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: Wrong Answer)

import java.util.*;
public class Main {
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
        int ans = 0;
		int [] [] [] in = new int [n][n][2];       
        for(int i=0;i<n;i++){in[i][0][0]=sc.nextInt();
                             for(int j=1;j<=in[i][0][0];j++){
                               for(int k=0;k<2;k++){in[i][j][k] = sc.nextInt();}
                             }
                            }
        int max = 1;
        for(int i=0;i<n;i++){max*=2;}
        max--;
        int [] check = new int [n];  
        for(int i=0;i<=max;i++){int p = i; int count = 0; int bool = 0;
                                for(int j=0;j<n;j++){check[j]=p%2;
                                                     p/=2;
                                                     count+=(1+check[j])%2;
                                                    }
                                for(int j=0;j<n;j++){
                                  if(check[j]==0){
                                  for(int k=1;k<=in[j][0][0];k++)
                                {if(check[in[j][k][0]-1]!=in[j][k][1])
                                {bool=1;}
                                }                                
                                }
                                }
                               if(bool==0){ans=Math.max(ans,count);}
                               }        
		System.out.println(ans);
}
}