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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int main()
{
    int n;
    cin >> n;
    vector<vector<int>> g(n);
    vector<int> lf;
    for(int i = 0; i < n; i++)
    {
        int leaf = 0;
        for(int j = 0; j < n; j++)
        {
            char a;
            cin >> a;
            if(a == '1')
            {
                g[i].push_back(j);
                leaf++;
            }
        }
        if(leaf == 1)
        {
            lf.push_back(i);
        }
    }
    int ans = 0;
    for(int v : lf)
    {
        queue<int> que;
        que.push(v);
        vector<int> d(n);
        d[v] = 1;
        bool flag = true;
        while(que.size())
        {
            int pos = que.front();
            que.pop();
            for(int nx : g[pos])
            {
                if(d[nx] == 0)
                {
                    d[nx] = d[pos] + 1;
                    que.push(nx);
                }
                else if(d[nx] == d[pos]-1 || d[nx] == d[pos]+1) continue;
                else
                {
                        flag = false;
                        break;
                }
            }
            if(!flag)
            {
                break;
            }
        }
        int tmp = 0;
        for(int i = 0; i < n; i++)
        {
            tmp = max(tmp, d[i]);
        }
        if(!flag) tmp = 0;
        ans = max(ans, tmp);
    }
    if(ans == 0) cout << -1 << endl;
    else cout << ans << endl;
}