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.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)) // Modify to 0 origin
//    }
//    testimonies
//  }
  val sc = new Scanner(System.in)
  val n = sc.nextInt
  val testimoniesPerPerson: Array[Map[Int, Boolean]] = Array.fill(n){
    Array.fill(sc.nextInt)(sc.nextInt - 1 -> (sc.nextInt == 1)).toMap
  }


  def solve(): Int = {
    val valids = (0 until (1 << n)).map{ i =>
      val valid = (0 until n)
        .filter{j => isBitOn(i, j)}
        .forall{j => isHonestPerson(j, i)}
      if (valid) Integer.bitCount(i) else 0
    }
    valids.max
  }

  def isHonestPerson(personIndex: Int, bit: Int): Boolean =
    testimoniesPerPerson(personIndex).forall { case (x, y) =>
      y == isBitOn(bit, x)
    }

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

  def countHonest(bit: Int): Int=
    Integer.bitCount(bit)


  println(solve())

}