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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

#define rep(i, a) for (int i = 0; i < (a); i++ )
#define FOR(i, m, n) for(int i = m; i < n; i++)
 
typedef pair<int,int> P;
typedef long long ll;
 
const int INF = 1001001001;
const int MOD = 1000000007;
const double PI=acos(-1);

int main(){
    int N;
    cin >> N;
    vector<vector<P>> A(N,vector<P>(0));

    rep(i,N) {
        int a;
        cin >> a;
        rep(j,a) {
            int x,y;
            cin >> x >> y;
            x--;
            A[i].push_back(make_pair(x,y));
        }
    }
    
    int ans = 0;
    for (int bit = 0; bit< 1<<N; bit++) {
        bool t = true;
        rep(i,N) {
            if (bit & 1<<i) {
                rep(j,A[i].size()) {
                    int x,y;
                    x = A[i][j].first;
                    y = A[i][j].second;
                    if (y==0 and bit & 1<<x) {
                        t = false;
                    }
                    else if (y==1 and !(bit & 1<<x)) {
                        t = false;
                    }
                }
            }
        }
        int tmp;
        if (t) {
            tmp = 0;
            rep(i,N) {
                if (bit & 1<<i) {
                    tmp++;
                }
            }

        }
        ans = max(ans,tmp);
    }
    cout << ans << endl;
}