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>
#define rep(i, n) for(int i=0; i<(n); ++i)
#define rep2(i, s, n) for(int i=s; i<(n); ++i)

using namespace std;

typedef long long ll;

static const int INTINF = (2147483647);
static const ll LLINF = (9223372036854775807);
static const int MAX = 100001;
static const ll MOD = 1000000007;

//--global--//

int n;
vector<pair<int, int> > vv[16];
int cnt = 0;
//----------//

void solve(vector<int> v){
    int res = 0;
    rep(i, v.size()){
        //if(i)cout << " ";
        //cout << v[i];
        if(v[i]){
            int u = i;
            rep(j, vv[u].size()){
                pair<int, int> f = vv[u][j];
                int neU = f.first;
                int neT = f.second;
                //cout << "->" << neU << ", " << neT;
                if(v[neU] != neT){
                    return;
                }
            }
            res++;
        }
    }
    //cout << endl;
    cnt = max(res, cnt);
}

void cir(vector<int> v){
    if(v.size() == n){
        solve(v);
        return;
    }
    v.push_back(0);
    cir(v);
    v.erase(v.end()-1);
    v.push_back(1);
    cir(v);
    return;
}

int main(int argc, const char * argv[]) {
    //提出時、消す----//
    //--------------//

    cin >> n;
    rep(i, n){
        int a;
        cin >> a;
        rep(j, a){
            int x, y;
            cin >> x >> y;
            x--;
            vv[i].push_back(make_pair(x, y));
        }
    }

    vector<int> v;
    cir(v);

    cout << cnt << endl;


    return 0;

}