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 <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<int,int>;

int main() {
    int N;
    cin >> N;
    vector<vector<int>> Z(N,vector<int>(N, 2));
    for (int i = 0; i < N; i++)
    {
        int A;
        cin >> A;
        while (A--)
        {
            int x, y;
            cin >> x >> y;
            --x;
            Z[i][x] = y;
        }
        Z[i][i] = 1;
    }
    cerr << Z[0][0] << Z[0][1] << Z[0][2] << endl;
    int honest = 0;
    vector<vector<int>> hlist(N);
    for (int i = 0; i < N; i++)
    {
        int temp = 0;
        bool hflg = true;
        for (int j = 0; j < N; j++)
        {
            if (Z[i][j] == 1 && Z[j][i] == 0)
            {
                hflg = false;
                break;
            }
            
            if ((Z[i][j] == 1 || Z[i][j] == 2) &&(Z[j][i] == 1 || Z[j][i] == 2))
            {
                temp++;
                hlist[i].push_back(j);
            }
            
        }
        if (!hflg)
        {
            continue;
        }
        
        vector<bool> flg(hlist[i].size(), true);
        for (int k = 0; k < hlist[i].size(); k++)
        {
            if (!flg[k])
            {
                continue;
            }
            
            for (int l = 0; l < hlist[i].size(); l++)
            {
                if (!flg[l])
                {
                    continue;
                }
                
                if (Z[hlist[i][k]][hlist[i][l]] == 0 && Z[i][hlist[i][l]] == 1)
                {
                    temp = 0;
                    break;
                }
                if (Z[hlist[i][k]][hlist[i][l]] == 0 && Z[i][hlist[i][l]] == 2)
                {
                    temp--;
                    flg[l] = false;
                }
                if (Z[i][hlist[i][l]] == 2 && Z[i][hlist[i][k]] == 1 && Z[hlist[i][k]][hlist[i][l]] == 2 && Z[hlist[i][l]][hlist[i][k]] == 0)
                {
                    cerr << "i=" << i << " l=" << l << " k=" << k << endl;
                    temp--;
                    flg[l] = false;
                }
                
                
            }
            if (temp == 0)
            {
                break;
            }
            
            
        }
        honest = max(honest, temp);
    }
    cout << honest << endl;
  return 0;
}