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<vector>
#include<algorithm>
#include<cmath>
#include<bitset>
using namespace std;



int main(){
	int n;
	cin  >> n;
	vector<vector<pair<int,int> > > data(n);
	
	for(int i=0;i<n;++i){
		int m;
		cin >> m;
		vector<pair<int,int> > vm(m);
		for(int j=0;j<m;++j){
			cin >> vm[j].first >> vm[j].second;
			vm[j].first--;
		}
		data[i] = vm;
	}
	
	int res = 0;
	for(int i=0;i<pow(2,n);++i){
		bitset<15> bs(i);
		bool f = true;
		for(int j=0;j<n;++j){
			//cout << "--"<<j <<endl;
			if(bs[j]){
				for(int k=0;k<data[j].size();++k){
//					cout <<k<< "--"<<endl;
					int target = data[j][k].first;
					int tf = data[j][k].second;
//					cout << target <<" "<< bs[target] << " " << tf<<" "<<(bs[target] != tf)  <<endl;
					if(bs[target] != tf){
						f = false;
						break;
					}
				}
			}
		}
		//cout << bs << " " << f <<endl;
		if(f){
			res = max(res,(int)bs.count());
			//cout << bs <<" "<<bs.count()<<  endl;
		}
		
	}
	
	cout << res <<endl;
	
	return 0;
}