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 Python (3.4.3) (judge verdict: Runtime Error)

#include<bits/stdc++.h>
using namespace std;using ll=long long;using vi=vector<int>;using vvi=vector<vi>;using vl=vector<ll>;using vvl=vector<vl>;using P=pair<int,int>;using PL=pair<ll,ll>;using vp=vector<P>;using vpl=vector<PL>;
template<typename T>constexpr auto inf=numeric_limits<T>::max()/2;constexpr int INF=inf<int>,MOD=1000000007;constexpr ll LINF=inf<ll>;
#define _ol3(_1,_2,_3,name,...)name
#define _rep(i,n)_repi(i,0,n)
#define _repi(i,a,b)for(int i=a,i##_l=(b);i<i##_l;++i)
#define REP(...)_ol3(__VA_ARGS__,_repi,_rep,)(__VA_ARGS__)
#define REPR(i,n)for(int i=n-1;i>=0;--i)
#define REPA(i,v)REP(i,(v).size())
#define all(v)(v).begin(),(v).end()
#define rall(v)(v).rbegin(),(v).rend()
#define bit(n)(1ll<<(n))
#define F first
#define S second
#define endl '\n'
#define cho(n,a,b)cout<<((n)?a:b)<<endl
void YES(int n){cho(n,"YES","NO");}void Yes(int n){cho(n,"Yes","No");}void Poss(int n){cho(n,"Possible","Impossible");}
void _print(ostream&){}template<class T,class...U>void _print(ostream&s,const T&t,const U&...u){s<<t<<(sizeof...(u)?' ':'\n');_print(s,u...);}
template<class...T>void print(const T&...t){_print(cout,t...);}template<class...T>void dprint(const T&...t){_print(cerr,t...);}
#ifndef LOCAL
struct osd{template<class T>osd&operator<<(const T&t){return*this;}};osd cer_;
#define dprint(...)
#define cerr cer_
#endif
template<class T>bool chmax(T&a,const T&b){bool x=a<b;x?a=b:b;return x;}template<class T>bool chmin(T&a,const T&b){bool x=a>b;x?a=b:b;return x;}
template<class T,class U>ostream&operator<<(ostream&o,const pair<T,U>&p){return o<<p.F<<' '<<p.S;}
template<class T,class U>istream&operator>>(istream&i,pair<T,U>&p){return i>>p.F>>p.S;}
template<class T>class iterable{static false_type c(string v);template<class U>static auto c(U v)->decltype(all(v),true_type());static false_type c(...);public:const static bool value=decltype(c(declval<T>()))::value;};
template<class T,typename=enable_if_t<iterable<T>::value>>ostream&operator<<(ostream&o,const T&v){for(auto&&i:v)o<<i<<' ';return o;}
template<class T>istream&operator>>(istream&i,vector<T>&v){for(T&j:v)i>>j;return i;}
template<class T>vector<T>&operator<<(vector<T>&v,const T&t){v.push_back(t);return v;}
template<class T>set<T>&operator<<(set<T>&v,const T&t){v.insert(t);return v;}
template<class T>multiset<T>&operator<<(multiset<T>&v,const T&t){v.insert(t);return v;}

// 解説を読んだ.がんばってとけなかったのになんでレートが落ちたのか考えると悔しさのようなものを感じている
// いや天才解法じゃん?なんでRecommendationsでProabilityが58%なんですか,キレている

int main(){
    cin.tie(0);ios::sync_with_stdio(0);
    cout<<setprecision(10);
    int n;
    cin>>n;
    vector<set<int>> es(n);
    vvi g(n, vi(n,INF));
    REP(i,n){
        string s;
        cin>>s;
        REP(j,n){
            g[i][j]=s[j]-'0';
            if(s[j]=='1'){
                es[i].insert(j);
            }
        }
    }
    REP(i,n)dprint(es[i]);
    vi c(n, -1);
    queue<int> q;
    q.push(0);
    c[0]=0;
    while(q.size()){
        dprint(c);
        const int x=q.front();
        q.pop();
        for(int a:es[x]){
            if(c[a]!=-1){
                if(c[a]==c[x]){
                    cout<<-1<<endl;
                    return 0;
                }
                continue;
            }
            c[a]=1-c[x];
            q.push(a);
        }
    }
    int ma=0;
    REP(k,n){
        REP(i,n){
            REP(j,n){
                chmin(g[i][j], g[i][k]+g[k][j]);
            }
        }
    }
    REP(i,n){
        dprint(g[i]);
        REP(j,n)if(g[i][j]!=INF)chmax(ma, g[i][j]);
    }
    cout<<ma+1<<endl;
}