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[] a = new int[n];
        int[][] x = new int[n][];
        int[][] y = new int[n][];
        for (int i=0;i<n;i++){
            a[i]=sc.nextInt();
            x[i] = new int[a[i]];
            y[i] = new int[a[i]];
            for (int j=0;j<a[i];j++){
                x[i][j] = sc.nextInt();
                y[i][j] = sc.nextInt();
            }
        }
        int max = 0;
        for (int i=0;i<Math.pow(2,n);i++){
            String s = Integer.toBinaryString(i);
            while (s.length()<n){
                s = "0"+s;
            }
            int[] tf = new int[n];
            int count = 0;
            for (int j=0;j<n;j++){
                if (s.charAt(j)=='0')tf[j]=0;
                if (s.charAt(j)=='1'){
                    tf[j]=1;
                    count++;
                }
            }
            boolean b = true;
            for (int j=0;j<n&&b;j++){
                if (tf[j]==1){
                    for (int k=0;k<a[j]&&b;k++){
                        if (y[j][k]!=tf[x[j][k]-1])b=false;
                    }
                }
            }
            if (b){
                max = Math.max(max,count);
            }

        }
        System.out.println(max);
    }
}