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 int long long
using namespace std;

signed main() {
  int N;
  cin >> N;
  
  vector<vector<int>> hito(N, vector<int>(N));
  for( int i = 0; i < N; i++ ){
    for( int j = 0; j < N; j++ ){
      hito.at(i).at(j) = -1;
    }
  }
  
  int A, X, Y;
  for( int i = 0; i < N; i++ ){
    cin >> A;
    for( int j = 0; j < A; j++ ){
      cin >> X >> Y;
      hito.at(i).at(X-1) = Y;
    }
  }
  
  int ans = 0;
  for (int bit = 0; bit < (1<<N); ++bit){
    int cnt = 0;
    for (int i = 0; i < N; ++i) {
      if (bit & (1<<i)) { // i が bit に入るかどうか
        bool fg = true;
        for( int j  = 0; j < N; j++ ){
          //if( hito.at(i).at(j) == 1 && hito.at(j).at(i) == 0 ) fg = false;
          if (bit & (1<<j)){
            if( hito.at(i).at(j) == 0 ) fg = false;
            if( hito.at(j).at(i) == 0 ) fg = false;
          }
        }
        if( fg == true ) cnt++;
      }
    }
    ans = max( ans, cnt );
  }
  cout << ans << endl;

}