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.Scanner;
class Main{
  public static void main(String[] args){
    Scanner sc = new Scanner(System.in);
   int n = sc.nextInt();
   int a;
   int[][] clue = new int[n][n];
   for(int i = 0; i < n; i++){
     for(int j = 0; j < n; j++){
       clue[i][j] = -1; 
     }
   }
   for(int i = 0; i < n; i++){
     a = Integer.parseInt(sc.next());
     for(int j = 0; j < a; j++){
       clue[i][Integer.parseInt(sc.next()) - 1] = Integer.parseInt(sc.next());
     }
   }

   int honest;
   int max = 0;
   boolean bl;

   for(int i = 0; i < (1<<n); i++ ){
     bl = true;
     for(int j = 0; j < n; j++){

       honest = ((i >> j) & 1);//no problem
       if(honest == 1){
         for(int k = 0; k < n;k++){
           if(clue[j][k] != -1 && clue[j][k] != ((i >> k) & 1)){
             bl = false;
             break;
           }
         }
       }
       if(!bl){
         break;
       }
    }
    if(!bl){
      continue;
    }
    max = Math.max(max,Integer.bitCount(i));//problem
  }
  System.out.println(max);
}
}