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>
using namespace std;

#define rep(i, n) for(int i = 0; i < (n); ++i)

typedef long long ll;
typedef pair<int, int> P;

const int MAX = 1e5 + 5;
const int INF = 1001001001;
const int MOD = 1e9 + 7;

int main(){
    int N;
    cin >> N;

    vector<vector<P>> A;
    rep(i, N) {
        int a;
        cin >> a;
        vector<P> v;
        rep(j, a) {
            int x, y;
            cin >> x >> y;
            x--;
            v.push_back(P(x, y));
        }
        A.push_back(v);
    }

    int ans = 0;
    for (int i = 0; i < (1 << N); ++i) {
        bool ok = true;
        rep(j, N) {
            if ((i >> j) & 1) {
                for (auto p : A[j]) {
                    if (p.second != (i >> p.first & 1)) ok = false;
                }
            }
        }
        if (ok) {
            int res = 0;
            int X = i;
            while (X > 0) {
                res += X % 2;
                X /= 2;
            }
            ans = max(ans, res);
        }
    }

    cout << ans << endl;
}