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;

using N = pair<int, int>;
using Data = vector<vector<N>>;

int main() {
  int n;
  cin>>n;
  Data data(n);
  for(int i=0; i<n; i++) {
    int a;
    cin>>a;
    for(int j=0; j<a; j++) {
      N claim;
      cin>>claim.first>>claim.second;
      data[i].push_back(claim);
    }
  }

  int ans = 0;
  for(int bit=0; bit<(1<<n); bit++) {
    bool ok = true;
    int num = __builtin_popcount(bit);
    for(int i=0; i<n; i++) {
      if(((bit>>i)&1)==1) {  //人i+1が正直者と仮定するなら、そいつの言い分は検証する必要がある

        for(N check : data[i]) {
          if(((bit>>(check.first-1))&1) != check.second) ok=false;
        }
      }
    }
    if(ok && num>ans) ans=num;
  }

  cout<<ans<<endl;

  return 0;
}