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: Wrong Answer)

#[allow(unused_imports)]
use std::cmp::*;
use std::collections::HashMap;
#[allow(unused_imports)]
use std::collections::*;
// https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8

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
    };

    // 3
    // 1 2
    // 1 3
    // 1 4
    // のようなタイプ
    ($next:expr, [ $t:tt ]) => {{
        let len = read_value!($next, usize);
        (0..len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    }};

    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}

fn main() {
    input! {
        m: usize,
        // [type]だけなら一行目の数字を読み込んでその行数分指定した型を読み込む
        v: [[(usize,usize)];m]
    }

    // v[m][n] m番目の人物のn個目の情報 .0の人物が .1である

    let bit = 1 << m;

    let mut max_honestman = 0;

    // 各ビットが、1ならその人物がkind,0ならunkindを表している
    for i in 0..bit {
        // a番目の人物がtrueかどうかは i & 1 << aでいける
        // 情報が矛盾しないかどうかはO(n^2)でいける
        let mut is_contradict = false;

        // 仮定した内容を保存する
        let mut assume: HashMap<usize, usize> = HashMap::new();
        for j in 0..m {
            let j_assume = i >> (m - j - 1) & 1;

            // 仮定した情報を入れる
            if !assume.contains_key(&j) {
                assume.insert(j + 1, j_assume);
            } else {
                let last_info = assume.get(&j).unwrap();
                if last_info != &1 {
                    is_contradict = true;
                    break;
                }
            }
            if j_assume == 1 {
                // 正直者

                // 仮定した人が持つ情報に基づいて生成された仮定を入れる
                let info_length = v[j].len();
                for k in 0..info_length {
                    if !assume.contains_key(&v[j][k].0) {
                        assume.insert(v[j][k].0, v[j][k].1);
                    } else {
                        let last_info = assume.get(&v[j][k].0).unwrap();
                        if last_info != &(v[j][k].1) {
                            is_contradict = true;
                            break;
                        }
                    }
                }
            } else {
                // 嘘つき

                // 仮定した人が持つ情報に基づいて生成された仮定を入れる
                let info_length = v[j].len();
                for k in 0..info_length {
                    if !assume.contains_key(&v[j][k].0) {
                        assume.insert(v[j][k].0, 1 - v[j][k].1);
                    } else {
                        let last_info = assume.get(&v[j][k].0).unwrap();
                        if last_info == &(v[j][k].1) {
                            is_contradict = true;
                            break;
                        }
                    }
                }
            }

            if is_contradict {
                break;
            }
        }

        if !is_contradict {
            let mut count = 0;
            for c in 0..m {
                if i >> (m - c - 1) & 1 == 1 {
                    count += 1;
                }
            }

            if count > max_honestman {
                max_honestman = count;
            }
        }
    }

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