Problem:
Score : 500 points

Problem Statement

There are N cards placed face down in a row. On each card, an integer 1 or 2 is written.

Let A_i be the integer written on the i-th card.

Your objective is to guess A_1, A_2, ..., A_N correctly.

You know the following facts:

For each i = 1, 2, ..., M, the value A_{X_i} + A_{Y_i} + Z_i is an even number.

You are a magician and can use the following magic any number of times:

Magic: Choose one card and know the integer A_i written on it. The cost of using this magic is 1.

What is the minimum cost required to determine all of A_1, A_2, ..., A_N?

It is guaranteed that there is no contradiction in given input.

Constraints

All values in input are integers.

2 \leq N \leq 10^5

1 \leq M \leq 10^5

1 \leq X_i < Y_i \leq N

1 \leq Z_i \leq 100

The pairs (X_i, Y_i) are distinct.

There is no contradiction in input. (That is, there exist integers A_1, A_2, ..., A_N that satisfy the conditions.)

Input

Input is given from Standard Input in the following format:

N M
X_1 Y_1 Z_1
X_2 Y_2 Z_2
\vdots
X_M Y_M Z_M

Output

Print the minimum total cost required to determine all of A_1, A_2, ..., A_N.

Sample Input 1

3 1
1 2 1

Sample Output 1

2

You can determine all of A_1, A_2, A_3 by using the magic for the first and third cards.

Sample Input 2

6 5
1 2 1
2 3 2
1 3 3
4 5 4
5 6 5

Sample Output 2

2

Sample Input 3

100000 1
1 100000 100

Sample Output 3

99999

Solution in Rust (1.15.1) (judge verdict: Runtime Error)

use std::io::stdin;

fn solve(matrix: Vec<Vec<Option<u64>>>,num_node: u32) -> u32 {
    assert_eq!(matrix.len(), matrix[0].len());

    // let num_node = matrix.len();

    let mut visited: Vec<bool> = vec![false; num_node as usize];
    let mut count = 0;

    while let Some(root_index) = (0..num_node as usize).find(|&i| !visited[i]) {
        count += 1;
        let mut index_stack: Vec<u32> = Vec::new();
        visited[root_index] = true;
        index_stack.push(root_index as u32);

        while let Some(index) = index_stack.pop() {
            let indices: Vec<usize> = visited
                .iter()
                .enumerate()
                .filter(|&(_, &c)| !c)
                .map(|(i, _)| i)
                .collect();

            for i in indices.into_iter() {
                if let Some(_) = matrix[index as usize][i] {
                    if !visited[i] {
                        index_stack.push(i as u32);
                        visited[i] = true;
                    }
                }
            }
        }
    }
    return visited.iter().filter(|&&c| !c).count() as u32 + count;
}

fn main() {
    let mut input = String::new();
    let _ = stdin().read_line(&mut input).unwrap();
    let n_m = input
        .trim()
        .split(" ")
        .map(|s| s.to_string().parse::<u32>().unwrap())
        .collect::<Vec<u32>>();

    let num_node = n_m[0];
    let m = n_m[1];

    let mut graph: Vec<Vec<Option<u64>>> = vec![vec![None; num_node as usize]; num_node as usize];

    for (x, y, z) in (1..(m+1)).map(|_| {
        let mut input = String::new();
        let _ = stdin().read_line(&mut input).unwrap();
        let input: Vec<String> = input.trim().split(" ").map(|s| s.to_string()).collect();
        assert_eq!(3, input.len());
        (
            input[0].parse::<u32>().unwrap(),
            input[1].parse::<u32>().unwrap(),
            input[2].parse::<u64>().unwrap(),
        )
    }) {
        let (i, j) = ((x - 1) as usize, (y - 1) as usize);
        graph[i][j] = Some(z);
        graph[j][i] = Some(z);
    }

    println!("{}", solve(graph, num_node));
}