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 main()
{
  // 入力
  int N;
  cin >> N;
  vector<vector<pair<int, int>>> data(N);
  for (int i = 0; i < N; i++)
  {
    int A;
    cin >> A;
    for (int j = 0; j < A; j++)
    {
      int x, y;
      cin >> x >> y;
      x--;
      data.at(i).push_back(make_pair(x, y));
    }
  }

  int M = 0;
  // 正直者の組み合わせを全探索
  // O((2^N)N^2)のアルゴリズム
  for (int i = 0; i < (1 << N); i++)
  {
    bool ok = true;
    for (int j = 0; j < N; j++)
    {
      if (!(i & (1 << j)))
      {
        continue;
      }
      for (int k = 0; k < data.at(j).size(); k++)
      {
        if ((i & (1 << data.at(j).at(k).first)) ^ data.at(j).at(k).second)
        {
          ok = false;
        }
      }
    }
    if (ok)
    {
      M = max(M, __builtin_popcount(i));
    }
  }
  cout << M << endl;
}