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: Wrong Answer)

import util.control.Breaks._

object Main extends App {
  val n = io.StdIn.readLine().toInt
  val testimony = Array.fill(n)(Array.fill(n)(-1))
  var max = 0

  for (i <- 0 until n) {
    val a = io.StdIn.readLine().toInt
    for (_ <- 0 until a) {
      val Array(x, y) = io.StdIn.readLine().split(' ').map(_.toInt)
      testimony(i)(x-1) = y
    }
  }

  for (b <- 0 until 1 << n) {
    breakable {
      var count = 0
      for (i <- 0 until n; if (b & (1 << i)) != 0) {
        testimony(i).indices.withFilter(testimony(i)(_) != -1).foreach {j =>
          val x = (1 << j) & b
          if (testimony(i)(j) == 1 && x == 0) break()
          if (testimony(i)(j) == 0 && x != 0) break()
          count += 1
        }
      }
      if (max < count) max = count
    }
  }

  println(max)
}