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.*;
public class Main{
	public static void main(String[] args){
		Scanner sc=new Scanner(System.in);
		int n=sc.nextInt();
		ArrayList<Integer>[] peo=new ArrayList[n];
		for(int i=0; i<n; i++){
			peo[i]=new ArrayList<>();
			int a=sc.nextInt();
			for(int j=0; j<a; j++){
				int x=sc.nextInt()-1;
				int y=sc.nextInt();
				peo[i].add(x);
				peo[i].add(y);
			}
		}
		int ans=0;
		for(int i=1; i<(1<<n); i++){
			int c=i;
			int index=0;
			int count=0;
			Boolean[] px=new Boolean[n];
			while(index!=n){
				if((c & 1)==1){
					if(px[index]!=null && !px[index]){
						count=0;
						break;
					}
					px[index]=true;
					count++;
					for(int j=0; j<peo[index].size(); j+=2){
						Boolean kp=false;
						if(peo[index].get(j+1)==1){
							kp=true;
						}else{
							kp=false;
						}
						if(px[peo[index].get(j)]!=null && px[peo[index].get(j)]!=kp){
							count=0;
							break;
						}
						px[peo[index].get(j)]=kp;
					}
					if(count==0){
						break;
					}
				}else{
					if(px[index]!=null && px[index]){
						count=0;
						break;
					}
					px[index]=false;
				}
				index++;
				c>>=1;
			}
			ans=Math.max(ans,count);
		}
		System.out.println(ans);
	}
}