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 Rust (1.15.1) (judge verdict: Accepted)

use std::io::{self, BufRead};

fn main() {
    let mut tuples = Vec::new();

    let stdin = io::stdin();
    let mut lines = stdin.lock().lines();
    let n = lines.next().unwrap().unwrap().parse::<usize>().unwrap();
    for i in 1..n + 1 {
        let a = lines.next().unwrap().unwrap().parse::<usize>().unwrap();
        for _ in 0..a {
            let x_y = lines.next().unwrap().unwrap();
            let mut x_y = x_y.split_whitespace().map(|s| s.parse::<usize>().unwrap());
            let x = x_y.next().unwrap();
            let y = x_y.next().unwrap();
            tuples.push((i, x, y));
        }
    }

    let mut ans = 0;
    for bit in 0..1 << n {
        let mut people = vec![0; n + 1];
        let mut sum = 0;
        for i in 0..n {
            if bit & 1 << i != 0 {
                people[i + 1] = 1;
                sum += 1;
            }
        }

        let valid = !tuples
            .iter()
            .any(|t| people[t.0] == 1 && people[t.1] != t.2);

        if valid && sum > ans {
            ans = sum;
        }
    }

    println!("{}", ans);
}