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: Wrong Answer)

#include<iostream>
#include<vector>
#include<map>
#include<algorithm>
#include<math.h>
#include<string>
#include <iomanip>
#include<deque>
#include<queue>
#include<stack>
#include<set>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef int itn;
const ll LINF = 1e18;
const int INF = 1e9;

//マクロ定義
#define vvint(vec,n,m,l) vector<vector<int>> vec(n, vector<int>(m,l));	// lで初期化
#define vvll(vec,n,m,l) vector<vector<ll>> vec(n,vector<ll>(m,l));
#define vint vector<int>
#define pint pair<int,int>
#define rep(i,a) for(int i=0;i<(a);i++)
#define all(x) (x).begin(),(x).end()
#define debug system("pause")				//デバッグ用
#define ret return 0

template<class T>bool chmax(T & a, const T & b) { if (a < b) { a = b; return 1; } return 0; }
template<class T>bool chmin(T& a, const T& b) { if (b < a) { a = b; return 1; } return 0; }

#define keta setprecision(15) //15桁表示

int main(void)
{
	cin.tie(0);
	ios::sync_with_stdio(false);

	int n;
	cin >> n;
	vector<int> a(n);
	vector<vector<int>> x(n, vector<int>()), y(n, vector<int>());
	for (int i = 0; i < n; i++)
	{
		int tm;
		cin >> a[i];
		for (int j = 0; j < a[i]; j++)
		{
			cin >> tm;
			x[i].push_back(tm);
			cin >> tm;
			y[i].push_back(tm);
		}
	}

	int bin = pow(2, n);
	int tmp = 0;
	int sirabe = 0;
	bool ans = true;
	int sum = 0;
	for (int i = 0; i <= bin; i++)
	{
		tmp = i;
		while (sirabe != n)
		{
			for (int j = 0; j < a[sirabe]; j++)
			{
				if (tmp % 2 == 0) break;
				if ((tmp >> (x[sirabe][j] - 1)) % 2 == y[sirabe][j])
					continue;
				else
				{
					ans = false;
					break;
				}
			}
			tmp /= 10;
			sirabe++;
			if (!ans)
				break;
		}
		int tm = 0;
		if (ans)
		{
			tmp = i;
			while (tmp != 0)
			{
				tm += tmp &1;
				tmp = tmp >> 1;
			}
			sum = max(sum, tm);
		}
		ans = true;
		sirabe = 0;
	}
	cout << sum << endl;
	return 0;
}