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<iostream>
#include<vector>
using namespace std;
struct tree{
    vector<int> par;
    tree(int n):par(n){
        for(int i=0;i<n;i++){
            par[i] = i;
        }
    }
    int root(int x){
        if(par[x] == x) return x;
        return par[x] = root(par[x]);
    }
    void unite(int x,int y){
        int rx = root(x);
        int ry = root(y);
        if(rx == ry) return;
        par[rx] = ry;
    }
    bool same(int x,int y){
        int rx = root(x);
        int ry = root(y);
        return rx == ry;
    }
};

int main(){
    int n,m;
    cin >> n >> m;
    tree ut(n);
    for(int i=0;i<m;i++){
        int x,y,z;cin >> x >> y>>z;x--;y--;
        ut.unite(x,y);
    }
    bool par[n];
    for(int i=0;i<n;i++){
        par[i] = false;
    }
    int ans = 0;
    for(int i=0;i<n;i++){
        if(par[ut.root(i)]){

        }
        else{
            ans ++;
            par[ut.root(i)] = true;
        }
    }
    cout << ans << endl;
    return 0;
}