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.*;

public class Main{
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        ArrayList< ArrayList<int[]>> people = new ArrayList<ArrayList<int[]>>();

        for(int i = 0; i < N; i++){
            int A = sc.nextInt();
            ArrayList<int[]> say = new ArrayList<int[]>();
            for(int j = 0; j < A; j++){
                say.add(new int[]{sc.nextInt() - 1, sc.nextInt()});
            }
            people.add(say);
        }

        boolean[] honest = new boolean[N];
        int max = Integer.MIN_VALUE;

        for (int i = 0; i < (Math.pow(2 ,N)); i++) {
            Arrays.fill(honest, false);
            int count = 0;

            for (int j = 0; j < N; j++) {
                if ((1 & i >> j) == 1){
                    honest[j] = true;
                    count++;
                }
            }

            boolean flag = true;
            for(int j = 0; j < N; j++){
                if(honest[j]){
                    for(int k = 0; k < people.get(j).size(); k++){
                        if(people.get(j).get(k)[1] == 1 && !honest[ people.get(j).get(k)[0] ])
                            flag = false;
                        if(people.get(j).get(k)[1] == 0 && honest[ people.get(j).get(k)[0] ])
                            flag = false;
                    }
                }
            }
            if(flag)max = Math.max(max, count);
        }
        System.out.println(max);
    }
}