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;
// Welcome to my source code!

int main() {
    int n;
    cin >> n;
    int a[n];
    pair<int, int> xy[n][n];
    for (int i = 0; i < n; i++) {
        cin >> a[i];
        for (int j = 0; j < a[i]; j++) {
            int x, y;
            cin >> x >> y;
            xy[i][j] = pair<int, int>(x - 1, y);
        }
    }
    int ans = 0;
    for (int bit = 0; bit < (1 << n); bit++) {
        int cnt = 0;
        bool flag = true;
        for (int i = 0; i < n; i++) {
            if (!(bit & (1 << i))) continue;
            for (int j = 0; j < a[i]; j++) {
                if (bit & (1 << xy[i][j].first)) {
                    if (xy[i][j].second == 0) flag = false;
                    else cnt++;
                } else {
                    if (xy[i][j].second == 1) flag = false;
                }
                if (!flag) {
                    cnt = 0;
                    break;
                }
            }
            if (!flag) break;
        }
        ans = max(ans, cnt);
    }
    cout << ans << endl;
}