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>

using namespace std;

#define ll long long
#define F first
#define S second
#define ii pair < int , int >

const int N=15;
vector<ii> v[N];
int a[N];
int ans=0;
bool vis[N];
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=0; i<n; i++)
    {
        int sz;
        scanf("%d",&sz);
        for(int j=0; j<sz; j++)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            ii z;
            z.F=x-1;
            z.S=y;
            v[i].push_back(z);
        }
    }
    int ans=0;
    for(int msk=0; msk<pow(2,n); msk++)
    {
        bool q=1;
        for(int i=0; i<n; i++)
        {

            if((1<<i) & msk)
                for(auto x:v[i])
                {
                    if(x.S==1 && ((1<<x.F)&msk)==0){
                        q=0;
                        //cout<<msk<<" "<<i<<endl;
                    }
                    if(x.S==0 && (1<<x.F & msk))
                        q=0;
                }
            if(!q)
                break;
        }
        if(!q)
            continue;
        ans=max(ans,__builtin_popcount(msk));
    }
cout<<ans;

}