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.Scanner;
public class Main{
	public static int []syoujiki(int[]a,int[][]b,int[][]c) {
		int[] d = new int[a.length];
		for(int i = 0; i < a.length; i++) {
			d[i] = -1;
		}
		
		for(int i = 0; i < a.length; i++) {
			for(int j = 0; j < a[i]; j++) {
				
				if(d[b[i][j]-1]== 0 && c[i][j] == 1 ) {
					d[b[i][j]-1] =0;		
					continue;
				}
				
				d[b[i][j]-1] = c[i][j];
		
			}
		}
		
		
		
		return d;
	}
	
	

	public static void main(String[] args) {
		Scanner stdIn = new Scanner(System.in);
		int n = stdIn.nextInt(); //人数
		int [] a = new int[n];
		int [] []b = new int[n][n]; 
		int [] []c = new int[n][n];
		
		for(int i = 0; i < n; i++) {
			a[i] = stdIn.nextInt();
			for(int j = 0; j < a[i]; j++) {
				b[i][j] = stdIn.nextInt(); //ex)b[i][j] == 3であると、3人目の人が対象
				c[i][j] = stdIn.nextInt(); //b[i][j]個目の証言が正直か不親切か
				}
			}
		
			int [] d = syoujiki(a,b,c);
			
			int cnt = 0;
			for(int i = 0; i < d.length; i++) {
				if(d[i] ==1) {
					cnt++;
				}
			}
			System.out.println(cnt);
		
		
	}

}