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<iostream>
#include<vector>
#include<algorithm>
#include<map>
#include<cstring>
#include<math.h>
#include<bitset>
#include<queue>
#include<set>
#include<iomanip>
#include<math.h>
#include<assert.h>
// #include<bits/stdc++.h>
using namespace std;
typedef long long ll;
constexpr long long int INFLL = 1001001001001001LL;
constexpr int INFI = 1000000007;
const int mod = 1000000007;

struct mint {
  ll x; // typedef long long ll;
  mint(ll x=0):x((x%mod+mod)%mod){}
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod-a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) {
    (x *= a.x) %= mod;
    return *this;
  }
  mint operator+(const mint a) const {
    mint res(*this);
    return res+=a;
  }
  mint operator-(const mint a) const {
    mint res(*this);
    return res-=a;
  }
  mint operator*(const mint a) const {
    mint res(*this);
    return res*=a;
  }
  mint pow(ll t) const {
    if (!t) return 1;
    mint a = pow(t>>1);
    a *= a;
    if (t&1) a *= *this;
    return a;
  }

  // for prime mod
  mint inv() const {
    return pow(mod-2);
  }
  mint& operator/=(const mint a) {
    return (*this) *= a.inv();
  }
  mint operator/(const mint a) const {
    mint res(*this);
    return res/=a;
  }
};

// mint N;

struct combination {
  vector<mint> fact, ifact;
  combination(int n):fact(n+1),ifact(n+1) {
    assert(n < mod);
    fact[0] = 1;
    for (int i = 1; i <= n; ++i) fact[i] = fact[i-1]*i;
    ifact[n] = fact[n].inv();
    for (int i = n; i >= 1; --i) ifact[i-1] = ifact[i]*i;
  }
  mint operator()(int n, int k) {
    if (k < 0 || k > n) return 0;
    return fact[n]*ifact[k]*ifact[n-k];
  }
};

// combination c(N);
// mint ans = c(N,n);
// cout << ans.x << endl;

// -------------------------------------------------------------

int main(){
  int N;
  cin >> N;
  vector<vector<int>> A(N,vector<int> (N,0));
  for(int i=0; i<N; i++){
    int a;
    cin >> a;
    for(int j=0; j<a; j++){
      int x,y;
      cin >> x >> y;
      x--;
      if(y == 1){
        A[i][x] = 2;
      }
      if(y == 0){
        A[i][x] = 1;
      }
    }
  }
  int ans=0;
  for(int bit=0; bit<(1<<N); bit++){
    vector<int> h(N,0);
    bool can=true;
    for(int i=0; i<N; i++){
      if(bit & (1<<i)){
        h[i] = 2;
      }else{
        h[i] = 1;
      }
    }
    for(int i=0; i<N; i++){
      if(bit & (1<<i)){
        for(int j=0; j<N; j++){
          if(A[i][j] == 1){
            if(h[j]==0 || h[j]==1){
              h[j] = 1;
            }else{
              can = false;
            }
          }
          if(A[i][j] == 2){
            if(h[j]==0 || h[j]==2){
              h[j] = 2;
            }else{
              can = false;
            }
          }
        }
      }
    }
    if(can){
      ans = max(ans,__builtin_popcount(bit));
    }
  }
  cout << ans << endl;
  return 0;
}