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.ArrayList;
import java.util.Arrays;
import java.util.Scanner;

public class Main{
	public static void main(String[] args){
		Scanner scan = new Scanner(System.in);
		int n = scan.nextInt();
		int i,j,k,l,m,a,x,y;

		int pt = (int)Math.pow(2, n) - 1;
		ArrayList<Integer[]> list = new ArrayList<>();
		for(i = 0; i < n; i++) {
			Integer[] temp = new Integer[n];
			Arrays.fill(temp, -1);
			a = scan.nextInt();
			for(j = 0; j < a; j++) {
				x = scan.nextInt();
				y = scan.nextInt();
				temp[x - 1] = y;
			}
			list.add(temp);
		}

		int[] ptList = new int[n];
		Integer[] temp;
		int cnt;
		int max = Integer.MIN_VALUE;
		boolean f;

		for(i = 0; i <= pt; i++) {
			int[] ans = new int[n];
			Arrays.fill(ans, -1);
			for(j = 0; j < n; j++) {
				ptList[j] = (1 & i >> j);
			}
			cnt = 0;
			f = true;
			for(k = 0; k < n; k++) {
				temp = list.get(k);
				for(l = 0; l < n; l++) {
					if(temp[l] != -1) {
						if(ptList[k] == 0) {
							if(temp[l] == 0) {
								if(ans[l] != -1 && ans[l] == 0) {
									f = false;
									break;
								}
								ans[l] = 1;
							}else {
								if(ans[l] != -1 && ans[l] == 1) {
									f = false;
									break;
								}
								ans[l] = 0;
							}
						}else {
							if(ans[l] != -1 && ans[l] != temp[l]) {
								f = false;
								break;
							}
							ans[l] = temp[l];
						}
					}
				}
			}
			for(m = 0; m < n; m++) {
				if(ans[m] == -1)
					continue;
				if(ans[m] != ptList[m]) {
					f = false;
					cnt = 0;
					break;
				}
			}
			if(f) {
				for(m = 0; m < n; m++)
					if(ans[m] == 1)
						cnt++;
			}else {
				cnt = 0;
			}

			if(cnt > max)
				max = cnt;
		}
		System.out.println(max);
	}
}