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 namespace std;

#define fastio ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define pb push_back
#define mp make_pair
#define all(c) (c).begin(),(c).end()
#define M 1000000007
#define int long long
#define pr(...) dbs(#__VA_ARGS__, __VA_ARGS__)
template <class T> void dbs(string str, T t) {cerr << str << " : " << t << "\n";}
template <class T, class... S> void dbs(string str, T t, S... s) {int idx = str.find(','); cerr << str.substr(0, idx) << " : " << t << ", "; dbs(str.substr(idx + 1), s...);}
template <class S, class T>ostream& operator <<(ostream& os, const pair<S, T>& p) {return os << "(" << p.first << ", " << p.second << ")";}
template <class T>ostream& operator <<(ostream& os, const vector<T>& p) {os << "[ "; for (auto& it : p) os << it << " "; return os << "]";}
template <class T>ostream& operator <<(ostream& os, const set<T>& p) {os << "[ "; for (auto& it : p) os << it << " "; return os << "]";}
template <class S, class T>ostream& operator <<(ostream& os, const map<S, T>& p) {os << "[ "; for (auto& it : p) os << it << " "; return os << "]";}
template <class T> void prc(T a, T b) {cerr << "["; for (T i = a; i != b; ++i) {if (i != a) cerr << ", "; cerr << *i;} cerr << "]\n";}
// Use pr(a,b,c,d,e) or cerr<<anything or prc(v.begin(),v.end()) or prc(v,v+n)


int32_t main()
{
	fastio;
	int n;
	cin >> n;
	vector <vector <int> > good(n), bad(n);
	for(int i=0;i<n;i++){
		int x, y, m;
		cin >> m;
		for(int j=0;j<m;j++){
			cin >> x >> y;
			x--;
			if(y==1)
				good[i].push_back(x);
			else
				bad[i].push_back(x);
		}
	}

	int mask = (1<<n);
	int ans = 0;

	for(int i=0;i<mask;i++){
		int cur = i;
		bool poss = true;

		vector <int> face(n, -1);
		for(int j=0;j<n;j++){
			if((1<<j)&(cur)){
				face[j] = 1;
			}
			else
				face[j] = 0;
		}

		for(int j=0;j<n;j++){
			if((1<<j)&(cur)){
				for(auto z:good[j]){
					if(face[z]==0)
						poss = false;
					// face[z] = 1;
				}
				for(auto z:bad[j]){
					if(face[z]==1)
						poss = false;
					// face[z] = 0;
				}
			}
		}
		
		if(poss){
			int cnt = __builtin_popcount(cur);
			ans = max(ans, cnt);
		}
	}
	cout << ans << endl;
	return 0;
}