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>
typedef long long ll;
typedef long double ld;
using namespace std;

int N;
vector<vector<pair<int, int>>> P;

int kthbit(int A, int K) { return (A >> K) & 1; }

bool possible(int n){
    vector<int> isHonest(N, -1);
    for(int i=0; i<N; i++){
        isHonest[i] = kthbit(n, i);
    }

    for(int i=0; i<N; i++){
        if(isHonest[i] == 0) continue;

        for(auto p : P[i]){
            if(isHonest[p.first] != p.second){
                return false;
            }
        }
    }
    return true;
}

int cntBit(int n){
    if(n < 2) return n;
    else return n%2 + cntBit(n/2);
}

int main(){
    cin >> N;
    P.resize(N);
    for(int i=0; i<N; i++){
        int A;  cin >> A;
        P[i].reserve(A);
        for(int a=0; a<A; a++){
            int x, y;  cin >> x >> y;
            x--;
            P[i].push_back({x, y});
        }
    }

    int ans = 0;
    for(int i=0; i<(1<<N); i++){
        if(possible(i)){
            ans = max(ans, cntBit(i));
        }
    }
    cout << ans << endl;
}