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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define rep3(i, s, n, a) for (int i = (s); i < (int)(n); i += a)
#define rep2(i, s, n) rep3(i, s, n, 1)
#define rep(i, n) rep2(i, 0, n)

using namespace std;
using ll = long long;
using ull = unsigned long long;
using P = pair<int, int>;

int main(){
    int n;
    cin >> n;
    vector<vector<int>> x(n), y(n);
    rep(i, n){
        int a;
        cin >> a;
        rep(j, a){
            int xx, yy;
            cin >> xx >> yy;
            x[i].push_back(xx);
            y[i].push_back(yy);
        }
    }
    int total = 1<<n, tmp=0, ans=0;
    bool flag=true;
    rep(i, total){
        rep(j, n){
            if((i>>j)&1){
                tmp++;
                int len = x[j].size();
                rep(k, len){
                    if(((i>>(x[j][k]-1))&1)!=y[j][k]){
                        flag = false;
                        break;
                    }
                }
                if(!flag){
                    break;
                }
            }
        }
        if(flag){
            // cout << i << endl;
            ans = max(ans, tmp);
        }
        flag = true;
        tmp = 0;
    }
    cout << ans << endl;
    return 0;
}