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 C++ (GCC 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
struct unionfind{
  int par[100010]; // 親ノード  必要に応じて要素数を変えよう
  int rank[100010]; // ランク　必要に応じて要素数を変えよう
  unionfind(int n){
    init(n);
  }
  void init(int n){
    for(int i=0;i<=n;i++){
      par[i]=i;
      rank[i]=1;
    }
  }
  int root(int x) {
    if (par[x] == x) {
      return x;
    }
    else {
      return par[x] = root(par[x]);
    }
  }
  int size(int x){
    return rank[root(x)];
  }
  bool same(int x,int y) {
    return root(x) == root(y);
  }
  void unite(int x,int y) {
    x=root(x);
    y=root(y);
    if(x==y){
      return;
    }
    if(rank[x]<rank[y]){
      swap(x,y);
    }
    rank[x]=rank[x]+rank[y];
    par[y]=x;
  }
};
int main(){
  int n,m;
  cin >> n >> m;
  unionfind tree(n);
  int i;
  for(i=0;i<m;i++){
    int x,y,z;
    cin >> x >> y >> z;
    x--,y--;
    tree.unite(x,y);
  }
  int ans=0;
  vector<int> g(100010,0);
  for(i=0;i<n;i++){
    if(g[tree.root(i)]==0){
      ans++,g[tree.root(i)]=1;
    }
  }
  cout << ans << endl;
}