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(void){
    int n;

    cin >> n;
    int ans = 0;
    vector<vector<int>> y(n,vector<int>(n,2));

    for(int i=0; i<n; i++){
        int a;
        cin >> a;
        for(int j=0; j<a;j++){
            int x;
            cin >> x;
            cin >> y[i][x-1];
        }
    }

    int power = 1;
    for(int i=0; i<n; i++){
        power *= 2;
    }

    vector<int> doubt(n, 0);
    for(int a=0; a<power; a++){
        bool ok = true;
        int ok_ans = 0;
        int tmp = a;
        for(int i = 0; i<n; i++){
            if(tmp%2){
                doubt[i] = 1;
                ok_ans++;
            }else
                doubt[i] = 0;
            
            tmp/=2;
        }
        for(int i=0; i<n; i++){
            if(doubt[i]==0)
                continue;
            for(int j=0; j<n; j++){
                if(y[i][j] != 2 && y[i][j] != doubt[j]){
                    ok = false;
                    i=n;
                    break;
                }
            }
        }
        if(ok){
            ans = max(ans, ok_ans);
        }
    }

    cout << ans << endl;

}