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 <string.h>
#include <stdbool.h>
#include <stdint.h>
#include <stdlib.h>
#include <limits.h>
#include <math.h>
int acs(const void *a, const void *b){return *(int*)a - *(int*)b;} /* 1,2,3,4.. */
int des(const void *a, const void *b){return *(int*)b - *(int*)a;} /* 8,7,6,5.. */
int cmp_char(const void *a, const void *b){return *(char*)a - *(char*)b;} /* a,b,c,d.. */
int cmp_str(const void *a, const void *b){return strcmp(*(const char **)a, *(const char **)b);} /* aaa,aab.. */
#define min(a,b) (a < b ? a: b)
#define max(a,b) (a > b ? a: b)
#define rep(i, l, r) for (int i = l; i < r; i++)
#define MAX 100001
#define MOD 1000000007

typedef long long int lli;

int counter(int x) {
  if(x == 0) return 0;
  return counter(x >> 1) + (x & 1);
}

int main(void) {
  int n;
  int xy[16][16][2];
  int a[16];
  scanf("%d", &n);
  rep(i, 0, n) {
    scanf("%d", &a[i]);
    rep(j, 0, a[i]) {
      scanf("%d %d", &xy[i][j][0], & xy[i][j][1]);
    }
  }
  
  // bit探索
  int ans = 0;
  rep(t, 0, 1 << n) {
    int ok = 1;
    rep(p, 0, n) {
      if ((t & (1 << p)) == 0) continue;
      rep(i, 0, a[p]) {
        if (((t >> (xy[p][i][0] - 1)) & 1) != xy[p][i][1]) 
          ok = 0;
      }
    }
    if (ok) ans = max(ans, counter(t));
  }
  printf("%d\n", ans);
  return 0;
}