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.awt.Point;
import java.util.*;
import java.io.*;
import java.util.function.*;

class Main {
    static Scanner scanner = new Scanner();
    static int mod = 1000000007;
    static long inf = 0xfffffffffffffffL;

    void solve() {
        int n = scanner.nextInt();
        List<Point>[] list = new List[n];
        for (int i = 0; i < n; i++) {
            int a = scanner.nextInt();
            list[i] = new ArrayList();
            for (int j = 0; j < a; j++) {
                int x = scanner.nextInt() - 1;
                int y = scanner.nextInt();
                list[i].add(new Point(x, y));
            }
        }
        int max = 0;
        l: for (int i = 0; i < 1 << n; i++) {//1=本当
            for (int j = 0; j < n; j++) {
                if (((i >> j) & 1) == 1) {
                    for (Point p : list[j]) {
                        if (((i >> p.x) & 1) != p.y) {
                            continue l;
                        }
                    }
                }
            }
            max = Math.max(max, Integer.bitCount(i));
        }
        System.out.println(max);
    }

    public static void main(String[]$) {
        try {
            new Main().solve();
            System.out.flush();
        } catch (Throwable t) {
            System.out.flush();
            t.printStackTrace();
        }
    }
}

class System {
    static PrintWriter out = new PrintWriter(java.lang.System.out);

    static void exit() {
        java.lang.System.exit(0);
    }
}

class Scanner {
    private BufferedReader reader = new BufferedReader(new InputStreamReader(java.lang.System.in), 0x10000);
    private StringTokenizer tokenizer;

    String next() {
        try {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                tokenizer = new StringTokenizer(reader.readLine());
            }
        } catch (IOException ignored) {
        }
        return tokenizer.nextToken();
    }

    int nextInt() {
        return Integer.parseInt(next());
    }

    long nextLong() {
        return Long.parseLong(next());
    }

    char[][] nextCharArray(int n, int m) {
        char[][] a = new char[n][m];
        for (int i = 0; i < n; i++) {
            a[i] = next().toCharArray();
        }
        return a;
    }

    int[] nextIntArray(int n) {
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = nextInt();
        }
        return a;
    }

    int[] nextIntArray(int n, IntUnaryOperator op) {
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = op.applyAsInt(nextInt());
        }
        return a;
    }

    long[] nextLongArray(int n) {
        long[] a = new long[n];
        for (int i = 0; i < n; i++) {
            a[i] = nextLong();
        }
        return a;
    }

    long[] nextLongArray(int n, LongUnaryOperator op) {
        long[] a = new long[n];
        for (int i = 0; i < n; i++) {
            a[i] = op.applyAsLong(nextLong());
        }
        return a;
    }
}