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 <stdlib.h>
//#include <string.h>
#include <stdbool.h>
//#include <limits.h>
//#include <math.h>

#define max(a,b) (a > b ? a : b)

int main() {
  int n;
  scanf("%d", &n);
  int x, y, a, b[n][n], ans = 0;
  for (int i=0; i<n; i++)
    for (int j=0; j<n; j++)
      b[i][j] = 2;
  for (int i=0; i<n; i++) {
    scanf("%d", &a);
    for (int j=0; j<a; j++) {
      scanf("%d %d", &x, &y);
      b[i][x-1] = y;
    }
  }
  for (unsigned int i=0; i<(unsigned int)(1 << n); i++) {
    bool c = false;
    for (int j=0; j<n; j++) {
      for (int k=0; k<n; k++) {
        if (j == k) continue;
        if (b[j][k] == 2) continue;
        if (((i>>j)&1) == 1) {
          if (b[j][k] != ((i>>k)&1)) {
            c = true;
            break;
          }
        }
      }
      if (c) break;
    }
    if (c) continue;
    int cnt = 0;
    for (int j=0; j<n; j++) {if (((i>>j)&1) == 1) cnt++;}
    ans = max(ans, cnt);
  }
  printf("%d\n", ans);
  return 0;
}