Problem:
Score : 300 points

Problem Statement

There are N people numbered 1 to N. Each of them is either an honest person whose testimonies are always correct or an unkind person whose testimonies may be correct or not.

Person i gives A_i testimonies. The j-th testimony by Person i is represented by two integers x_{ij} and y_{ij}. If y_{ij} = 1, the testimony says Person x_{ij} is honest; if y_{ij} = 0, it says Person x_{ij} is unkind.

How many honest persons can be among those N people at most?

Constraints

All values in input are integers.

1 \leq N \leq 15

0 \leq A_i \leq N - 1

1 \leq x_{ij} \leq N

x_{ij} \neq i

x_{ij_1} \neq x_{ij_2} (j_1 \neq j_2)

y_{ij} = 0, 1

Input

Input is given from Standard Input in the following format:

N
A_1
x_{11} y_{11}
x_{12} y_{12}
:
x_{1A_1} y_{1A_1}
A_2
x_{21} y_{21}
x_{22} y_{22}
:
x_{2A_2} y_{2A_2}
:
A_N
x_{N1} y_{N1}
x_{N2} y_{N2}
:
x_{NA_N} y_{NA_N}

Output

Print the maximum possible number of honest persons among the N people.

Sample Input 1

3
1
2 1
1
1 1
1
2 0

Sample Output 1

2

If Person 1 and Person 2 are honest and Person 3 is unkind, we have two honest persons without inconsistencies, which is the maximum possible number of honest persons.

Sample Input 2

3
2
2 1
3 0
2
3 1
1 0
2
1 1
2 0

Sample Output 2

0

Assuming that one or more of them are honest immediately leads to a contradiction.

Sample Input 3

2
1
2 0
1
1 0

Sample Output 3

1

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#include <unordered_map>

using namespace std;
typedef long long int ll;
typedef long double ld;
#define rep(i, n) for(ll i = 0; i < (ll)(n); i++)
#define repr(i, n) for (ll i = (ll)(n) - 1; i >= 0; i--)
#define all(x) (x).begin(),(x).end()
#define MOD 1000000007
#define INF (1LL << 62)

#define PI (acos(-1))
#define print(x) cout << x << endl
 
ll gcd(ll a, ll b) { return b ? gcd(b,a%b) : a;}
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
ll countDigit(ll n) { return floor(log10(n) + 1); } 
typedef pair <ll,ll> P;
static const ll dy[4] = {0,1,0,-1}, dx[4] = {1,0,-1,0};

template <typename T> 
ostream& operator<<(ostream& os, const vector<T>& v) 
{ 
    for (int i = 0; i < (int)v.size(); ++i) { 
        os << v[i]; 
        if (i != (int)v.size() - 1) 
            os << " "; 
    } 
    return os; 
} 

ll nCrModp(ll n, ll r, ll p)
{
    ll C[r+1];
    memset(C, 0, sizeof(C));

    C[0] = 1;

    for (ll i = 1; i <= n; i++)
    {
        for (ll j = min(i, r); j > 0; j--)
            C[j] = (C[j] + C[j-1])%p;
    }
    return C[r];
}

vector<ll> divisor(ll n) {
  vector<ll> ret;
  for(ll i = 1; i * i <= n; i++) {
    if(n % i == 0) {
      ret.push_back(i);
      if(i * i != n) ret.push_back(n / i);
    }
  }
  sort(begin(ret), end(ret));
  return (ret);
}

int main()
{
    ll N;
    cin >> N;
    vector<vector<P>> v(N, vector<P>());
    rep(i, N) {
        ll A;
        cin >> A;
        rep(j, A) {
            ll x, y;
            cin >> x >> y;
            v[i].push_back(P(x-1, y));
        }
    }
    ll ma = 0;
    rep(i, 1 << N) {
        vector<ll> tmp(N, 0);
        ll m = 0;
        rep(j, N) {
            tmp[j] = (i & (1 << j)) > 0;
            m += tmp[j];
        }
        bool flag = false;
        rep(j, N) {
            if (tmp[j] == 0) continue;
            for(P p : v[j]) {
                if(tmp[p.first] != p.second) flag = true;
            }
        }
        if (!flag) {
            ma = max(ma, m);
        }
    }
    print(ma);
    return 0;
}