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: Accepted)

#include <bits/stdc++.h>
using namespace std;
 
int main() {
  int n,kanou,ans;
  ans=0;
  cin>>n;
  vector<int> itirann(n);
  vector<vector<pair<int,int>>> vec(n,vector<pair<int,int>>(n));
  for(int i=0; i<n; i++){
    cin>>itirann.at(i);
    for(int j=0; j<itirann.at(i); j++){
      cin>>vec.at(i).at(j).first;
      cin>>vec.at(i).at(j).second;
    }
  }
  for (int bit = 0; bit < (1<<n); ++bit){
    kanou=1;
    for(int i=0; i<n; i++){
      //hitosyuukai
      if(bit & (1<<i) ){//syouzikikadouka
        for(int j=0; j<itirann.at(i); j++){//kyogihantei
          if(vec.at(i).at(j).second==1){
            if(!(bit & (1<<vec.at(i).at(j).first-1) ) ){
              kanou=0;
              
            }
          }
          else{
            if(bit & (1<<vec.at(i).at(j).first-1) ){
              kanou=0;
            }
          }
        }
      }
    }
    if(kanou==1){
      
      int k=0;
      int m;
      m=bit;
      while(m>0){
        k=k+m%2;
        m=m/2;
        
      }
      
      ans=max(ans,k);
    }
  }
  cout<<ans;
}