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 {
	static int N, A[];
	static List X;
	static boolean mujun;
	public static void main(String[] args) {	
		Scanner sc = new Scanner(System.in);
		N = sc.nextInt();
		A = new int[N];

		X = new ArrayList<ArrayList<int[]>>();
		for(int i=0;i<N;i++) {
			A[i]=sc.nextInt();
			List tmp = new ArrayList<int[]>();
			for(int j=0;j<A[i];j++) {
				int x = sc.nextInt() - 1; //人だけ０スタートにする
				int y = sc.nextInt();
				int t[] = {x, y};
				tmp.add(t);
			}	
			X.add(tmp);
		}
		sc.close();
		int ans = 0;
		for(int bit=0;bit<(1<<N);bit++) {
			//0bbitが全員のフラグを表す
			mujun = false;
			for(int i=0;i<N;i++) {
				if(((bit>>i)&1)==1) {
					List tmp = (List)X.get(i);
					for(int j=0;j<A[i];j++) {
						int judge[] = (int[])tmp.get(j); 
						int toPerson = judge[0];
						int toHonest = judge[1];
						if(((bit>>toPerson)&1)!=toHonest) {
							mujun = true;
							break;
						}
					}
				}
			}
			if(mujun)continue;
			ans = Math.max(ans,Integer.bitCount(bit));
		}
		
		
		System.out.println(ans);
	}
	
	
}