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<string.h>
#include<stdlib.h>
#include<stdbool.h>
int n,a[16],x[16][16],y[16][16],i,j;

int max(int a,int b){
  return (a>b)?a:b;
}

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

void solve(){
  scanf("%d",&n );
  for(i=0;i<n;i++){
    scanf("%d",&a[i] );
    for (j = 0; j < a[i]; j++) {
      scanf("%d %d",&x[i][j],&y[i][j]);
    }
  }
  int ans = 0;
  for(int bits=1;bits<(1<<n);bits++){
    bool ok=true;
    for (i = 1; i <= n; i++) {
      if(!(bits&(1<<(i-1)))) continue;
      for(j=1;j<=a[i];j++){
        if(((bits>>(x[i][j]-1))&1)^y[i][j]) ok=false;
      }
    }
    if(ok) ans=max(ans,counter(bits));
  }
  printf("%d\n",ans );
  return;
}

int main(){
  solve();
  return 0;
}