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 Scala (2.11.7) (judge verdict: Accepted)

import java.io.{BufferedReader, InputStreamReader}

object Main {
  def main(args: Array[String]): Unit = {
    val in = new BufferedReader(new InputStreamReader(System.in))
    val n = in.readLine().toInt

    val evs =
      (0 until n).map { _ =>
        val a = in.readLine().toInt
        (0 until a).map { _ =>
          in.readLine().split("\\s+").map(_.toInt) match {
            case Array(x, y) => Array(x - 1, y)
          }
        }
      }

    System.out.println(
      (0 until (1 << n)).map { i =>
        if (evs.zipWithIndex.filter {
          case (ev, evIndex) => ((i >> evIndex) & 1) == 1
        }.forall {
          case (ev, evIndex) =>
            ev.forall { e =>
              ((i >> e(0)) & 1) == e(1)
            }
        }) {
          countBit(i)
        } else {
          0
        }
      }.max
    )
  }

  def countBit(n: Int): Int = {
    if (n == 0) {
      0
    } else {
      countBit(n / 2) + n % 2
    }
  }
}