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 <algorithm>
#include <cmath>
#include <iomanip>
#include <iostream>
#include <vector>
#define rep(i, n) for (int i = 0; i < (n); i++)
typedef long long ll;
using namespace std;

int main(void) {
  ios::sync_with_stdio(false);
  int N;
  cin >> N;
  vector<int> A(N);
  vector<vector<int>> X(N + 1);
  vector<vector<int>> Y(N + 1);
  int loop = 1;
  rep(i, N) {
    cin >> A[i];
    rep(j, A[i]) {
      int x, y;
      cin >> x >> y;
      x--;
      X[i].push_back(x);
      Y[i].push_back(y);
    }
    loop *= 2;
  }

  int ans = 0;
  for (int i = loop - 1; i >= 0; i--) {
    vector<int> TrueFalse(N, -1);
    bool flag = true;
    rep(j, N) {
      bool Honest = i & (1 << j);
      if (TrueFalse[j] != -1) {
        if (TrueFalse[j] == 0 && Honest) {
          flag = false;
          break;
        } else if (TrueFalse[j] == 1 && !Honest) {
          flag = false;
          break;
        }
      }

      // cout << "j=" << j << " Honest=" << Honest << endl;
      // rep(k, N) { cout << "TrueFalse=" << TrueFalse[k] << endl; }
      if (Honest) {
        rep(k, A[j]) {
          if (TrueFalse[X[j][k]] != -1 && TrueFalse[X[j][k]] != Y[j][k]) {
            flag = false;
            break;
          }
          TrueFalse[X[j][k]] = Y[j][k];
        }
      } else {
        rep(k, A[j]) {
          if (TrueFalse[X[j][k]] != -1 && TrueFalse[X[j][k]] == Y[j][k]) {
            flag = false;
            break;
          }
          TrueFalse[X[j][k]] = (Y[j][k]) ? 0 : 1;
        }
      }
      if (!flag) break;
    }
    if (flag) {
      int tmp = 0;
      for (int j = 0; j < N; j++) {
        if (i & (1 << j)) tmp++;
      }
      ans = max(ans, tmp);
    }
  }
  cout << ans << endl;
  return 0;
}