる点を着色
    colors[v] = color
    #今の頂点から行けるところをチェック
    for to in ab[v]:
        #同じ色が隣接してしまったらFalse
        if colors[to] == color:
            return False
        #未着色の頂点があったら反転した色を指定し、再帰的に調べる
        if colors[to] == 0 and not dfs(to, -color):
            return False
    #調べ終わったら矛盾がないのでTrue
    return True

#2部グラフならTrue, そうでなければFalse
def is_bipartite():
    return dfs(0,1) # 頂点0を黒(1)で塗ってDFS開始

if not is_bipartite():
    print(-1)
    exit()

#N,M=map(int,input().split())

for i in range(n+1):
	mat[i][i]=0

for i in range(1,n+1):
	for x in range(1,n+1):
		for y in range(1,n+1):
			mat[x][y] = min(mat[x][y], mat[x][i] + mat[i][y])

ans = 0

for i in range(1,n+1):
    for j in range(i+1,n+1):
        ans = max(ans,mat[i][j])
        
print(ans+1)