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)

fn read_stdin() -> String {
    use std::io::Read;
    let mut input = String::new();
    std::io::stdin().read_to_string(&mut input).unwrap();
    input
}

fn scan<'a, I: Iterator<Item = &'a str>, T: std::str::FromStr>(tokens: &mut I) -> T {
    tokens.next().unwrap().parse().ok().unwrap()
}

fn main() {
    let input = read_stdin();
    let tokens = &mut input.split_whitespace();
    let n: usize = scan(tokens);
    let mut solver = Solver::new(n);
    for i in 0..n {
        let a: usize = scan(tokens);
        for _ in 0..a {
            let x: usize = scan(tokens);
            let y: usize = scan(tokens);
            solver.add_edge(i, x - 1, y);
        }
    }
    println!("{}", solver.solve());
}

#[derive(Debug)]
struct Solver {
    n: usize,
    edges: Vec<Vec<(usize, usize)>>,
}

impl Solver {
    fn new(n: usize) -> Solver {
        Solver {
            n: n,
            edges: vec![Vec::new(); n],
        }
    }
    fn add_edge(&mut self, i: usize, j: usize, t: usize) {
        self.edges[i].push((j, t));
    }
    fn solve(&mut self) -> usize {
        let state = vec![2; self.n];
        self.dfs(0, 0, &state)
    }
    fn dfs(&mut self, i: usize, count: usize, state: &Vec<usize>) -> usize {
        if i == self.n {
            count
        } else if state[i] == 0 {
            self.dfs(i + 1, count, state)
        } else if state[i] == 1 {
            let mut state1 = state.clone();
            let mut ok = true;
            for &(j, t) in &self.edges[i] {
                if state1[j] == t {
                    // ok
                } else if state1[j] == 1 - t {
                    ok = false;
                    break;
                } else {
                    state1[j] = t;
                }
            }
            if ok {
                self.dfs(i + 1, count + 1, &state1)
            } else {
                0
            }
        } else {
            let mut state0 = state.clone();
            state0[i] = 0;
            let tmp = self.dfs(i + 1, count, &state0);
            let mut state1 = state.clone();
            state1[i] = 1;
            let mut ok = true;
            for &(j, t) in &self.edges[i] {
                if state1[j] == t {
                    // ok
                } else if state1[j] == 1 - t {
                    ok = false;
                    break;
                } else {
                    state1[j] = t;
                }
            }
            if ok {
                std::cmp::max(tmp, self.dfs(i + 1, count + 1, &state1))
            } else {
                tmp
            }
        }
    }
}