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();
		int[][] m = new int[N][N];
		for(int i=0; i<N; i++) {
			Arrays.fill(m[i], -1);
			int A = sc.nextInt();
			for(int j=0; j<A; j++) {
				int x = sc.nextInt()-1;
				int y = sc.nextInt();
				m[i][x] = y;
			}
		}
		
		for(int i=N; i>0; i--) {
			int comb = (1<<i)-1;
			while(comb<(1<<N)) {
				boolean ok = true;
				for(int j=0; j<N; j++) {
					if((comb&(1<<j))!=0) {
						for(int k=0; k<N; k++) {
							if(m[j][k]==0 && (comb&(1<<k))!=0) {
								ok = false;
								break;
							}
							if(m[j][k]==1 && (comb&(1<<k))==0) {
								ok = false;
								break;
							}
						}
					}
					if(!ok)
						break;
 				}
				
				if(ok) {
					System.out.println(i);
					sc.close();
					return;
				}
			
				int x = comb & -comb;
				int y = comb+x;
				comb = ((comb & ~y)/ x>>1) | y;
			}
		}

		System.out.println(0);
		sc.close();
	}
}