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 <bits/stdc++.h>

#define REP(i, n) for(int i = 0; i < (int)(n); i++)
#define FOR(i, m, n) for(int i = m;i < n;i++)
#define SORT(v, n) sort(v, v+n)
#define pb(a) push_back(a)
#define eb(a) emplace_back(a)
#define vi vector<int>
#define ddvi vector<vector<int>>

using namespace std;
typedef long long ll;
typedef pair<int, int> P;

int main(){
cin.tie(0);
ios::sync_with_stdio(false);
// ここに処理を記述

	int N;
	cin >> N;

	vector<vector<P>> A(15,vector<P>(0));
	int temp = 0;
	int x,y;
	for(int i=0; i<N; i++){
		cin >> temp;
		REP(j,temp){
			cin >> x >> y;
			A[i].eb(make_pair(x-1,y));
			// cout << A[i][j].first << A[i][j].second << "\n";
		}
	}

	int ans=0;

	FOR(i,0,(1<<N)+1){
		bool flg = true;
		int honest = 0;

		// cout << bitset<15>(i) << "\n";

		REP(j,N){
			REP(k,A[j].size()){
				int bit = (i >> A[j][k].first) & 1;
				// int bit = i & (1 << A[j][k].first);
				if(i & (1<<j)){
					if(bit != A[j][k].second) flg = false;
				}
				else{
					if(bit == A[j][k].second) flg = false;
				}
				if(!flg){
				// cout << bitset<15>(bit) << " ";
				// cout << bitset<15>(i) << "/" << j << "says" << A[j][k].first << "is" << A[j][k].second << "\n";
				break;
				}
			}
			if(!flg) break;
		}
		if(flg){
			REP(l,N){
				if(i & (1<<l)){
					honest++;
				}
			}
			ans = max(ans, honest);
		}
	}

	cout << ans << "\n";

return 0;
}