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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;

public class Main
{
	

	public static void main(String[] args)
	{
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();

		boolean[][] s = new boolean[200][200];
		int[] num = new int[200];

		for(int i = 0 ; i < n ; i++){
			String tmp = sc.next();
			for(int j = 0 ; j < n ; j++){
				s[i][j] = (tmp.charAt(j) == '1');
			}
			num[i] = 0;
		}

		int ans = -1;

		for(int k = 0 ; k < n ; k++){
			LinkedList<Integer> que = new LinkedList<Integer>();
			int kans = 1;

			for(int i = 0 ; i < n ; i++){
				num[i] = 0;
			}

			que.add(k);
			num[k] = 1;
			
			while(que.size() > 0){
				Integer vNum = que.poll();
				for(int i = 0 ; i < n ; i++){
					if(num[i] == 0){
						if(s[vNum][i]){
							num[i] = num[vNum]+1;
							kans = num[i];
							que.add(i);
						}
					}
				}
			}

			for(int i = 0 ; i < n && ans != -1 ; i++){
				for(int j = i+1 ; j < n ; j++){
					if(s[i][j]){
						if(num[i] - num[j] != 1 && num[j] - num[i] != 1){
							kans = -1;
							break;
						}
					}
				}
			}

			if(ans < kans){
				ans = kans;
			}
		}

		System.out.println(ans);
	}
}