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 {

	static int N;
	static int[] color;
	static List<Integer>[] e;
	
	static boolean dfs(int v, int c) {
		color[v] = c;
		for(int i : e[v]) {
			if(color[i] == c)
				return false;
			if(color[i] == 0 && !dfs(i, -c))
				return false;
		}
		return true;
	}
	
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int INF  = Integer.MAX_VALUE / 2;
		N = sc.nextInt();
		color = new int[N];
		e = new ArrayList[N];
		int[][] d = new int[N][N];
		for(int i = 0; i < N; i++) {
			e[i] = new ArrayList<>();
			String S = sc.next();
			for(int j = 0; j < N; j++) {
				if(i == j)
					continue;
				if(S.charAt(j) == '1') {
					e[i].add(j);
					d[i][j] = 1;
				}
				else
					d[i][j] = INF;
			}
		}
		if(!dfs(0, 1)) {
			System.out.println(-1);
			return;
		}
		int dmax = 0;
		for(int k = 0; k < N; k++)
			for(int i = 0; i < N; i++)
				for(int j = 0; j < N; j++) {
					d[i][j] = Math.min(d[i][j], d[i][k] + d[k][j]);
					if(d[i][j] != INF)
						dmax = Math.max(dmax, d[i][j]);
				}
		System.out.println(dmax + 1);
	}

}