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>
#define l_ength size
const int inf = (1<<30);
const int mod = 1000000007;
using ll = long long;
using namespace std;

vector<int> g[256];
int dist[256];
int couleur[256];

void init(){
	for( int i = 0; i < 256; ++i ) dist[i] = -1;
}

bool dfs( int p, int f ){
	couleur[p] = f;
	for( auto nextp : g[p] ){
		if( couleur[nextp] == f ) return false;
		if( couleur[nextp] == 0 && !dfs(nextp, (-1)*f) ) return false;
	}
	return true;
}

int main(){
	init();
	int n; cin >> n;
	for( int i = 0; i < n; ++i ){
		string s; cin >> s;
		for( int j = i; j < n; ++j ){
			if( s[j] == '1' ){
				g[i].push_back(j);
				g[j].push_back(i);
			}
		}
	}
	if( !dfs(0, 1) ){
		cout << -1 << endl;
		return 0;
	}
	int ans = 0;
	for( int s = 0; s < n; ++s ){
		queue<int> q;
		q.push(s);
		dist[s] = 1;
		while( !q.empty() ){
			int p = q.front(); q.pop();
			for( auto nextp : g[p] ){
				if( dist[nextp] == -1 ){
					q.push(nextp);
					dist[nextp] = dist[p]+1;
				}
			}
		}
		for( int i = 0; i < n; ++i ) ans = max(ans, dist[i]);
		init();
	}
	cout << ans << endl;
}