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 main(){
    int n;
    scanf("%d",&n);
    int a[n];
    int x[n][n], y[n][n];
    for(int i = 0; i < n; i++){
        scanf("%d",&a[i]);
        for(int j = 0; j < a[i]; j++)
            scanf("%d%d",&x[i][j],&y[i][j]);
    }
    int ans = 0;
    for(int bit = 1; bit < (1 << n); bit++){
        int flag = 1;
        int count = 0;
        for(int i = 0; i < n && flag; i++){
            if(bit & (1 << i)){
                count ++;
                for(int j = 0; j < a[i] && flag; j++){
                    if((bit & (1 << (x[i][j] - 1))) != (y[i][j] << (x[i][j] - 1)))
                        flag = 0;
                }
            }
        }
        if(flag && ans < count)
            ans = count;
    }
    printf("%d\n",ans);
    return 0;
}