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;

const int N = 1000;
const int inf = 1000 * 1000 * 1000 + 7;

int n;
string st[N];
vector <int> pesar[N];
bool mark[N];
int dis[N], KeepMax[N];

void input() {
    cin >> n;
    for (int i = 0; i < n; i++)
        cin >> st[i];

    for (int i = 0; i < n; i++)
        dis[i] = inf;

    return;
}

void BFS() {
    queue <int> Q;

    Q.push(0) ;
    mark[0] = true;
    dis[0] = 0;
    while (!Q.empty()) {
        int v = Q.front() ;
        Q.pop() ;

        for (int i = 0; i < n; i++)
            if (st[v][i] == '1')
                if (!mark[i]) {
                    mark[i] = true;
                    dis[i] = dis[v] + 1;
                    pesar[v].push_back(i) ;
                    Q.push(i) ;
                }
                else if (dis[i] % 2 == dis[v] % 2) {
                    cout << -1 << endl;
                    exit(0) ;
                }
    }
    return;
}

int DFS(int v) {
    if (pesar[v].empty())
        return dis[v];

    int res = -1;
    for (int i = 0; i < pesar[v].size(); i++)
        res = max(DFS(pesar[v][i]), res) ;

    return res;
}

int main() {
    input() ;
    BFS() ;

    int ans = 0;
    for (int i = 0; i < pesar[0].size() ; i++)
        ans += DFS(pesar[0][i]) ;

    if (pesar[0].size() == 1)
        ans++;

    cout << ans << endl;
    return 0;
}