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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;

public class Main{
    
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        
        int n = Integer.parseInt(sc.next());
        int m = Integer.parseInt(sc.next());
        
        UnionFind uf = new UnionFind(n);
        
        for(int i=0; i<m; i++){
            int x = Integer.parseInt(sc.next())-1;
            int y = Integer.parseInt(sc.next())-1;
            int z = Integer.parseInt(sc.next());
            uf.unite(x,y);
        }
        
        System.out.println(uf.groupCount());
    }
}

class UnionFind{
    int n;
    int[] par;
    int[] size;
    
    public UnionFind(int n){
        this.n = n;
        par = new int[n];
        for(int i=0; i<n; i++){
            par[i] = i;
        }
        size = new int[n];
        Arrays.fill(size, 1);
    }
    
    public int find(int x){
        if(par[x] == x){
            return x;
        }else{
            par[x] = find(par[x]);
            return par[x];
        }
    }
    
    public boolean unite(int x, int y){
        x = find(x);
        y = find(y);
        if(x==y) return false;
        
        if(size[x]<size[y]){
            int tmp = x;
            x = y;
            y = tmp;
        }
        size[x] += size[y];
        par[y] = x;
        return true;
    }
    
    public boolean same(int x, int y){
        return find(x)==find(y);
    }
    
    public int size(int x){
        return size[find(x)];
    }
    
    public int groupCount(){
        Set<Integer> s = new HashSet<>();
        for(int i=0; i<n; i++){
            s.add(this.find(i));
        }
        return s.size();
    }
}