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 <vector>
#include <algorithm>
#include <string.h>
#include <iomanip>
#include <map>
#include <cstdlib>
#include <cctype>
#include <ctype.h>
#define ll long long
using namespace std;
int main()
{
    int n;
    cin >> n;
    int a[n];
    int x[20][20];
    int y[20][20];
    for (int i = 0; i < n; i++)
    {
        cin >> a[i];
        for (int j = 0; j < a[i]; j++)
        {
            cin >> x[i][j] >> y[i][j];
        }
    }
    int ans = 0;
    for (int bit = 0; bit < (1 << n) - 1; bit++)
    {
        int jud = 1;
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < a[i]; j++)
            {
                if(bit &(1<<i))
                {
                    if(!(bit&(1<<x[i][j])))
                    {
                        if(y[i][j]==1)
                        {
                            jud = 0;
                        }
                    }
                }
                
            }
        }
        if (jud == 1 && ans <= __builtin_popcount(bit))
        {
            ans = __builtin_popcount(bit);
        }
    }
    cout << ans << endl;

    return 0;
}