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<iostream>
#include<vector>
#include<utility>
#include<queue>
using namespace std;
int main(){
  int N;
  cin >> N;
  vector<vector<pair<int,int>>> ZAITHON(N);
  priority_queue<int> Q;
  for(int i=0;i<N;i++){
    int A;
    cin >> A;
    for(int j=0;j<A;j++){
      int x,y;
      cin >> x >> y;
      ZAITHON.at(i).push_back(make_pair(x,y));
    }
  }
  for(int bit=0;bit<(1<<N);bit++){
    int C=__builtin_popcount(bit);
    bool el=true;
    for(int i=0;i<N;i++){
      if(!(bit& (1<<i))) continue;
      
      for(int j=0;j<ZAITHON.at(i).size();j++){
        int x=ZAITHON.at(i).at(j).first;
        int y=ZAITHON.at(i).at(j).second;
        if(y==1 && !(bit >> (x-1))) el=false;
        if(y==0 &&  (bit >> (x-1))) el=false;
      }
    }
    if(el) Q.push(C);
  }
  
  cout << Q.top() << endl;
    
}