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 5.4.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
//#include <stdlib.h>
//#include <limits.h>
//#include <math.h>
#include <stdbool.h>

#define N 100001

typedef struct UnionFindTree{
  int parent;
  int size;
  bool reveal;
} uft;

uft t[N];
int ans = 0;

void init(int n) {
  for (int i=0; i<n; i++) {
    t[i].parent = i;
    t[i].size = 1;
    t[i].reveal = false;
  }
}

int root(int a) {
  if (t[a].parent == a) return a;
  else return root(t[a].parent);
}

void unite(int a, int b) {
  int ra=root(a), rb=root(b);
  if (t[ra].size < t[rb].size) {
    t[rb].parent = ra;
    t[ra].size += t[rb].size;
  } else {
    t[ra].parent = rb;
    t[rb].size += t[ra].size;
  }
}

void dfs(int a) {
  if (t[a].reveal == true) return ;
  if (t[a].parent == a) {
    t[a].reveal = true;
    ans++;
    return ;
  }
  dfs(t[a].parent);
  t[a].reveal = true;
  return ;
}

int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  init(n);
  int x, y, z;
  for (int i=0; i<m; i++) {
    scanf("%d%d%d", &x, &y, &z);
    x--;
    y--;
    unite(x, y);
  }
  for (int i=0; i<n; i++) {
    if (t[i].reveal) continue;
    dfs(i);
  }
  printf("%d\n", ans);
  return 0;
}