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>

using namespace std;

typedef long long ll;

mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());

int main() {
#ifdef LOCAL
    freopen("input.txt", "r", stdin);
#endif
    ios::sync_with_stdio(0);
    cin.tie(0);
    int n;
    cin >> n;
    vector <pair <int, int> > t[n];
    for (int i = 0; i < n; i++) {
        int k;
        cin >> k;
        for (int j = 0; j < k; j++) {
            int x, y;
            cin >> x >> y;
            x--;
            t[i].push_back({x, y});
        }
    }
    int ans = 0;
    for (int msk = 0; msk < (1 << n); msk++) {
        bool bad = false;
        vector <int> honest(n, false);
        for (int j = 0; j < n; j++) {
            if (msk & (1 << j)) {
                honest[j] = true;
            }
        }
        for (int i = 0; i < n; i++) {
            for (auto p : t[i]) {
                int x = p.first;
                int y = p.second;
                if (honest[i]) {
                    if (y != honest[x]) {
                        bad = true;
                    }
                } else {
                    if (y == honest[x]) {
                        bad = true;
                    }
                }
            }
        }
        if (!bad) {
            ans = max(ans, __builtin_popcount(msk));
        }
    }
    cout << ans << '\n';
    return 0;
}