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: Wrong Answer)

import java.util.*;

public class Main{
  public static void main(String[] args){
    Scanner sc = new Scanner(System.in);
    int n=sc.nextInt();
    int[] a = new int[n];
    int[][] x = new int[n][n];
    int[][] y = new int[n][n];
    
    for(int i=0;i<n;i++){//証人ごと
      a[i]=sc.nextInt();
      for(int j=0;j<a[i];j++){//証言ごと
        x[i][j]=sc.nextInt();
        y[i][j]=sc.nextInt();
      }
    }
    
    int map =0;
    int ans=0;
    /*
          for(int i2=0;i2<n;i2++){
            System.out.println(i2);
            for(int j2=0;j2<a[i2];j2++){
              System.out.println(x[i2][j2]+" "+y[i2][j2]);
            }
          }
    */
    
    for(;map<Math.pow(n,2);map++){//正直者,嘘つきのパターン
    //for(map=2;map<3;map++){
      
      Boolean[] honest = new Boolean[n];
      Boolean ok = true;//矛盾がないか
      int tempAns=0;//現パターンの正直者の人数
       
      for(int i=0;i<n;i++){
        if((map>>i)%2==1){
          honest[i] = true;
          tempAns++;
        }else honest[i] = false;
      }
      
      for(int i=0;i<n;i++){//証人ごと
        
        for(int j=0;j<a[i];j++){//証言ごと
          //Boolean temp =  honest[x[i][j]]==(y[i][j]==1);//test
          //System.out.println(honest[i] +" "+ i+" "+j+" " + temp);//test
          

          
          if(!honest[i])break;
          if(honest[x[i][j]-1]==(y[i][j]==1)){
          }else{
            ok = false;
            break;
          }
        }
        
        if(!ok)break;
      }
      
      //System.out.println(map + " " + ok);
      if(ok)ans = Math.max(ans,tempAns);
    }
    
    System.out.println(ans);
    
  }
}