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++ (Clang 10.0.0) (judge verdict: Wrong Answer)

#include <iostream>
#include <vector>
using namespace std;

#define ll long long int
#define rep0(i,n) for(ll i=0;i<=(ll)(n);i++)
#define rep1(i,n) for(ll i=1;i<=(ll)(n);i++)
#define repd(i,n) for(ll i=n-1;i>=0;i--)

int main(){
    int n;
    cin >> n;
    int x[n+1][n+1];
    int y[n+1][n+1];
    int a[n+1];
    bool all_not_told = true;
    rep1(i, n){
        cin >> a[i];
        if(a[i] > 0) all_not_told = false;
        rep1(j, a[i]){
            cin >> x[i][j] >> y[i][j];
        }
    }
    if(all_not_told){
        cout << n << endl;
        return 0;
    }
    
    

    int max_honest_count = 0;
    // {0, 1, ..., n-1} の部分集合の全探索
    for(int bit = 0; bit < (1<<n); ++bit){
        if(bit == 0) continue;//全員真偽不明
        
        bool ishonest[n+1];

        int honest_count = 0;
        rep0(i,n-1){
            if (bit & (1<<i)) { // i が bit に入るかどうか
                ishonest[i+1] = true;
                honest_count++;
            }
        }    

        bool contradiction = false;
        rep1(i,n){
            if(ishonest[i]){
                rep1(j,a[i]){
                    if(y[i][j] == 1){   //honest
                      if(!ishonest[x[i][j]]){
                          contradiction = true;
                          break;
                      }
                    }else{  //not know whether honest or a lier
                        if(ishonest[x[i][j]]){
                          contradiction = true;
                          break;
                      }
                    }
                }
            }
        }
        if(contradiction) continue;
        if(honest_count > max_honest_count){
            max_honest_count = honest_count;
        }
    }
    cout << max_honest_count << endl;
    return 0;
}