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<math.h>

int xy[300][10],a[20],n;

void input(){
  scanf("%d",&n);
  for(int i=0;i<n;i++){
    scanf("%d",&a[i]);
    for(int j=0;j<a[i];j++){
      scanf("%d",&xy[i][j]);
    }
  }
}

void make_bit(int a[],int p){
  int q=pow(2,n-1);
  for(int i=0;i<n;i++){
    a[i]=(int)p/q;
    p%=q;
    q/=2;
  }
}

void solve(){
  //alp=all patern;
  int ans=0;
  int alp=pow(2,n);
  for(int i=0;i<alp;i++){
    int bits[n],ok=1;
    make_bit(bits,i);
    for(int j=0;j<n;j++){
      if(bits[j]==1){
        for(int k=0;k<a[j];k++){
          if(bits[xy[k][0]-1]!=xy[k][1]){
            ok=0;
          }
        }
      }
    }
    int sum=0;
    if(ok){
      for(int j=0;j<n;j++){
        sum+=bits[j];
      }
    }
    ans=ans<sum?sum:ans;
  }
  printf("%d\n",ans);
}

int main(){
  input();
  solve();
}