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 <iomanip>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <algorithm>
#include <cstdio>
#include <utility>
#include <string>
#include <cmath>
#include <cstdlib>
#include <cstring>

using namespace std;

typedef uint64_t u64;
typedef int64_t s64;
typedef uint32_t u32;
typedef int32_t s32;
typedef vector<s32> vs32;
typedef vector<u32> vu32;
typedef vector<s64> vs64;
typedef vector<u64> vu64;

const double PI=3.14159265358979323846;

#define MAX(x, y) ((x) < (y) ? (y) : (x))
#define MIN(x, y) ((x) > (y) ? (y) : (x))

#define rep(i, N) for(int i = 0; i < N; ++i)

#define CEIL(x, y) (((x) + (y) - 1) / (y))
#define MOD 1000000007ULL

vector< vs32 > t;
vector< vs32 > temp(15, vector<int>(15, -1));
int ans = 0;

int num(int x)
{
    int ret = 0;
    while (x)
    {
        ret += x & 1;
        x >>= 1;
    }
    return ret;
}

void traverse(int i, int N)
{
    rep (j, N)
    {
        if (t[i][j] != -1)
        {
            temp[i][j] = t[i][j] ^ 1;
        }
        else
        {
            temp[i][j] = -1;
        }
    }
}

void solve(int m, int N)
{
    if (N == 0)
    {
        return;
    }

    rep (i, N)
    {
        if ((m & (1 << i)) == 0)
        {
            traverse(i, N);
        }
        else
        {
            rep (j, N)
            {
                temp[i][j] = t[i][j];
            }
        }
    }

    int ttt = 0;
    rep (j, N)
    {
        for (int i = j + 1; i < N; ++i)
        {
            if (temp[i][j] != -1)
            {
                if (temp[j][i] != -1 && temp[i][j] != temp[j][i])
                {
                    if (num(m) == N) return;
                    solve(m + 1, N);
                    return;
                }
                if (temp[i][j] == 0) ttt -= 1;
            }
        }
    }

    int tt = num(m);
    ans = MAX(tt + ttt, ans);
    if (num(m) == N) return;
    solve(m + 1, N);
}

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);

    int N;
    cin >> N;

    t.resize(N, vector<int>(N, -1));

    int A, x, y;
    rep (i, N)
    {
        cin >> A;
        rep (j, A)
        {
            cin >> x >> y;
            --x;
            t[i][x] = y;
        }
    }

    solve(0, N);
    cout << ans << "\n";
    return 0;
}