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: Wrong Answer)

#include <stdio.h>
int main(void)
{
  int N;
  int syougen[16][16];
  int i,j,search;
  int answer = 0;
  scanf("%d",&N);
  
  for(i=1;i<N;i++)
  {
    int A;
    scanf("%d",&A);
    for(j=0;j<A;j++)
    {
      int x,y;
      scanf("%d %d",&x,&y);
      syougen[i][x-1] = (y==1 ? 1:-1);
    }
  }
  
  for(search=0;search<(1<<N);search++)
  {
    int score = 0;
    int ok =1;
    for(i=0;ok && i<N;i++)
    {
      if((search >> i)&1) 
      {
        score++;
        for(j=0;ok && j<N;j++)
        {
          if(syougen[i][j]!=0)
          {
            if(((search>>j)&1) !=(syougen[i][j]==1))
              ok = 0;
          }
        }
      }
    }
    if(ok && score > answer)  answer=score;
  }
  printf("%d\n",answer);
   
  return 0;
}