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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
int main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int n;
    cin >> n;
    vector <vector <int> > G(n);
    for(int i = 0; i < n; ++i){
        string v;
        cin >> v;
        for(int j = 0; j < n; ++j){
            if(v[j] == '1'){
                G[i].push_back(j);
                G[j].push_back(i);
            }
        }
    }
    int best = 0, resBest = 0, trt = 0;
    bool isBip = 1;
    for(int g = 0; g < n; ++g)
    {
        queue <int> Q;
        vector <int> vis(n);
        vector <int> dist(n);
        vector <int> color(n);
        Q.push(g);
        while(!Q.empty()){
            int v = Q.front();
            Q.pop();
            vis[v] = 1;
            for(int u : G[v]){
                if(!vis[u]){
                    dist[u] = dist[v] + 1;
                    if(dist[u] > dist[best]){
                        best = u;
                    }
                    vis[u] = 1;
                    Q.push(u);
                    color[u] = dist[u]&1;
                }
                if((color[u] == color[v]) && vis[u] && vis[v])isBip = 0;
            }
        }
        if(dist[best] > resBest){
            trt = g;
            resBest = dist[best];
        }
    }
        queue <int> Q;
        vector <int> vis(n);
        vector <int> dist(n);
        Q.push(trt);
        while(!Q.empty()){
            int v = Q.front();
            Q.pop();
            vis[v] = 1;
            for(int u : G[v]){
                if(!vis[u]){
                    dist[u] = dist[v] + 1;
                   if(dist[u] > dist[best]){
                        best = u;
                    }
                    vis[u] = 1;
                    Q.push(u);
                }
            }
        }

    if(isBip){
        cout << dist[best] + 1 << '\n';
    }
    else cout <<"-1\n";
}