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.*;
import static java.lang.System.*;
import java.math.*;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt(); //人数	
		//証言リスト。n+1番目にはnの証言がTesimonyクラスとして格納されている
		ArrayList<Tesimony>[] tesList = new ArrayList[n + 1];
		for (int i = 0; i < n + 1; i++)
			tesList[i] = new ArrayList<>();
		for(int i = 0; i < n; i++) {
			int a = sc.nextInt();
			 for (int j = 0; j < a; j++) {
				 int to = sc.nextInt();
				 int isTrust = sc.nextInt();
				 Tesimony tes = new Tesimony(to, isTrust);
				 tesList[i + 1].add(tes);
			 }
		}
		int ans = 0;

		for (int bit=0; bit< Math.pow(2, n); bit++) {
			boolean flag = true;
			//全員に対して証言チェック。証言者が「不親切」ならスルー
			for(int i = 0; i < n; i++) {
				if ((1&bit>>i) == 0)  continue;
				//証言の矛盾がないかチェック。矛盾があればループを抜ける
				for(int j = 0; j < tesList[i+ 1].size(); j++) {
					Tesimony tes = tesList[i+1].get(j);
					//矛盾があった場合
					if ( (1&bit>>(tes.to - 1) )  != tes.isTrust) {
						flag = false;
						break;
					}			
				}
			}
			if (flag) {
				int cnt = 0;
				for(int k = 0; k < n; k++) {
					if ((1&bit>>k) == 1) cnt++;
				}
				if (ans < cnt) ans = cnt;
			}
		}
		out.println(ans);	
	}
	//証言。証言対象と真偽を持つ
	static class Tesimony {
		int to;
		int isTrust;
		
		Tesimony(int to, int isTrust){
			this.to = to;
			this.isTrust = isTrust;
		}
		
	}
}