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 (Clang 3.8.0) (judge verdict: Accepted)

#include <stdio.h>
#include <stdint.h>

int n;
int a[15];
int x[15][15];
int y[15][15];

int check(char *testcase) {
  int count = 0;
  for (int i = 0; i < n; i++) {
    if (testcase[i]) {
      for (int j = 0; j < a[i]; j++) {
        int p = x[i][j];
        if (testcase[p] != y[i][j]) return 0;
      }
      count++;
    }
  }
  return count;
}

int main() {
  scanf("%d", &n);
  for (int i=0; i<n; i++) {
    int k = 0;
    scanf("%d", &k);
    a[i] = k;
    for (int j=0; j<k; j++) {
      scanf("%d %d", &(x[i][j]), &(y[i][j]));
      x[i][j]--;
    }
  }

  int len = 1 << n, max = 0;
  for (int i = 0; i < len; i++) {
    char testcase[15];
    int l = i;
    for (int j = 0; j < n; j++) {
      testcase[j] = (l & 1); l >>= 1;
      //printf("%d ", testcase[j]);                                             
    }
    int r = check(testcase);
    //printf(" -> %d\n", r);                                                    
    if (max < r) max = r;
  }
  printf("%d\n", max);
  return 0;
}