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: Accepted)

#include<stdio.h>

int power(int x, int y) {
    int sum = 1;
    for (int i = 0; i < y; i++) sum *= x;
    return sum;
}

int main() {
    int A[15], N, x[15][14], y[15][14], i, j, k, which[15], honests, ans=0, success;

    scanf("%d", &N);

    for (i = 0; i < N; i++) {
        scanf("%d", A+i);
        for (j = 0; j < A[i]; j++) {
            scanf("%d %d", x[i]+j, y[i]+j);
            x[i][j]--;
        }
    }

    for (k = 0; k < power(2, N); k++) {
        honests = 0;
        for (i = 0; i < N; i++) {
            which[i] = (k>>i) & 1;
            honests += which[i];
        }
        success = 1;
        for (i = 0; i < N; i++) {
            if (which[i]) {
                for (j = 0; j < A[i]; j++) {
                    if (which[x[i][j]] != y[i][j]) success = 0;
                }
            }
        }
        if (success && ans < honests) ans = honests;
    }

    printf("%d", ans);
}