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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;

struct UnionFind{

    vector<int> par;
    vector<int> sizes;

    UnionFind(int n):par(n),sizes(n,1){
        for(int i=0;i<n;i++){
            par[i]=i;
        }
    }
    
    int find(int x) { //xの根を得る
        if (x == par[x]) return x;
        return par[x] = find(par[x]); //根を張り替えながら再帰的に根ノードを探す
    }

    void unite(int x, int y) { //xとyの集合を併合
        x = find(x);
        y = find(y);
        if (x == y) return;
        if (sizes[x] < sizes[y]) swap(x, y);
        par[y] = x;
        sizes[x] += sizes[y];
    }

    bool same(int x, int y) {
        return find(x) == find(y);
    }

    int size(int x) {
        return sizes[find(x)];
    }
};

int main(){
    int N,M; cin >> N >> M;
    UnionFind uf(N);
    int ans=0;
    bool used[N];
    int x[M],y[M];
    for(int i=0;i<M;i++){
        int z; cin >> x[i] >> y[i] >> z;
        x[i]--;y[i]--;
        uf.unite(x[i],y[i]);
    }
    set<int> se;
    for(int i=0;i<N;i++){
        se.insert(uf.find(i));
    }
    cout << se.size() << endl;
}