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;

public class Main {
	private static Integer N;

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		N = Integer.parseInt(sc.next());
		int[][] x = new int[N][];
		int[][] y = new int[N][];
		int[] A = new int[N];
		for(int i = 0; i < N; i++) {
			A[i] = Integer.parseInt(sc.next());
			x[i] = new int[A[i]];
			y[i] = new int[A[i]];
			for(int j = 0; j < A[i]; j++) {
				x[i][j] = Integer.parseInt(sc.next()) - 1;
				y[i][j] = Integer.parseInt(sc.next());
			}
		}
		sc.close();

		int res = 0;
		for(int bit = 0; bit < (1 << N); bit++) {
			if(judge(bit, A, x, y)) {
				int count = 0;
				for(int i = 0; i < N; i++) {
					if((bit & (1 << i)) != 0) count++;
				}
				res = Math.max(res, count);
			}
		}
		System.out.println(res);

	}

	//整数bitに対応する「決め打ち」が整合しているかどうか判定
	public static boolean judge(int bit, int[] A, int[][] x, int[][] y) {
		for(int i = 0; i < N; i++) {
			if((bit & (1 << i)) == 0) {
				continue;
			}
			for(int k = 0; k < A[i]; k++) {
				int a = x[i][k];
				int b = y[i][k];

				//b = 0なのに「正直者」ならダメ
				if(b == 0 && ((bit & (1 << a)) != 0)) return false;
				//b = 1なのに「不親切」ならダメ
				if(b == 1 && ((bit & (1 << a)) == 0)) return false;
			}
		}
		return true;
	}
}