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"
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;
typedef pair<P,ll> Pll;
typedef pair<P,P> PP;
#define rep(i,a,n) for(ll i = a;i < n;i++)
#define repr(i,a,n) for(ll i = n-1;i >= a;i--)
#define LINF (ll)1e17
#define INF 1e9
#define MOD (ll)1e9+7
#define sMOD 1e5
#define fs first
#define sc second
#define setprecision(d) cout << fixed << setprecision(d);

ll gcd(ll a,ll b){
	if(!b) return a;
	return gcd(b,a%b);
	
}

ll lcm(ll a,ll b){
	return a * b / gcd(a,b);
}





signed main(){
	ll n,ans = 0;
	cin >> n;

	vector<vector<ll>> x(n),y(n);

	rep(i,0,n){
		ll a;
		cin >> a;

		rep(j,0,a){
			ll b,c;
			cin >> b >> c;
			x[i].push_back(b);
			y[i].push_back(c);
		}
	}

	rep(i,0,1 << n){
		ll temp = i;
		ll check[n] = {};
		ll temp_array[n] = {};
		bool flag = true;

		rep(j,0,n){
			temp_array[j] = (temp >> j) & 1;
			check[j] = temp_array[j];
		}

		rep(j,0,n){
			rep(k,0,x[j].size()){
				if(temp_array[j] && check[x[j][k]-1] != y[j][k]){ 
					flag = false;
					break;

				}
				else if(!temp_array[j] && check[x[j][k]-1] == y[j][k]){ //j人目が嘘つきと仮定

					flag = false;
					break;

				}
			}
		}

		if(flag){
			ll memo = 0;
			rep(j,0,n){
				memo += temp_array[j];
			}
			ans = max(ans,memo);
		}

	}

	cout << ans << endl;
	
    return 0;
}