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>
#include<math.h>

int main(void)
{
	int i,j,n;
	scanf("%d",&n);
	int k;
	
	int num;
    int n_test;
    int person, honest;
    int sum=0;
    int max=0;

	num = (int)pow(2,n);
	
	int a[num][n];
    int judge[num];

    for(i=0;i<num;i++)
        judge[i]=1;

	for(i=0;i<num;i++){
        for(j=0;j<n;j++){
          	if(i%(int)pow(2,j+1)>=pow(2,j))
	        	a[i][n-1-j]=1;
          	else
              	a[i][n-1-j]=0;
		}
	}

for(i=0;i<n;i++){
    scanf("%d",&n_test);

    for(j=0;j<n_test;j++){
        scanf("%d%d",&person,&honest);

        for(k=0;k<num;k++){
            if(a[k][i]==1){
                if(a[k][person-1]!=honest)
                    judge[k]=0;
            }
        }
    }
}

for(i=0;i<num;i++){
    if(judge[i]==1){
        for(j=0;j<n;j++)
            sum+=a[i][j];
        if(sum>max)
            max = sum;
            sum =0;
    }
}

printf("%d",max);

	
	return 0;
}