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>
#include <stdlib.h>
#include <string.h>
#include <math.h>

int main()
{
  int i,j,k;
  int N;
  scanf("%d",&N);
  
  int A[N+2];
  int x[N+2][N+2];
  int y[N+2][N+2];
  int true[N+2];
  int count;
  for(i=1;i<=N;i++){
    true[i] = 1;
  }
  
  for(i=1;i<=N;i++){
    scanf("%d",&A[i]);
    for(j=1;j<=A[i];j++){
      scanf("%d %d",&x[i][j],&y[i][j]);
    }
  }
  
  /*for(i=1;i<=N;i++){
    for(j=1;j<=A[i];j++){
      k = x[i][j];
      if( true[k] == y[i][j] ){
      }
      else{
        true[i] = 0;
      }
    }
  }*/
  
  for(i=1;i<=N;i++){
    if( true[i]==1 ){
      for(j=1;j<=A[i];j++){
        k = x[i][j];
        if( true[k] == y[i][j] ){
        }
        else{
          true[i] = 0;
          i = 1;
          break;
        }
      }
    }
  }
  
  count = 0;
  for(i=1;i<=N;i++){
    if( true[i]==1 ){
      count++;
    }
  }
  printf("%d\n",count);
  //printf("%d %d\n",x[1][1],y[1][1]);
 
  
  return 0; 
}