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 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <math.h>

typedef enum {false, true} bool;


bool trace(long long xy[][2], long long temp_or, long long temp_and, long long basis, int N, int num, int ctr, int index)
{
    static int i;
    
    
    if(num == ctr)
    {
        if((num > 1 && temp_or == temp_and) || (temp_or & temp_and) == basis)
            return true;
        else
            return false;
    }
    else
    {
        for(i=index; i<=N-(num-ctr); i++)
        {
            if(ctr == 0)
            {
                temp_or = xy[i][0];
                temp_and = xy[i][1];
                basis = 0;
            }
            if(trace(xy, temp_or | xy[i][0], temp_and & xy[i][1], basis | 1<<(N-i-1), N, num, ctr+1, i+1))
                return true;
        }
    }
    
    return false;
}

int main(void)
{
    char c;
    int N;
    int A, x, y;
    
    long long xy[15][2];
    
    int i, j;
    
    
    scanf("%d%c", &N, &c);
    
    
    for(i=0; i<N; i++)
    {
        scanf("%d%c", &A, &c);
        
        xy[i][0] = (long long)exp2l((long double)(N-i-1));
        xy[i][1] = (long long)exp2l((long double)N)-1;
        
        for(j=0; j<A; j++)
        {
            scanf("%d%c%d%c", &x, &c, &y, &c);
            
            if(y == 0)
                xy[i][1] -= (long long)exp2l((long double)(N-x));
            if(y == 1)
                xy[i][0] |= (long long)exp2l((long double)(N-x));
        }
    }
    
    for(i=N; i>0; i--)
    {
        if(trace(xy, 0, 0, 0, N, i, 0, 0))
            break;
    }
    
    
    printf("%d\n", i);
    
    
    //system("pause");
    
    return 0;
}