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)

#![allow(unused_imports)]
#![allow(dead_code)]

macro_rules! input_byline {
    ($($r:tt)* ) => {
        let stdin = std::io::stdin();
        let mut q = std::collections::VecDeque::new();
        let mut next = || {
            if q.is_empty() {
                let mut s = String::new();
                let _ = stdin.read_line(&mut s);
                for v in s.split_whitespace() {
                    q.push_back(v.to_string());
                }
            }
            q.pop_front().unwrap()
        };
        input_inner!{next, $($r)*}
    };
}
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        let mut next = || { iter.next().unwrap() };
        input_inner!{next, $($r)*}
    };
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}
macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};
    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}
macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };
    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };
    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };
    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };
    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}

use std::cmp::{max, min};
use std::collections::{BinaryHeap, HashMap};
 
type Graph = Vec<Vec<usize>>;
const M: usize = 1e9 as usize + 9;

fn main() {
    input_byline! {
        n: usize,
    }
 
    let mut v = vec![vec![]; n];
    for i in 0..n {
        input_byline! {
            a: usize,
        }
        for j in 0..a {
            input_byline! {
                x: usize1,
                y: usize,
            }
            v[i].push((x, y));
        }
    }
 
    let mut ans = 0;
    for i in 0..1 << n {
        let mut ok = true;
        for j in 0..n {
            if (i >> j) & 1 == 0 {
                continue;
            }
            for k in 0..v[j].len() {
                if (i >> v[j][k].0) & 1 != v[j][k].1 {
                    ok = false
                }
            }
        }
        if ok {
            ans = max(ans, (i as u32).count_ones());
        }
    }
    println!("{}", ans);
}