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;
using System.Collections.Generic;

namespace AtCoder
{
    public class Program
    {
        public static void Main(string[] args)
        {
            //入力部分,最初のN（人数）を読み取る
            var N = int.Parse(Console.ReadLine());
            var input_data = new List<List<int[]>>();

            //人数分の証言を読み取る
            for(int i = 0; i < N; i++)
            {
                var num_evidence = int.Parse(Console.ReadLine());
                var tmp = new List<int[]>();
                //特定の人の証言の数だけ読み取る
                for(int j= 0; j < num_evidence; j++)
                {
                    var evidence = Console.ReadLine().Split().Select(int.Parse).ToArray();
                    tmp.Add(evidence);
                }
                input_data.Add(tmp);
            }

            var max = 0;//正直者の最大値
            
            //仮定のパターンだけ繰り返す、1をNビットシフトさせた数。つまり2のN乗
            for (int i = 0; i < 1<<N ; i++)
            {
                var flag = true;//今回の仮定はすべてあってるかどうかを確認する
                for (int j = 0; j < N; j++)
                {
                    //仮定のパターンに対して１ビットずつ確認していく
                    if ((i >> j & 1) == 0) { continue; }//不誠実な人のパターンは無視
                    else
                    {
                        for(int k = 0; k < input_data[j].Count(); k++)
                        {
                            var num = input_data[j][k][0]-1;
                            if (((i >> num) & 1) != input_data[j][k][1])//パターンが一致しなかった
                            {
                                flag = false;
                            }
                            
                        }
                    }
                }
                if (flag)
                {
                    var c=0;//立っているビットの数、つまり正直者の数
                    for(int n = 0; n < N; n++)
                    {
                        if(((i >> n) & 1) == 1) { c++; }//１つずつずらして「1」が立っているビットの数を調べる
                    }
                    max = max > c ? max : c;
                }
                
            }
            Console.WriteLine(max);

        }
    }
}