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: Wrong Answer)

using System;
using System.Collections.Generic;

namespace cswork
{

    class Program
    {
        public Program() { }

        Scanner cin;

        void Omame()
        {
            cin = new Scanner();

            int n = cin.nextInt();

            List<List<int[]>> say = new List<List<int[]>>();
            for (int i = 0; i < n; i++)
            {
                int m = cin.nextInt();
                List<int[]> list = new List<int[]>();
                for (int j = 0; j < m; j++)
                {
                    int[] tmp = new int[2];
                    tmp[0] = cin.nextInt() - 1;
                    tmp[1] = cin.nextInt();
                    list.Add(tmp);
                }
                say.Add(list);
            }

            int ans = 0;
            // bit全探索（1:正直者、0:不親切）
            for (int i = 0; i < 1 << n; i++)
            {
                int count = 0;
                for (int j = 0; j < n; j++)
                {
                    // 不親切な人はスキップ
                    if ((i >> j & 1) == 0)
                    {
                        continue;
                    }

                    // 誠実な人の言うことがあっているか確認
                    for (int k = 0; k < say[j].Count; k++)
                    {
                        int trg = i >> say[j][k][0] & 1;
                        int isHonest = say[j][k][1];

                        if (trg != isHonest)
                        {
                            break;
                        }
                    }
                    count++;
                }
                ans = Math.Max(ans, count);
            }

            Console.WriteLine(ans);
        }

        static void Main(string[] args)
        {
            new Program().Omame();
        }
    }

    class Scanner
    {
        string[] s;
        int i;

        char[] cs = new char[] { ' ' };

        public Scanner()
        {
            s = new string[0];
            i = 0;
        }

        public string next()
        {
            if (i < s.Length) return s[i++];
            string st = Console.ReadLine();
            while (st == "") st = Console.ReadLine();

            s = st.Split(cs, StringSplitOptions.RemoveEmptyEntries);
            i = 0;
            return next();
        }

        public int nextInt()
        {
            return int.Parse(next());
        }

        public long nextLong()
        {
            return long.Parse(next());
        }

        public double nextDouble()
        {
            return double.Parse(next());
        }
    }
}