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;
typedef long long ll;
typedef pair<int,int> P;
typedef pair<ll,ll> Pl;
#define all(a) (a).begin(),(a).end()
#define rep(i,n) repi(i,0,n)
#define repi(i,a,b) for(ll i=(a);i<(b);i++)
#define rrep(i,n) for(ll (i)=(n)-1;(i)>=0;i--)
#define SHOW(x) cout << #x << ": " << x << endl;
const ll INF=1e18,MOD = 1e9+7;
const int iINF = 1e9;
int x[20][20],y[20][20],a[20],check[20];
int main(){
    int n,ans = 0;
    cin>>n;
    rep(i,n){
        cin>>a[i];
        rep(j,a[i]){
            cin>>x[i][j]>>y[i][j];
        }
    }
    rep(i,pow(2,n)){
        rep(j,n)check[j] = -1;
        int cnt = 0;
        bool flag = true;
        rep(j,n){
            if(i & (1 << j))check[j] = 1;
            else check[j] = 0;
        }
        rep(j,n){
            if(i & (1 << j)){
                rep(k,a[j]){
                    if(check[x[j][k]-1] == -1){
                        check[x[j][k]-1] = y[j][k];
                    }
                    else{
                        if(check[x[j][k]-1] != y[j][k]){
                            flag = false;
                            break;
                        }
                    }
                }
                if(!flag)break;
            }
        }
        rep(j,n)if(check[j] == 1)cnt++;
        if(flag)ans = max(ans,cnt);
    }
    cout<<ans<<endl;

    
}