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: Wrong Answer)

#include <algorithm>
#include <bitset>
#include <cmath>
#include <complex>
#include <deque>
#include <functional>
#include <iomanip>
#include <iostream>
#include <iterator>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>

typedef long long ll;

#define rep(i, n) for (ll i = 0; i < (ll)n; ++i)
#define INF 10e12
#define MOD 1000000000 + 7
#define MAX 51000
#define all(x) (x).begin(), (x).end()
#define MX(x) *max_element(all(x))
#define MN(x) *min_element(all(x))

using namespace std;

vector<ll> kumiawase(ll s, ll N)
{
    vector<ll> ans;
    rep(i, N)
    {
        if (s & (1 << i))
            ans.push_back(i);
    }
    return ans;
}
int main(void)
{
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(false);
    ll N;
    cin >> N;
    vector<vector<pair<ll, ll>>> vec(N);
    rep(i, N)
    {
        ll k;
        cin >> k;
        rep(j, k)
        {
            ll a, b;
            cin >> a >> b;
            vec[i].push_back({a, b});
        }
    }
    //全探索
    ll ans = 0;
    for (ll i = 1; i < (1 << N); ++i)
    {
        bool flag = true;
        vector<ll> state(N, -1);
        vector<ll> tmp = kumiawase(i, N);
        ll tadasii = tmp.size();
        ll fusinsetsu = N - tadasii;
        for (auto i : tmp)
        {
            for (auto j : vec[i])
            {
                if (state[j.first] == -1) //まだ判定されてない場合
                    state[j.first] = j.second;
                else if (state[j.first] != j.second)
                    flag = false;
            }
        }
        if (flag == true)
            ans = tadasii;
    }
    cout << ans << endl;
}