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.*;

public class Main{

    static long gcd(long a,long b){
        if(b==0)return a;
        return gcd(b,a%b);
    }

    static long lcm(long a,long b){
        return a/gcd(a,b)*b;
    }
    public static void main(String[] srgs){
        Scanner sc=new Scanner(System.in);

        int n=sc.nextInt();
        int[][][] in=new int[n][][];
        for(int i=0;i<n;i++){
            int m=sc.nextInt();
            in[i]=new int[m][2];
            for(int j=0;j<m;j++){
                in[i][j][0]=sc.nextInt();
                in[i][j][1]=sc.nextInt();
            }
        }
        int out=0,N=1<<n;
        for(int i=0;i<N;i++){
            int c=0,t=1;
            for(int j=0;j<n;j++)if((i>>j&1)==1){
                c++;
                for(int k=0;k<in[j].length;k++){
                    if((i>>(in[j][k][0]-1)&1)+in[j][k][1]==1)
                        t=0;
                }
            }
            System.out.println(i+" "+t+" "+c);
            if(t==1)out=Math.max(out,c);
        }
        System.out.println(out);
    }
}