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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;
public class Main {

  public static void main(String[] args) { 
      Scanner sc = new Scanner(System.in);
      int n = sc.nextInt();
      List<int[]>[] arr = new List[n];
      for(int i = 0; i < n; i++) {
        int aNum = sc.nextInt();
        List<int[]> list = new ArrayList<>();
        for(int j = 0; j < aNum; j++) {
          list.add(new int[]{ sc.nextInt() - 1, sc.nextInt() });
        }
        arr[i] = list;
      }
      int maxCnt = 0;
      loop:
      for(int i = 0; i < (1 << n); i++) {
        int curCnt = 0;
        for(int j = 0; j < n; j++) {
          if(((i >> j) & 1) == 1) { // j番めの人がtrue
            curCnt++;
            for(int[] t : arr[j]) {
              if(((i >> t[0]) & 1) != t[1])
                continue loop;
            }
          }
        }
        maxCnt = Math.max(maxCnt, curCnt);
      }
      System.out.println(maxCnt);
  }
}