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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.util.*;

public class Main {
    public static void main(String[] args) throws Exception {
        // Your code here!
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int h = n;
        int w = n;
        
        // board[i][j]: 迷路
        char[][] board = new char[h][w];
        for (int i = 0; i < h; i++) {
            String str = sc.next();
            for (int j = 0; j < w; j++) {
                board[i][j] = str.charAt(j);
            }
        }
        
        // to: 隣接リスト
        List<Integer>[] to = new List[n];
        for (int i = 0; i < n; i++) {
            to[i] = new ArrayList<Integer>();
        }
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (board[i][j] == '1') {
                    to[i].add(j);
                    to[j].add(i);
                }
            }
        }
        
        int ans = 0;
        for (int i = 0; i < n; i++) {
            // visited: 当該点を既に通ったか否か
            boolean[] visited = new boolean[n];
            visited[i] = true;
            
            // max: iから最も遠い点との距離
            int max = 0;
            int[] dist = new int[n];
            // 頂点0から順に、親から子へ初期値を配り最終値を求める
            Queue<int[]> que = new ArrayDeque<int[]>();
            que.add(new int[]{i, 0});
            while (!que.isEmpty()) {
                int[] cur = que.poll();
                int p = cur[0];
                int dis = cur[1];
                
                visited[p] = true;
                for (Integer q : to[p]) {
                    if (visited[q]) continue;
                    
                    if (dist[q] != 0 && dist[q] != dis + 1) {
                        System.out.println(-1);
                        return;
                    }
                    
                    if (dist[q] != 0 && dist[q] == dis + 1) continue;
                    
                    dist[q] = dis + 1;
                    que.add(new int[]{q, dist[q]});
                    max = Math.max(max, dist[q]);
                }
            }
            
            ans = Math.max(ans, max);
        }
        
        System.out.println(ans+1);
    }
}