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# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;
using static System.Console;
using static System.Math;

class Program
{
    static void Main()
    {
        var N = int.Parse(ReadLine());
        var graph = new List<int>[N];
        var uf = new BipartiteGraph(N);
        for (int i = 0; i < N; i++)
        {
            graph[i] = new List<int>(N);
        }
        for (int i = 0; i < N; i++)
        {
            var s = ReadLine();
            for (int j = i + 1; j < N; j++)
            {
                if (s[j] == '1')
                {
                    graph[i].Add(j);
                    graph[j].Add(i);
                    uf.UniteDiff(i, j);
                }
            }
        }
        var b = new bool[N];
        if (!uf.TryColoring(out b))
        {
            WriteLine(-1);
            return;
        }

        var ans = 0;
        var vis = new bool[N];
        var q = new Queue<int>(N * N);
        var p = new Queue<int>(N * N);
        for (int i = 0; i < N; i++)
        {
            
            Array.Clear(vis, 0, vis.Length);
            q.Clear();
            p.Clear();
            p.Enqueue(1);
            q.Enqueue(i);
            vis[i] = true;
            var t = 1;
            while (q.Count != 0)
            {
                var v = q.Dequeue();
                var d = p.Dequeue();
                
                t = Max(t, d);
                foreach (var u in graph[v])
                {
                    if (!vis[u])
                    {
                        vis[u] = true;
                        q.Enqueue(u);
                        p.Enqueue(d + 1);
                    }
                }
            }
            ans = Max(ans, t);
        }
        WriteLine(ans);
    }
}

public class UnionFind
{
    int[] parent;
    int count;
    public int Count => count;

    public UnionFind(int length)
    {
        parent = new int[length];
        for (int i = 0; i < parent.Length; i++)
        {
            parent[i] = -1;
        }
        count = length;
    }

    public int Root(int x) => parent[x] < 0 ? x : parent[x] = Root(parent[x]);

    public bool Same(int x, int y) => Root(x) == Root(y);

    public void Unite(int x, int y)
    {
        x = Root(x);
        y = Root(y);
        if (x == y) return;
        count--;
        if (parent[x] < parent[y])
        {
            parent[x] += parent[y];
            parent[y] = x;
        }
        else
        {
            parent[y] += parent[x];
            parent[x] = y;
        }
    }

    public int Size(int x) => -parent[Root(x)];
}


class BipartiteGraph : UnionFind//unitesomeができるから二部グラフではないが
{
    int V;
    public BipartiteGraph(int V) : base(V + V)
    {
        this.V = V;
    }

    public void UniteDiff(int x, int y)
    {
        Unite(x, y + V);
        Unite(x + V, y);
    }

    public void UnateSame(int x, int y)
    {
        Unite(x, y);
        Unite(x + V, y + V);
    }

    public bool IsBipartite()
    {
        for (int i = 0; i < V; i++)
        {
            if (Same(i, i + V)) return false;
        }
        return true;
    }

    public bool[] Coloring()
    {
        var rc = new int[V * 2];
        for (int i = 0; i < V; i++)
        {
            var a = Root(i);
            var b = Root(i + V);
            if (rc[a] == 0)
            {
                rc[a] = 1;
                rc[b] = -1;
            }
        }
        var ret = new bool[V];
        for (int i = 0; i < ret.Length; i++)
        {
            ret[i] = rc[Root(i)] == 1;
        }
        return ret;
    }

    public bool TryColoring(out bool[] ret)
    {
        var rc = new int[V * 2];
        ret = new bool[V];
        for (int i = 0; i < V; i++)
        {
            var a = Root(i);
            var b = Root(i + V);
            if (a == b) return false;
            if (rc[a] == 0)
            {
                rc[a] = 1;
                rc[b] = -1;
            }
        }

        for (int i = 0; i < ret.Length; i++)
        {
            ret[i] = rc[Root(i)] == 1;
        }
        return true;
    }
}