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<(int)(n); i++)
#define REP(i,m,n) for(ll i=(ll)(m);i<(ll)(n);i++)
using namespace std;
using ll = long long;
using P = pair<int, int>;

int main() {
  int n; cin >> n;
  vector<int> a(n);
  vector<vector<P>> man(n, vector<P>());
  rep(i,n) {
    cin >> a[i];
    rep(j, a[i]) {
      int x, y; cin >> x >> y;
      man[i].push_back(make_pair(x-1, y));
    }
  }

  int maxt = 0;
  for (int b = 1; b<(1<<n); b++) {
    bool flag = true;
    int cnt = 0;
    rep(i,n){
      if (!(1 & (b>>i))) continue;
      rep(j,a[i]) {
        int x = man[i][j].first;
        int y = man[i][j].second;
        if ((1 & (b>>x)) != y) { flag = false; break; }
        cnt++;
      }
      if (!flag) break;
    }
    if (flag) maxt = max(cnt, maxt);
  }
  cout << maxt << endl;
}