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 <iostream>

using namespace std;

int N, x[15][15], y[15][15];
int A[15];
int person[15];

int dfs(int n, int judgh){
  int flag=0, ans=0;
  if(n==N){
    for(int i=0; i<N; i++){
      if(flag == 1) break;
      for(int j=0; j<A[i]; j++){ // i番目の人のことについて検証(A[i]個の証言)
        if(person[i] == 0){ // i番目が嘘つき
          if(person[x[i][j]-1] == y[i][j]){ // i番目のj個目の証言が矛盾
            flag = 1;
            break;
          }
        }else{ // i番目は正直者
          if(person[x[i][j]-1] != y[i][j]){
            flag = 1;
            break;
          }
        }
      }
    }
    if(flag == 0){
      for(int i=0; i<N; i++){
        if(person[i] == 1) ans++;
      }
      return ans;
    }
    else return 0;
  }
  person[n] = judgh;
  int x0 = dfs(n+1, 0);
  int x1 = dfs(n+1, 1);
  if(x0 > x1) return x0;
  else return x1;
}

int main(){
  cin >> N;
  for(int i=0; i<N; i++){
    cin >> A[i];
    for(int j=0; j<A[i]; j++){
      cin >> x[i][j] >> y[i][j];
    }
  }
  int ans1 = dfs(0, 0), ans2 = dfs(0, 1);
  if(ans1 > ans2)   cout << ans1 << endl;
  else   cout << ans2 << endl;
  return 0;
}