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: Accepted)

#include <algorithm>
#include <bitset>
#include <cmath>
#include <ctime>
#include <iostream>
#include <list>
#include <map>
#include <math.h>
#include <numeric>
#include <queue>
#include <set>
#include <stack>
#include <stdio.h>
#include <string>
#include <tuple>
#include <vector>
 
using namespace std;
 
#define rep(i,n) for(int i = 0; i < (n); i++)
#define all(a) begin(a),end(a)
 
typedef long long ll;
typedef pair<int,int> P;
const int inf=1000000007;
const ll mod=1000000007;
const double PI=3.14159265358979323846;
 
int dx[] = {1,-1,0,0};
int dy[] = {0,0,1,-1};
 
ll gcd(ll a, ll b){
  if (b==0) return a;
  else return gcd(b,a%b);
}
 
int cans(bool f){
  if(f) cout << "Yes" << endl;
  else cout << "No" << endl;
  return 0;
}

  

int N;
vector<vector<P>> v;

bool judge(int bit){
  rep(i,N){
    
    if(!(bit & (1<<i))) continue;

    for(P itr : v[i]){
      int x = itr.first;
      int y = itr.second;

      if(y == 1 && !(bit & (1<<x))) return false;
      if(y == 0 &&  (bit & (1<<x))) return false;
    }
  }
  return true;
}

                           

int main(){
  cin >> N;
  v.resize(N);
  rep(i,N){
    int A;
    cin >> A;
    v[i].resize(A);
    rep(j,A){
      cin >> v[i][j].first >> v[i][j].second;
      v[i][j].first--;
    }
  }




  
  int ans = 0;
  for(int bit=0; bit < (1<<N); bit++){
    if(judge(bit)){
      int res = 0;
      rep(i,N) if(bit & (1<<i)) res++;
      ans = max(res,ans);
    }

  }
  cout << ans << endl;
  
  return 0;
}