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()
{
  int nn, aa, xx, yy, flag, count, ans = 0;
  int table[15][15] = {0};	//table[a][b]はa→bの証言	真偽不明は0だから、0で初期化
  scanf("%d", &nn);
  for(int ii = 0;ii < nn;ii++)
  {
    scanf("%d", &aa);
    {
      for(int jj = 0;jj < aa;jj++)
      {
        scanf("%d %d", &xx, &yy);
        table[ii][xx - 1] = yy;
      }
    }
  }
  
  for(int bit = 1;bit < (1 << nn);bit++)	//正直者の組み合わせ	2進数各桁について1で正直、0で不親切
  {
    flag = 1;
    count = 0;
    for(int ii = 0;ii < nn;ii++)
    {
      if(bit & (1 << ii))
      {
        count++;
        for(int jj = 0;jj < nn && flag;jj++)
        {
          if(ii == jj)
          {
            jj++;	//ii→iiへの評価の検証は意味がない
            count--;
          }
          if(table[ii][jj] && !table[jj][ii] && bit & (1 << jj))
          {
            flag = 0;
          }
        }
      }
    }
    if(flag && ans < count)
    {
      ans = count;
      printf("%d, %d\n\n", bit, count);
    }
  }
  
  printf("%d\n", ans);
  return 0;
}