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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
int ans;
int n;
struct T{
  int nxy;
  int x[19],y[19];
}t[19];
int gg(int x){
  int ans=0;
  while(x){
    ans+=(x&1);
    x>>=1;
  }
  return ans;
}
bool check(int x){
  for(int i=1;i<=n;i++)
    for(int j=1;j<=t[i].nxy;j++){
      bool flag1=((x>>(i-1))&1);
      bool flag2=((x>>(t[i].x[j]-1))&1);
      if(flag1^(t[i].y[j]==flag2))return 0;
    }
  return 1;
}
int main(){
  cin>>n;
  for(int i=1;i<=n;i++){
    cin>>t[i].nxy;
    for(int j=1;j<=t[i].nxy;j++)cin>>t[i].x[j]>>t[i].y[j];
  }
  for(int i=0;i<(1<<n);i++)if(check(i))ans=max(ans,gg(i));
  cout<<ans<<endl;
  return 0;
}