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::*;
use std::str::FromStr;
use std::cmp;

fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();

    token.parse().ok().expect("failed to parse token")
}

fn main() {
    let n: usize = read();
    let mut a: [usize; 20] = [0; 20];
    //有向グラフ -1：未証言、0：不親切な人と証言、1：正直者と証言
    let mut g: [[i16; 15]; 15] = [[-1; 15]; 15];

    for i in 0..n {
        a[i] = read();
        for _j in 0..a[i] {
            let x: usize = read();
            let y = read();
            let x = x - 1;
            g[i][x] = y;
        }
    }

    let mut ans = 0; 
    let mut bits: u16 = 0;
    while bits < (1 << n) { //2^n
        //1人につき正直者か不親切な人いずれかを1bitで割り当てる
        //3人なら3bitで表現でき、全パターン(bit)探索する場合
        //3bitは000,001,010,011,100,101,111の7パターン
        //これは、1<<3=0b1000=8となるので 
        // { i = 0 ;i < (1 << 3); i++}
        //でループさせれば全bit探索できる。
        //println!("bits:{:0>8b}, 1<<n:{:0>8b}", bits, 1<<n);

        // 1なら正直者とする
        //dの中に正直者と不親切の割当を入れる。 bitsの立っているパターンと同値
        let mut d = vec![0; n];
        for i in 0..n {
            if ((bits >> i) & 1) == 1 {
                d[i] = 1;
            }
        }
        //println!("bits:{:0>3b}, vec:{:?}", bits, d);

        let mut ok: bool = true;
        for i in 0..n {
            //証言に矛盾がないか判定する
            //正直者と仮定した人に対して証言をチェックする
            if d[i] == 1 {
                for j in 0..n {
                    if g[i][j] == -1 { continue; }
                    if g[i][j] != d[j]  { ok = false; } 
                }
            }
        }
        if ok { 
            ans = cmp::max(ans, bits.count_ones());
        }

        bits += 1;
    }
    println!("{}", ans);
}