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;

/**
 * HonestOrUnkind2
 */
public class Main {

    public static void main(String[] args) throws Exception {
        try (Scanner sc = new Scanner(System.in)) {
            int N = sc.nextInt();

            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() - 1;
                    Y[i][j] = sc.nextInt();
                }
            }

            int ans = 0;

            for (int i = 1; i < (1 << N); i++) {
                int ans2 = 0;

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

                for (int j = 0; j < N; j++) {
                    if (honest[j] == 1) {
                        ans2++;

                        for (int k = 0; k < A[j]; k++) {
                            if (honest[X[j][k]] != Y[j][k]) {
                                ans2 = -1;
                                j = N;
                                break;
                            }
                        }
                    }
                }

                ans = Math.max(ans, ans2);
            }

            System.out.println(ans);
        }
    }

}