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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
 
#define lli long long int
#define pb push_back
 
int m [100][100];
int q , n , a , b;
 
void backtracking (int i , int j , int *p , int pa , set <int> &p1 , set <int> &p2)
{
    if (i >= q)
    {
        if (pa < *p) *p = pa ;
 
        return;
    }
    if (j >= q) 
    {
        return ;
    }
 
    if(p1.find(i) == p1.end() && p2.find(j) == p2.end())
    {
        pa = pa + m[i][j];
        p1.insert(i);
        p2.insert(j);
        int jaux = 0 ;
 
        while(jaux < q)
        {
            backtracking(i + 1 , jaux , p , pa , p1 , p2);
            jaux++ ;
        }
 
        p1.erase(i);
        p2.erase(j);
        pa = pa - m[i][j];
    }
    
    backtracking(i , j + 1 , p , pa , p1 , p2);
 
}
int main()  
{
    cin >> q ;
   
    for (int i = 0 ; i < 100 ; i++)
    {
        for (int j = 0 ; j < 100 ; j++)
        {
            m[i][j] = 1;
        }
    }
 
    for (int i = 0 ; i < q ; i++)
    {
        cin >> n ;
 
        for (int j = 0 ; j < n ; j++)
        {
            cin >> a >> b ;
            m[i][a - 1] = b ;   
        }
    }
 
    int p = INT_MAX ;
    set <int> p1 ;
    set <int> p2 ;
 
    backtracking(0 , 0 , &p , 0 , p1 , p2);
 
    cout << p << endl ;
 
    return 0;  
}