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 Scala (2.11.7) (judge verdict: Accepted)

import java.util.Scanner

import scala.collection.mutable.ArrayBuffer


object Main {
  case class Edge(x: Int, y: Int, z: Long)
  
  def read() = {
    val sc = new Scanner(System.in)
    val n, m = sc.nextInt()
    //0-index化
    val edges = IndexedSeq.fill(m)(Edge(sc.nextInt() - 1, sc.nextInt() - 1, sc.nextInt()))

    (n, m, edges)
  }


  def solve2(n: Int, m: Int, edges: IndexedSeq[Edge]): Int = {
    var cnt = 0 // 連結成分数

    val nextList = IndexedSeq.fill(n)(ArrayBuffer.empty[Edge])
    for (edge <- edges) {
      val from = edge.x
      val to = edge.y
      nextList(from).append(edge)
      nextList(to).append(Edge(to, from, edge.z))
    }

    val visited = Array.fill(n)(false)
    for (i <- 0 until n) {
      if (!visited(i)) {
        visited(i) = true
        cnt += 1
        //BFS
        val open: java.util.Queue[Int] = new java.util.ArrayDeque[Int]()
        open.add(i)
        while (!open.isEmpty) {
          val current = open.remove()
          for (edge <- nextList(current)) {
            val next = edge.y
            if(!visited(next)) {
              open.add(next)
              visited(next) = true
            }
          }
        }

      }
    }
    cnt
  }

  def main(args: Array[String]): Unit = {
    val (n, m, edges) = read()
    println(solve2(n, m, edges))
  }
}