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 fi first
#define se second
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define rrep(i, n) for (int i = 1; i <= (n); ++i)
#define drep(i, n) for (int i = (n)-1; i >= 0; --i)
#define srep(i, s, t) for (int i = s; i < t; ++i)
#define rng(a) a.begin(), a.end()
#define maxs(x, y) (x = max(x, y))
#define mins(x, y) (x = min(x, y))
#define pb push_back
#define sz(x) (int)(x).size()
#define PQ(T) priority_queue<T, v(T), greater<T>>
#define bn(x) ((1 << x) - 1)
#define v(T) vector<T>
#define vv(T) v(v(T))
using namespace std;
typedef long long int ll;
typedef unsigned uint;
typedef unsigned long long ull;
typedef pair<int, int> P;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<P> vp;
const double eps = 1e-10;
const ll LINF = 1001002003004005006ll;
const int INF = 1001001001;

int g[15][15];

int main(){
    int n;
    cin >> n;
    rep(i, n)rep(j, n) g[i][j] = -1;
    rep(i, n){
        int a;
        cin >> a;
        rep(j, a){
            int x, y;
            cin >> x >> y;
            x--;
            g[i][x] = y;
        }
    } 

    int ans = 0;

    rep(i, 1<<n){
        vi d(n);
        rep(j, n){
            d[j] = ((i>>j)&1);
        }

        bool ok = true;
        rep(j, n){
            if(d[j]==0) continue;
            rep(k, n){
                if(g[j][k] == -1) continue;
                if(g[j][k] != d[k]) ok = false;
            }
        }
        if(ok) ans = max(ans, __builtin_popcount(i));
    }

    cout << ans << endl;
}





/*
int g[15][15];

int main(){
    //読み込み
    int n;
    cin >> n;
    rep(i, n)rep(j, n) g[i][j] = -1;
    rep(i, n){
        int m;
        cin >> m;
        rep(j, m){
            int a, x;
            cin >> a >> x;
            --a;
            g[i][a] = x;
        }
    }

    //答えの変数用意
    int ans = 0;

    //処理 大きなループとして2のn乗
    rep(i, 1<<n){
        //まずは参照しやすいように各0,1をベクタに入れる
        vi d(n);
        rep(j, n) if(i>>j & 1){
            d[j] = 1;
        }
        //今回のパターンで成功するかどうかを表す変数
        bool ok = true;

        //正直者の発言だけを見てあげる        
        rep(j, n){
            if(d[j]){
                //ある人に対して発言してなければcontinue
                //発言していればそれと実際があっているか確かめる
                rep(k, n){
                    if(g[j][k] == -1) continue;
                    if(g[j][k] != d[k]) ok = false;
                }
            }
        }
        //間違いがなければ正直者の人数を数えてmax
        if(ok) ans = max(ans, __builtin_popcount(i));
    }
    cout << ans << endl;
    return 0;
}

*/