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();
		List<List<Integer[]>> p = new ArrayList<>();
		int max = 0;
		for (int i = 0; i < N; i++) {
			int Ai = sc.nextInt();
			List<Integer[]> ops = new ArrayList<>();
			for (int j = 0; j < Ai; j++) {
				Integer[] op = new Integer[2];
				op[0] = sc.nextInt();
				op[1] = sc.nextInt();
				ops.add(op);
			}
			p.add(ops);
		}
		for (int i = (1 << N) - 1; 0 < i; i--) {
			boolean[] ans = new boolean[N];
			int count = 0;
			for (int j = 0; j < N; j++) {
				if (((i>>j)&1) == 1) {
					ans[j] = true;
					count++;
				}
			}
			boolean flg = true;
			for (int k = 0; k < N; k++) {
				if (ans[k]) {
					for (Integer[] opi : p.get(k)) {
						if ((ans[opi[0]-1] && opi[1] == 1)
								|| (!ans[opi[0]-1] && opi[1] == 0)) {
							continue;
						} else {
							flg = false;
							break;
						}
					}
				}
			}
			if (flg) {
				if (max < count) {
					max = count;
				}
			}
		}
		System.out.println(max);
	}
}