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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.util.*;
public class Main {
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		
		
		
        
        int [] b = new int[N];
		int[][] d = new int[N][N];
		int[][] e = new int[N][N];
		for(int i = 0; i < N; i++){
            b[i] = sc.nextInt();
          
            for(int j = 0; j < b[i]; j++){
                d[i][j] = sc.nextInt();
                d[i][j] --;
             
                e[i][j] = sc.nextInt();
              
            }
        }
        int [] c = new int[N];
        int f = 0;
        int g = 0;
        boolean h = false;
		for(int i = 0; i < (int)Math.pow(2, N ); i++){
              int l = 1;
			for(int j = 0; j < N; j++){
                c[j] = (i / l) % 2;
              l = l * 2;
            }
            h = true;
            for(int j = 0; j < N && h; j++){
                for(int k = 0; k < b[j] && h; k++){
                    if(c[j] == 1 && e[j][k] != c[d[j][k]] ){
                        h = false;
                      
                    }
                }
            }
            if(h){
              
                for(int j = 0; j < N; j++){
                    f += c[j];
                }
                if(f > g){
                    g = f;
                  
                }
                f = 0;
            }
            
		}
      
		System.out.println(g);
		
		
	}
}