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.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Main main = new Main();
        main.solve();
    }
    public void solve() {
        Scanner scan = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        int N  = scan.nextInt();
        List<Pair>[] list = new List[N];
        for (int i = 0; i < N; i++) {
            int A = scan.nextInt();
            list[i] = new ArrayList<Pair>();
            for (int j = 0; j < A; j++) {
                int x = scan.nextInt()-1;
                int y = scan.nextInt();
                Pair p = new Pair(x, y);
                list[i].add(p);
            }
        }
        int max = pow(N);
        int answer = 0;
        for (int pattern = 0; pattern < max; pattern++) {
            int current = 1;
            boolean[] judge = new boolean[N];
            Arrays.fill(judge, false);
            int count = 0;
            for (int i = 0; i < N; i++) {
                if (0 < (pattern & current)) {
                    judge[i] = true;
                    count += 1;
                }
                current *= 2;
            }
            boolean contradiction = false;
            for (int j = 0; j < N; j++) {
                List<Pair> pair = list[j];
                for (Pair p : pair) {
                    // 正直者
                    if (judge[j]) {
                        if (judge[p.x] && (p.y == 1)) {
                            continue;
                        }
                        if ((!judge[p.x]) && (p.y == 0)) {
                            continue;
                        }
                        contradiction = true;
                        break;
                    }
                }
                if (contradiction) {
                    break;
                }
            }
            if (!contradiction) {
                answer = Math.max(answer, count);
            }
        }
        System.out.println(answer);
    }
    private int pow(int N) {
        int i = 0;
        int pow = 1;
        while (i < N) {
            pow *= 2;
            i += 1;
        }
        return pow;
    }
    class Pair {
        int x;
        int y;
        Pair (int x, int y) {
            this.x = x;
            this.y = y;
        }
    }
}