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.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {

		try(Scanner sc = new Scanner(System.in)) {

			// 人数
			int n = sc.nextInt();
			List<Map<Integer, Integer>> list = new ArrayList<>();
			for(int i = 0 ; i < n ; i++) {
				Map<Integer, Integer> map = new HashMap<>();
				int a = sc.nextInt();
				for(int j = 0 ; j < a ; j++) {
					map.put(sc.nextInt(), sc.nextInt());
				}
				list.add(map);
			}
			// 全検索
			int ans = 0;
			ONE: for(int i = 0 ; i < 1 << n ; i++) {
				// 正直判定
				Integer[] judge = new Integer[n];
				int sum = 0;
				for(int j = 0 ; j < n ; j++) {
					if((1 & i >> j) == 1 || (judge[j] != null && judge[j] == 1)) {
						judge = isHonest(judge, list, j);
						if(judge == null) continue ONE;
					}
				}
				for(Integer j : judge) {
					if(j != null && j.intValue() == 1) {
						sum++;
					}
				}
				ans = Math.max(ans, sum);
			}

			System.out.println(ans);

		}

	}

	private static Integer[] isHonest(Integer[] judge, List<Map<Integer, Integer>> list, int listIndex) {
		for(Map.Entry<Integer, Integer> entry : list.get(listIndex).entrySet()) {
			if(judge[entry.getKey() - 1] == null && entry.getValue() == 1) {
				judge[entry.getKey() - 1] = entry.getValue();
				judge = isHonest(judge, list, entry.getKey() - 1);
				if(judge == null) {
					return null;
				}
			}
			else if(judge[entry.getKey() - 1] == null && entry.getValue() == 0) {
				judge[entry.getKey() - 1] = entry.getValue();
			}
			else if(judge[entry.getKey() - 1].intValue() != entry.getValue()) {
				return null;
			}
		}
		if(judge != null) {
			if(judge[listIndex] != null && judge[listIndex] != 0) {
				judge[listIndex] = 1;
			} else {
				return null;
			}
		}
		return judge;
	}

}