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: Runtime Error)

#include <bits/stdc++.h>
#include <boost/multiprecision/cpp_int.hpp>
using bint = boost::multiprecision::cpp_int;

int main()
{
    int N, M;
    std::cin >> N >> M;
    std::vector<int> Xs(M);
    std::vector<int> Ys(M);
    std::vector<int> odd(M);
    for (int i = 0; i < M; i++) {
        int z;
        std::cin >> Xs.at(i) >> Ys.at(i) >> z;
        odd.at(i) = z % 2;
    }
    std::vector<int> card(N, -1);
    int tree_num = 0;

    card.at(0) = 0;

    std::vector<int> indexes{};
    for (int i = 0; i < N; i++) {
        indexes.emplace_back(i);
    }

    while (true) {
        std::vector<int> indexes_new{};
        for (const auto& i : indexes) {
            if (card.at(Xs.at(i) - 1) == -1 and card.at(Ys.at(i) - 1) == -1) {
                indexes_new.emplace_back(i);
            } else if (card.at(Xs.at(i) - 1) == -1) {
                card.at(Xs.at(i) - 1) = card.at(Xs.at(i) - 1);
            } else if (card.at(Ys.at(i) - 1) == -1) {
                card.at(Ys.at(i) - 1) = card.at(Xs.at(i) - 1);
            }
        }
        if (indexes == indexes_new) {
            tree_num++;
            for (int i = 0; i < N; i++) {
                if (card.at(i) == -1) {
                    card.at(i) = tree_num;
                }
            }
        }

        indexes = indexes_new;
        bool is_all_decided = true;
        for (const auto& i : card) {
            if (i == -1) {
                is_all_decided = false;
                break;
            }
        }
        if (is_all_decided) {
            break;
        }
    }
    std::cout << tree_num << std::endl;
    return 0;
}