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>
using namespace std;
 
#define rep(i, n) for (int i = (int)(0); i < (int)(n); ++i)
#define reps(i, n) for (int i = (int)(1); i <= (int)(n); ++i)
#define rrep(i, n) for (int i = ((int)(n)-1); i >= 0; i--)
#define rreps(i, n) for (int i = ((int)(n)); i > 0; i--)
#define irep(i, m, n) for (int i = (int)(m); i < (int)(n); ++i)
#define ireps(i, m, n) for (int i = (int)(m); i <= (int)(n); ++i)
#define FOR(e, c) for (auto &e : c)
#define SORT(v, n) sort(v, v + n);
#define vsort(v) sort(v.begin(), v.end());
#define rvisort(v) sort(v.begin(), v.end(), greater<int>());
#define all(v) v.begin(), v.end()
#define mp(n, m) make_pair(n, m);
#define cout(d) cout<<d<<endl;
#define coutd(d) cout<<std::setprecision(10)<<d<<endl;
#define cinline(n) getline(cin,n);
#define replace_all(s, b, a) replace(s.begin(),s.end(), b, a);
// #define int long long
#define PI (acos(-1))
#define FILL(v, n, x) fill(v, v + n, x);
#define sz(x) int(x.size())
 
using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using pii = pair<int, int>;
using ul = unsigned long;
 
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
 
ll GCD(ll a, ll b) { return b ? GCD(b, a%b) : a; }
ll LCM(ll a, ll b) { return a / GCD(a, b)* b; }
 
const int dy[] = {0, 1, 0, -1, -1, 1, 1, -1};
const int dx[] = {1, 0, -1, 0, 1, 1, -1, -1};
 
inline bool inside(int y, int x, int H, int W) {
    return (y >= 0 && x >= 0 && y < H && x < W);
}
 
const int INF = 1e9;
const int MOD = 1e9+7;
const ll LINF = 1e18;

signed main()
{
  cin.tie( 0 ); ios::sync_with_stdio( false );
  int n; cin>>n;
  vi a(n);
  vector<vector<pii>> b(n);
  rep(i,n){
    cin>>a[i];
    rep(j,a[i]){
      int x,y; cin>>x>>y;
      x--; y;
      b[i].emplace_back(x,y);
    }
  }
  int ans=0;
  for(int bit=0; bit<(1<<n); bit++){
    bool ok=true;
    rep(i,n){
      if(bit&(1<<i)){
        rep(j,a[i]){
          int x=b[i][j].first;
          int y=b[i][j].second;
          if((bit & (1<<x))){
            if(y==0) ok=false;
          }else{
            if(x==1) ok=false;
          }
        }
      }
    }
    if(ok){
      int cnt=0;
      rep(i,n){
        if(bit&(1<<i)) cnt++;
        chmax(ans,cnt);
      }
    }
  }
  cout<<ans<<endl;
}