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 in = new Scanner(System.in);
		
		int n = in.nextInt();
		int[] a = new int[n];
		
		ArrayList<int[]>[] t = new ArrayList[n];
		for (int i = 0; i < n; i++)
			t[i] = new ArrayList<>();
		
		for (int i = 0; i < n; i++) {
			a[i] = in.nextInt();
			for (int j = 0; j < a[i]; j++) {
				int x = in.nextInt() - 1;
				int y = in.nextInt();
				t[i].add(new int[]{x, y});
			}
		}

		int max = 0;
		for (int i = 0; i < (1 << n); i++) {
			boolean valid = true;
			for (int j = 0; j < n; j++) {
				boolean p = ((1 << j) & i) > 0;
				for (int[] tt : t[j]) {
					boolean x = ((1 << tt[0]) & i) > 0;
					if (p && x && tt[1] == 0)
						valid = false;
					if (p && !x && tt[1] == 1)
						valid = false;
				}
			}
			if (valid)
				max = (int) Math.max(max, Integer.bitCount(i));
		}

		System.out.println(max);
	}
}