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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;

class Main{
  public static void main(String[] args){
    Scanner sc = new Scanner(System.in);
    int N = sc.nextInt();
    int ans = 0;
    int A[] = new int[N];
    int x[][] = new int[N][N];
    int y[][] = new int[N][N];
    for (int i=0;i<N;i++) {
      A[i] = sc.nextInt();
      for(int j=0;j<A[i];j++){
        x[i][j] = sc.nextInt();
        y[i][j] = sc.nextInt();
      }
    }
    //ここまで入力

    //bit全探索
    for(int bits=1; bits < (1 << N); bits++){
      boolean flag = true;
      for(int i=0; i<N; i++){
        //不親切な人の証言は考慮しない
        if((bits & (1 << i)) == 0) continue;
        for(int j=0; j<A[i]; j++){
          if((((bits >> x[i][j]-1) & 1) ^ y[i][j]) == 1) flag = false;
          //（bitsが101のとき、人１の証言が人2は1ですなど）正直者が、正直者だと証言した人が正直者ではない場合
          //正直者が、不親切な人だと証言した人が正直者だった場合
        }
      }

      if(flag == true){
        int cnt = 0;
        for(int j = 0; j < N; j++) {
          if ((bits >> j) % 2 == 1){
            cnt++;
            ans = Math.max(ans, cnt);
          }
        }
      }

    }
    System.out.print(ans);
  }
}