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 <algorithm>
#include <vector>

using namespace std;

int main() {
	int n; cin >> n;
	vector<vector<pair<int, int> > > xy(n);
	for (int i = 0; i < n; ++i) {
		int a; cin >> a;
		for (int j = 0; j < a; ++j) {
			int x, y; cin >> x >> y;
			xy[i].push_back(make_pair(x - 1, y));
		}
	}
	int ans = 0;
	for (int bit = 0; bit < (1 << n); ++bit) {
		vector<int> tmp(n, -1);
		int cnt = 0;
		for (int i = 0; i < n; ++i) {
			if (bit & (1 << i)) {
				++cnt;
				if (tmp[i] == 0) goto fal;
				tmp[i] = 1;
				for (auto e : xy[i]) {
					if (tmp[e.first] == (1 - e.second)) goto fal;
					tmp[e.first] = e.second;
				}
			} else {
				if (tmp[i] == 1) goto fal;
				tmp[i] = 0;
			}
		}
		ans = max(ans, cnt);
	fal:;
	}
	cout << ans << endl;

	return 0;
}