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.util.Scanner

import scala.annotation.tailrec

object Main {
  def main(args: Array[String]): Unit = {
    //nextLineとnextを同時に使ったら死ぬ
    val sc = new Scanner(System.in)

    val n = sc.nextInt()
    val testimony = Vector.fill(n) {
      val m = sc.nextInt()
      Vector.fill(m) {
        val a, b = sc.nextInt()
        (a, if (b == 1) true else false)
      }
    }

    val bits = bitSearch(n, Vector(true, false))

    val ans =
      bits
        .filter { bit =>
          bit.zipWithIndex.forall {
            case (b, i) =>
              if (!b) true
              else {
                testimony(i).forall {
                  case (x, y) => bit(x - 1) == y
                }
              }
          }
        }
        .map(_.count(_ == true))
        .max

    println(ans)
  }
  def debug(args: Any*): Unit = {
    Console.err.println(args.mkString(" "))
  }

  def printExecTime(process: => Unit): Unit = {
    val start = System.currentTimeMillis
    process
    debug("処理時間： " + (System.currentTimeMillis - start) + " ミリ秒")
  }

  def bitSearch[T](n: Int, choice: Vector[T]): Vector[Vector[T]] = {
    @tailrec
    def bitSearchLogic(n: Int,
                       choice: Vector[T],
                       acc: Vector[Vector[T]]): Vector[Vector[T]] = {
      if (n == 0) acc
      else {
        val newAcc =
          acc.flatMap(vec => {
            choice.map(c => vec :+ c)
          })
        bitSearchLogic(n - 1, choice, newAcc)
      }
    }

    val firstAcc = choice.map(i => Vector(i))
    bitSearchLogic(n - 1, choice, firstAcc)
  }
}