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 <iostream>
#include <fstream>
#include <bits/stdc++.h>
#include <map>
#define pb push_back
#define ll long long
#define pii pair<int,int>
#define pll pair<long,long>
using namespace std;
ifstream fin("f.in");
ofstream fout("f.out");
int N,ans,A[20];
int X[20][20],y[20][20];
//vector <dubla> v(30);

int counter(int x)
{
    if(x==0)
        return 0;
    else
        return counter(x>>1)+(x&1);
}
int solve(int x)
{
    int rez=0;
    bool aux=true;
    for(int j=1;j<=N;j++)
    {
        if(x&(1<<(j-1))==0)
            continue;
        for(int k=1;k<=A[j];k++)
        {
            if((x>>(X[j][k]-1)^1)==1)
            {
                aux=false;
                break;
            }
            if(((x>>(X[j][k]-1))&1)) aux=false;
            break;
        }
    }
     if(aux)
            return counter(rez);
    else
        return 0;

}
int main()
{
    int i,j,k;
    cin>>N;
    for(i=1;i<=N;i++)
    {
        cin>>A[i];
        for(j=1;j<=A[i];j++)
        {
            cin>>X[i][j]>>y[i][j];
        }
    }
    for(i=1;i<(1<<N);i++)
    {
        //solve(i);
        ans=max(ans,solve(i));
    }
    cout<<ans;
    fin.close();
    fout.close();
    return 0;
}