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: Accepted)

import java.util.*;

public class Main {

	private static int[][] a;
	private static int[] color;
	private static int dist[];
	private static boolean visited[];

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		a = new int[n][n];
		sc.nextLine();
		for (int i = 0; i < n; i++) {
			char[] s = sc.nextLine().toCharArray();
			for (int j = 0; j < s.length; j++) {
				if (i == j) {
					a[i][j] = 0;
				} else if (s[j] == '1') {
					a[i][j] = 1;
				} else {
					a[i][j] = 1000000;
				}
			}
		}

		color = new int[n];
		for (int i = 0; i < n; i++) {
			if (color[i] == 0) {
				if (!isBipartite(i, 1)) {
					System.out.println(-1);
					return;
				}
			}
		}

		//for (int i = 0; i < n; i++) {
		//	for (int j = 0; j < n; j++) {
		//		System.out.print(a[i][j]);
		//		System.out.print(" ");
		//	}
		//	System.out.println();
		//}


		for (int k = 0; k < n; k++) {
			for (int i = 0; i < n; i++) {
				for (int j = 0; j < n; j++) {
					if (a[i][k] >= 1000000 || a[k][j] >= 1000000) continue;
					if (a[i][j] > a[i][k] + a[k][j]) {
						a[i][j] = Math.min(a[i][j], a[i][k] + a[k][j]);
					}
				}
			}
		}

		//for (int i = 0; i < n; i++) {
		//	for (int j = 0; j < n; j++) {
		//		System.out.print(a[i][j]);
		//		System.out.print(" ");
		//	}
		//	System.out.println();
		//}

		int answer = 0;
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < n; j++) {
				if (a[i][j] < 1000000) {
					answer = Math.max(answer, a[i][j]);
				}

			}
		}

		System.out.println(answer + 1);
	}


	private static boolean isBipartite(int v, int c) {
		color[v] = c;
		for (int i = 0; i < a[v].length; i++) {
			if (a[v][i] != 1000000 && a[v][i] != 0) {
				if (color[i] == c) return false;
				if (color[i] == 0 && !isBipartite(i, -c)) return false;
			}
		}

		return true;
	}
}