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.Scanner;

public class Main {
    public static void main(String[] args) throws Exception {
        final Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int[][][] data = new int[N][][];
        for (int i = 0; i < N; i++) {
            int a = sc.nextInt();
            data[i] = new int[a][2];
            for (int j = 0; j < a; j++) {
                data[i][j][0] = sc.nextInt() - 1;
                data[i][j][1] = sc.nextInt();
            }
        }
        int ans = 0;
        for (int bit = 0; bit < Math.pow(2, N); bit++) {
            boolean f = true;
            for (int i = 0; i < N; i++) {
                if (getBit(bit, i) == 0) {
                    continue;
                }
                for (int j = 0; j < data[i].length; j++) {
                    int shougen = data[i][j][1];
                    if (shougen != getBit(bit, data[i][j][0])) {
                        f = false;
                    }
                }
            }
            if (f) {
                ans = Math.max(ans, Integer.bitCount(bit));
                System.out.println(bit);
            }
        }
        System.out.println(ans);
    }
    
    static int getBit(int bit, int digit) {
        if ((bit & (1 << (digit))) > 0) {
            return 1;
        }
        return 0;
    }
}