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;

void solve(int N,vector<vector<int>> &x,vector<vector<int>> &y,vector<vector<int>> &hito,int &ans){
	for(int i=0;i<hito.size();i++){
		bool pro =true;
		for(int k=0;k<N;k++){
			if(hito.at(i).at(k)==1)
        for(int j=0;j<x.at(k).size();j++)
          if(hito.at(i).at(x.at(k).at(j)) != y.at(k).at(j) ){pro =false;break;}
			if(!pro)break;
		}
		if(pro){
			int cu=0;
			for(int j=0;j<N;j++)if(hito.at(i).at(j)==1)cu++;
			ans=max(ans,cu);
		}
	}
}

int main(){
	int N; cin >> N;
	vector<vector<int>> x(N);
	vector<vector<int>> y(N);
	int a;
	int x1,y1;
	for(int i=0;i<N;i++){
		cin >> a;
		for(int j=0;j<a;j++){
			cin >> x1>> y1;
			x.at(i).push_back(x1-1);
			y.at(i).push_back(y1);
		}
	}
	int p =(1<<16);
	vector <vector<int>> hito(p,vector<int>(16,0));
	
	for(int t=0;t<(1<<16);t++){
		bitset<16> s(t);
		for(int i=0;i<N;i++){
			if(s.test(i)){
				hito.at(t).at(i)=1;
			}
		}
	}
	
	int ans=0;
	solve(N,x,y,hito,ans);

	cout<<ans<<endl;
}