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{
	static Scanner sc = new Scanner(System.in);
	static int n = 0;
	
	public static void main(String[] args){
		n = sc.nextInt();
		int[] binExps = buildBinExps(n);
		int[] a = new int[n];
		int[][] x = new int[n][];
		int[][] y = new int[n][];
		for(int i = 0; i < n; ++i){
			a[i] = sc.nextInt();
			x[i] = new int[a[i]];
			y[i] = new int[a[i]];
			for(int j = 0; j < a[i]; ++j){
				x[i][j] = sc.nextInt();
				y[i][j] = sc.nextInt();
			}
		}
		
		int ans = 0;
		for(int num = 0; num < binExps[n - 1]; ++num){
			int[] info = getBinary(num, n);
			int numHonest = count1(info);
			for(int i = 0; i < n; ++i){
				if(info[i] != 1) continue;
				for(int j = 0; j < a[i]; ++j){
					if(info[x[i][j] - 1] != y[i][j]){
						numHonest = -1;
						break;
					}
					
					//if(info[x[i][j] - 1] == 1 && y[i][j] == 1) ++numHonest;
				}
				if(numHonest < 0) break;
			}
			ans = (numHonest > ans) ? numHonest : ans;
		}
		System.out.println(ans);
	}
	
	static int[] buildBinExps(int w){
		int[] res = new int[w];
		for(int i = 0; i < w; ++i){
			res[i] = (i == 0) ? 1 : 2 * res[i - 1];
		}
		return res;
	}
	
	static int[] getBinary(int num, int w){
		int[] res = new int[w];
		for(int i = 0; i < w; ++i){
			res[i] = (num % 2 == 1) ? 1 : 0;
			num /= 2;
		}
		return res;
	}
	
	static int count1(int[] info){
		int res = 0;
		for(int i = 0; i < n; ++i){
			if(info[i] == 1) ++res;
		}
		return res;
	}
}