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 IOS std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
template <class T>
struct rbtree: tree<T,null_type,less<T>, rb_tree_tag,tree_order_statistics_node_update>{};
#define IN freopen("in.txt", "r", stdin);
#define endl '\n'
#define mp make_pair
#define pb push_back
#define lowb lower_bound
#define eb emplace_back
#define fi first
#define se second
#define all(x) x.begin(),x.end()
#define sz(x) (int)x.size()
#define mem0(a) memset((a), 0, sizeof(a))
#define mem(a, b) memset((a), (b), sizeof(a))
typedef unsigned long long ull;
typedef long long ll;
typedef vector<int> VI;
typedef vector<ll> VL;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const int inf = 0x3f3f3f3f;

const int N = 20;
int res[N];
int c[N];
int a[N][N];

int main(){
    IOS;
    //IN;
    int n; cin>>n;
    for(int i = 1; i <= n; i++){
        cin>>c[i];
        for(int j = 1; j <= c[i]; j++){
            int y, z; cin>>y>>z;
            a[i][y] = z;
        }
    }
    int ans = 0;
    for(int i = 1; i <= (1<<n)-1; i++){
        bool f = 0;
        int cnt=0;
        for(int j = 1; j <= n; j++)res[j] = 0;
        for(int j = 0; j < n; j++){
            if(i&(1<<j)){
                cnt++;
                res[j+1]++;
            }
        }
        for(int j = 1; j <= n; j++){
            if(!res[j])continue;
            for(int z = 1; z <= n; z++){
                if(z==j)continue;
                if((a[j][z] == 0 && res[z]) || (a[j][z] == 1 && !res[z])){
                    //cout<<"2==="<<z<<endl;
                    f=1;
                    break;
                }
            }
            if(f)break;
        }
        //cout<<"1===="<<cnt<<endl;
        if(!f)ans = max(ans, cnt);
    }
    cout<<ans;

    return 0;
}