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 <iostream>
#include <vector>
#include <string>
#include <queue>
#include <algorithm>

using namespace std;

class Unionfind {
    vector<int> par; //親
    vector<int> rank; //深さ
public:
    void init(int n)//初期化
    {
        par = vector<int>(n);
        rank = vector<int>(n);
        for (int i = 0; i < n; i++) {
            par[i] = i;
            rank[i] = 0;
        }
    }

    int root(int x)//根を求める
    {
        if (par[x] == x) return x;
        else return par[x] = root(par[x]);
    }

    void unite(int x, int y)//集合を併合
    {
        x = root(x);
        y = root(y);
        if (x == y) return;
        else {
            if (rank[x] < rank[y]) par[x] = y;
            else {
                par[y] = x;
                if (rank[x] == rank[y]) rank[x]++;
            }
        }
    }

    bool issame(int x, int y)//同じ集合に属しているか
    {
        return root(x) == root(y);
    }
};

int main() {
    int n;
    cin >> n;
    vector<vector<int>> graph(n);
    Unionfind uf;
    uf.init(n);
    for (int i = 0; i < n; i++) {
        string s;
        cin >> s;
        for (int j = 0; j < n; j++) {
            if (s[j] == '1') {
                graph[i].push_back(j);
                if (i != 0 && j != 0)uf.unite(i, j);
            }
        }
    }

    bool possible = true;
    vector<int> dis(n, -1);
    dis[0] = 1;
    queue<vector<int>> que;
    que.push({0, 1});
    while (!que.empty() && possible) {
        auto now = que.front();
        que.pop();

        if (dis[now[0]] >= 0) {
            if (dis[now[0]] % 2 != now[1] % 2)possible = false;
            dis[now[0]] = min(dis[now[0]], now[1]);
        } else {
            dis[now[0]] = now[1];
        }

        for (auto nextv:graph[now[0]]) {
            if (dis[nextv] < 0 || dis[nextv] > now[1] + 1) que.push({nextv, now[1] + 1});
        }
    }

    vector<int> ret(n + 10);
    for (int i = 1; i < n; i++) {
        int x = uf.root(i);
        ret[x] = max(ret[x], dis[i]);
    }
    sort(ret.begin(), ret.end(), greater<int>());

    if (ret[1] > 0)ret[1]--;
    cout << (possible ? ret[0] + ret[1] : -1) << endl;
    return 0;
}