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();
        Map<Integer,List<Integer>> t = new HashMap<>();
        Map<Integer,List<Integer>> f = new HashMap<>();
        for(int i = 0;i < n;i++){
            int a = sc.nextInt();
            List<Integer> tt = new ArrayList<>();
            List<Integer> tf = new ArrayList<>();
            for(int j = 0;j < a;j++){
                int x = sc.nextInt();
                int y = sc.nextInt();
                if(y == 1){
                    tt.add(x - 1);
                }else{
                    tf.add(x - 1);
                }
            }
            t.put(i,tt);
            f.put(i,tf);
        }
        int result = 0;
        for(int i = 0;i < (1 << n);i++){
            boolean[] trust = new boolean[n];
            int ti = 0;
            for(int j = 0;j < n;j++){
                if((1 & i >> j) == 1){
                    trust[j] = true;
                    ti++;
                }else{
                    trust[j] = false;
                }
            }
            boolean flag = true;
            NEXT:
            for(int j = 0;j < n;j++){
                boolean trustable = trust[j];
                if(!trustable){
                    continue;
                }
                for(int tp : t.get(j)){
                    if(!trust[tp]){
                        flag = false;
                        break NEXT;
                    }
                }
                for(int fp : f.get(j)){
                    if(trust[fp]){
                        flag = false;
                        break NEXT;
                    }
                }
            }
            if(flag){
                if(result < ti){
                    result = ti;
                }
            }
        }
        System.out.println(result);
    }
}