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: Runtime Error)


import scala.io.StdIn._
object Main extends App {

  val n: Int = readInt()
  val testimoniesPerPerson: Seq[Seq[(Int, Int)]] = (1 to n).map { _ =>
    val numOfTestimonies = readInt()
    val testimonies: Seq[(Int, Int)] = Seq.fill(numOfTestimonies) {
      val testimony: Array[Int] = readLine().split(" ").map(_.toInt)
      (testimony(0) - 1, testimony(1))
    }
    testimonies
  }


  def solve(): Int = {
    (0 until (1 << n)).foldLeft(0) { case (acc, i) =>
      val isAllTrue: Boolean = (0 until i).collect { case j if isBitOn(i, j) =>
        testimoniesPerPerson(j).forall { case (x, y) =>
          y ==  ((i >> x) & 1)
        }
      }.forall(identity)
      if(isAllTrue) Math.max(acc, countHonest(i)) else acc
    }
  }

  def isBitOn(bit: Int, digit: Int): Boolean =
    ((bit >> digit) & 1) != 0

  def countHonest(bit: Int): Int=
    (0 until n).foldLeft(0) { case (acc, digit) =>
      if(isBitOn(bit, digit)) acc + 1 else acc
    }


  println(solve())

}