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>

#define N 15

int main(){
  const int n, a[N], x[N][N-1], y[N][N-1];
  scanf("%d",&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]);
//      printf("%d %d\n",y[i][j], x[i][j]);
    }
  }
  
  int isHonest[n], sumHonest, maxSumHonest=0, isContra=0;
  for(int k=0; k<pow(2,n); k++){
    isContra=0;
    for(int i=0; i<n; i++){
      isHonest[i] = (k>>i)&1;   // decide whether person i is honest or not
    }
    for(int i=0; i<n; i++){
      if(isHonest[i]==1){
//        for(int l=0; l<n; l++) { printf("%d ",isHonest[l]); } printf("\n");
        for(int j=0; j<a[i]; j++){
//          printf("%d %d\n",y[i][j], x[i][j]);
          if(y[i][j]!=isHonest[x[i][j]-1]){  // if 'honest i' isn't honest 
            isContra=1;
//            printf("%d ",isContra);
//            for(int l=0; l<n; l++) { printf("%d ",isHonest[l]); } printf("\n");
            break;
          }
        }
        if(isContra==1){
//          for(int l=0; l<n; l++) { printf("%d ",isHonest[l]); } printf("\n");
          break;
        }
      }
    }
    if(isContra==0){
//      for(int l=0; l<n; l++) { printf("%d ",isHonest[l]); } printf("\n");
      sumHonest=0;
      for(int i=0; i<n; i++){
        sumHonest += isHonest[i];
        if(sumHonest>maxSumHonest){
          maxSumHonest = sumHonest;
        }
      } 
    }
  }
  printf("%d",maxSumHonest);
  
  return 0;
}