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: Accepted)

#include <iostream>
using namespace std;

int N;
int A[20];
int x[20][20];
int y[20][20];

int counter(int x) {
    if(x == 0) return 0;
    return counter(x >> 1) + (x & 1);
}

void solve() {
    int ans = 0;
    // 人数分 bit を立てる
    // 人数分の真偽パターン
    for(int bits = 1; bits < (1 << N); bits++) {
        bool ok = true;
        //cerr << bits << endl;
        for(int i = 1; i <= N; i++) {
            // i-1 はbitshift数を表す
            // 1 を 0 bitshiftすると 1 001
            // 1 を 1 bitshiftすると 2 010
            // 1 を 2 bitshiftすると 4 110
            // & は両方が1のときのみ1とする
            if(!(bits & (1 << (i-1)))) continue;
            // 真実を証言している場合のみ検証
            for(int j = 1; j <= A[i]; j++) {
                // ^ : XOR
                // 右bitシフトしている
                // bitですでにその人が真か偽か仮定しているのでそれと反して(XOR)いたら矛盾
                //cerr << "\t" << i << "," << j  << ":" << (bits >> (x[i][j]-1)) << endl;
                if(((bits >> (x[i][j]-1)) & 1) ^ y[i][j]) ok = false;
            }
        }
        if(ok) ans = max(ans, counter(bits));
    }
    cout << ans << endl;
    return;
}

int main() {
    cin >> N;
    for(int i = 1; i <= N; i++) {
        cin >> A[i];
        for(int j = 1; j <= A[i]; j++) {
            cin >> x[i][j] >> y[i][j];
        }
    }
    solve();
    return 0;
}