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 <bits/stdc++.h>

using vp = std::vector<std::pair<int, bool>>;

int N;
std::vector<vp> x;

bool able(int);

int main()
{
	scanf("%d", &N);
	x.resize(N);
	for (auto& e: x)
	{
		int A;
		scanf("%d", &A);
		e.resize(A);
		for (auto& f: e)
		{
			int tmp;
			scanf("%d%d", &f.first, &tmp);
			f.first--;
			f.second = tmp == 1;
		}
	}
	int max{};
	for (int i{}; i < 1 << N; i++)
		if (able(i))
		{
			int num{};
			for (int j{}; j < N; j++)
				if (i >> j & 1)
					num++;
			max = std::max(max, num);
		}
	printf("%d\n", max);

	return 0;
}

bool able(int set)
{
	for (int i{}; i < N; i++)
	{
		if (~set >> i & 1) continue;
		for (auto& e: x[i])
			if ((set >> e.first & 1) && !e.second)
				return false;
			else if ((~set >> e.first & 1) && e.second)
				return false;
	}
	return true;
}