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: Runtime Error)

import java.util.Scanner
object Main extends App {
  val sc = new Scanner(System.in)
  val n, m = sc.nextInt
  val uf = new UnionFind(n)
  (1 to m).foreach { i =>
    val x, y, _ = sc.nextInt - 1
    uf.union(x, y)
  }
  val (vs, trees) = (0 to n - 1).map(uf.find(_)._1).partition(_ == -1)
  println(vs.size + trees.groupBy(identity).size)
}

class UnionFind(size: Int) {
  val Undefined = -1
  val parents = Array.fill(size)(Undefined)

  /** @retrun (root, rank) */
  def find(x: Int, rank: Int = 0): (Int, Int) = {
    parents(x) match {
      case Undefined => parents(x) = x; (x, rank)
      case p if x == p => (p, rank)
      case p => {
        val (newRoot, newRank) = find(p, rank + 1)
        parents(x) = newRoot
        (newRoot, newRank)
      }
    }
  }

  def union(x: Int, y: Int): Unit = {
    val (xRoot, xRank) = find(x)
    val (yRoot, yRank) = find(y)
    val newRoot = if (xRank > yRank) yRoot else xRoot
    if (xRoot != yRoot) {
      parents(xRoot) = newRoot
      parents(yRoot) = newRoot
    }
    parents(x) = newRoot
    parents(y) = newRoot
  }
}