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>

int main(void)
{
    char c;
    int N;
    int A, x, y, xy[16][16];
    int i, j;
    int ctr = 0;
    
    
    scanf("%d%c", &N, &c);
    
    
    for(i=1; i<=N; i++)
    {
        for(j=1; j<=N; j++)
        {
            xy[i][j] = -1;
        }
    }
    
    for(i=1; i<=N; i++)
    {
        scanf("%d%c", &A, &c);
        
        for(j=1; j<=A; j++)
        {
            scanf("%d%c%d%c", &x, &c, &y, &c);
            
            xy[i][x] = y;
        }
    }
    
    
    for(i=1; i<=N; i++)
    {
        for(j=i+1; j<=N; j++)
        {
            if(xy[i][j] == xy[j][i] && xy[i][j] == 1)
            {
                ctr++;
            }
        }
    }
    
    
    if(ctr == 1)
        ctr++;
    else if(ctr == 0)
    {
        for(i=1; i<=N; i++)
        {
            for(j=i+1; j<=N; j++)
            {
                if(xy[i][j] == xy[j][i] && xy[i][j] == 0)
                {
                    ctr++;
                }
            }
        }
        
        if(ctr == (N-1)*N/2)
            ctr = 1;
        else
            ctr = 0;
    }
    
    
    printf("%d\n", ctr);
    
    
    //system("pause");
    
    return 0;
}