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 (Clang 3.8.0) (judge verdict: Accepted)

#include <stdio.h>

int main(void){
    int N, A[20], B[20], x[20][20], y[20][20], i, j, k, p, t, n, X, M;
    scanf("%d", &N);
    for (i=1; i<=N; i++) {
        scanf("%d", &A[i]);
        for (j=1; j<=A[i]; j++) scanf("%d%d", &x[i][j], &y[i][j]);
    }
    p=1;
    for (i=1; i<=N; i++) p*=2;
    n=p;
    M=0;
    for (i=0; i<=n-1; i++) {
        for (j=0; j<=N-1; j++) {
            p=1;
            for (k=1; k<=j; k++) p*=2;
            B[N-j]=i/p%2;
        }
        
        /*
        printf("%d: ", i);
        for (j=1; j<=N; j++) printf("%d ", B[j]);
        printf("\n");
        */
        
        t=1;
        for (j=1; j<=N; j++) {
            if (B[j]==0) continue;
            for (k=1; k<=A[j]; k++) {
                // printf("(%d %d)\n", B[x[j][k]], y[j][k]);
                if ((B[x[j][k]] != y[j][k])) {
                    t=0;
                    break;
                }
            }
            if (t==0) break;
        }
        
        if (t==1) {
            X=0;
            for (j=1; j<=N; j++) X+=B[j];
            // printf("!%d: %d\n", i, X);
            if (X>M) M=X;
        }
    }
    printf("%d", M);
    return 0;
}