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 stdin = io::stdin();
    let mut lines = stdin.lock().lines();
    let n: usize = lines.next().unwrap().unwrap().parse().unwrap();

    let mut honest: Vec<Vec<usize>> = vec![Vec::new(); n];
    let mut dishonest: Vec<Vec<usize>> = vec![Vec::new(); n];
    for i in 0..n {
        let a: usize = lines.next().unwrap().unwrap().parse().unwrap();
        for _ in 0..a {
            let xs: Vec<usize> = lines.next().unwrap().unwrap().split(' ').map(|x| x.parse().unwrap()).collect();
            let (x, y) = (xs[0] - 1, xs[1]);
            if y == 0 {
                dishonest[i].push(x);
            } else {
                honest[i].push(x);
            }
        }
    }

    let mut ans = 0;

    for i in 0..((2 as u32).pow(n as u32)){
        let mut ok = true;
        for j in 0..n {
            // print!("i & (1<<j) = {}, ", i & (1<<j));
            // print!("i = {}, ", i);
            // print!("j = {}, ", j);
            // println!("1<<j = {}", 1<<j);
            if i & (1<<j) != 0 {
                let p1 = honest[j].iter().all(|&h| i & (1 << h) != 0);
                let p2 = dishonest[j].iter().all(|&d| i & (1 << d) == 0);
                if !p1 || !p2 {
                    ok = false;
                    break;
                }
            }
        }
        if ok {
            let tmp = (0..n).filter(|&k| i & (1 << k) != 0).count();
            if ans < tmp {
                ans = tmp;
            }
        }
    }

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