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;

int main() {
    int N;
    cin >> N;
    vector<vector<pair<int, int>>> A(N);
    for (int i = 0; i < N; i++) {
        int K; cin >> K;
        for (int j = 0; j < K; j++) {
            int x, y; cin >> x >> y; x--;
            A[i].emplace_back(x, y);
        }
    }
    
    int ans = 0;
    for (int bit = 0; bit < (1<<N); bit++) {
        int res = 0;
        vector<bool> say(N, false);
        for (int i = 0; i < N; i++) {
            if (bit & (1<<i)) { say[i] = true; res++; }
        }
        
        bool ok = true;
        for (int i = 0; i < N; i++) {
            if (say[i] == false) continue;
            for (int j = 0; j < A[i].size(); j++) {
                int x, y; tie(x, y) = A[i][j];
                if (say[x] != y) ok = false;
            }
        }
        
        if (ok == true) ans = max(ans, res);
    }
    
    cout << ans << endl;
    return 0;
}