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 C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Linq;


public class A
{
    public static void Main(string[] args)
    {
        int t = 1;
        while (t > 0)
        {
            int n = int.Parse(Console.ReadLine());
            var mm = new int[n, n];
            for(int i = 0; i < n; i++)
            {
                int tt = int.Parse(Console.ReadLine());
                for(int j = 0; j < tt; j++)
                {
                    var arr = Console.ReadLine().Trim().Split().Select(x => int.Parse(x)).ToArray();
                    mm[i, arr[0] - 1] = (arr[1] == 1 ? 1 : -1);
                }
            }
            int max = 0;
            for(int i = 0; i <= Math.Pow(2,n); i++)
            {
                int honestCount = 0;
                bool ok = true;
                for(int j = 0; j < n; j++)
                {
                    if(((1 << j) & i) != 0)
                    {
                        honestCount++;
                        for(int k = 0; k < n; k++)
                        {
                            if(mm[j,k] == 1 && ((1 << k) & i) == 0)
                            {
                                ok = false;
                                break;
                            }
                            if(mm[j,k] == -1 && ((1 << k) & i) > 0)
                            {
                                ok = false;
                                break;
                            }
                        }
                    }
                }
                if (ok)
                    max = Math.Max(honestCount, max);
            }
            Console.WriteLine(max);
            t--;
        }
    }
}