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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>

using namespace std;

#define ll long long
const int N = 18;

int data[N],ans = 0,a[N],b[N][N][2];
int n;

bool check(){
	for(int i = 1; i <= n; i++){
		if(!data[i]) continue;
		for(int j = 1; j <= a[i]; j++)
			if((b[i][j][1] && !data[b[i][j][0]]) || (!b[i][j][1] && data[b[i][j][0]]))
				return false;
	}
	return true;
}

void dfs(int now,int sum){
	if(now > n){
			if(check()) ans = max(ans,sum);
			return;
	}
	data[now] = 1;
	dfs(now+1,sum+1);
	data[now] = 0;
	dfs(now+1,sum);
}

void in(){
	scanf("%d",&n);
	for(int i = 1; i <= n; i++){
		scanf("%d",&a[i]);
		for(int j = 1; j <= a[i]; j++)
			scanf("%d%d",&b[i][j][0],&b[i][j][1]);
	}
	dfs(1,0);
}

int main(){
	in();
	printf("%d",ans);
}