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>
#define rep(i,n) for(int (i)=0;(i)<(n);(i)++)
#define all(x) (x).begin(),(x).end()
#define INF 10000000
using ll = long long ;
ll gcd(ll a, ll b) {
    if(b == 0) return a;
    return gcd(b, a % b);
}
ll lcm(ll a,ll b){
    ll g = gcd(a,b);
    return a / g * b;
}
template< int id > struct CompTupleBy
{
  template<class T> 
  bool operator()( const T& a, const T& b ) const {
    return std::get<id>(a) < std::get<id>(b);
  }
};

using namespace std;

int counter(int x){
  if (x==0) return 0;
  return counter(x>>1)+(x&1);
}
int main(){
  cin.tie(0);
    ios::sync_with_stdio(false);

    int n;
    int x[20][20];
    int y[20][20];
    int A[20];
    cin >> n;
  rep(i,n){
    cin >> A[i];
    rep(j,A[i]){
      cin >> x[i][j] >> y[i][j];
    }
  int ans = 0;
  for(int bits = 1; bits < (1 << n); bits++) {
        bool ok = true;
        for(int i = 1; i <= n; i++) {
            if(!(bits & (1 << (i-1)))) continue;
            for(int j = 1; j <= A[i]; j++) {
                if(((bits >> (x[i][j]-1)) & 1) ^ y[i][j]) ok = false;
            }
        }
        if(ok) ans = max(ans, counter(bits));
    }
  cout << ans << endl;

  }

        
}