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.Scanner;

public class Main {

	public static int times(int x, int n) {
		int num = 1;
		for(int i = 0; i < n; i++) {
			num = num*x;
		}
		return num;
	}
	public static void digits(int x, int r, int[] d) {
		int digits = 1;
		//String dchar = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ";

		do {
				d[digits++] = (x % r);
				x /= r;

		}while(x != 0);
	}

	public static void main(String[] args) {
		// TODO 自動生成されたメソッド・スタブ
		Scanner sca = new Scanner(System.in);
		int n = sca.nextInt();
		int x[][] = new int[n+1][n+1];
		int y[][] = new int[n+1][n];
		int ai[] = new int[n+1];
		for(int i = 1; i < n+1; i++) {
			ai[i] = sca.nextInt();
			for(int j = 1; j < ai[i] + 1; j++) {
				x[i][j] = sca.nextInt();
				y[i][j] = sca.nextInt();
			}
		}
		int big = 0;
		int digi[] = new int[n+1];
		for(int i = 1; i < n+1; i++) {
			digi[i] = 0;
		}
		int count;
		for(count = 0; count < times(2, n); count++) {
			digits(count, 2, digi);
			boolean flag = true;
			for(int i = 1; i < n+1; i++) {
				if(flag == false) {
					break;
				}
				for(int j = 1; j < ai[i] + 1; j++) {
					if (digi[x[i][j]] == y[i][j] && digi[i] == 1) {
						// ループを続ける
					}
					//else if(digi[x[i][j]] != y[i][j] && digi[i] == 1 || digi[x[i][j]] == y[i][j] && digi[i] == 0){
					else if(digi[x[i][j]] != y[i][j] && digi[i] == 1){
						flag = false;
						break;
					}
				}
			}
			if(flag == true) {
				int big_t = 0;
				for(int i = 1; i < n+1; i++) {
					if(digi[i] == 1) {
						big_t++;
					}
				}
				if(big < big_t) {
					big = big_t;
				}
			}
		}
		System.out.println(big);
		sca.close();
	}
}