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: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <memory.h>
#include <limits.h>
#include <math.h>

typedef struct d_t {
    int x, y;
} DATA;

int qsort_data_u(const void *a, const void *b)
{
    return (*(DATA *) a).x - (*(DATA *) b).x;
}

int main(int argc, char *argv[])
{
    int n, m;
    int x, y, z;
    int root[100000];
    int remap[100000];
    DATA d[100000];
    int i;
    int p;
    int ans;

    scanf("%d %d", &n, &m);

    for (i = 0; i < m; i++) {
        scanf("%d %d %d", &x, &y, &z);
        x--;
        y--;
        // x < y (制約)
        d[i].x = x;
        d[i].y = y;
        // zは要らない
    }

    for (i = 0; i < n; i++) {
        root[i] = i;
        remap[i] = i;
    }

    qsort(d, m, sizeof(DATA), qsort_data_u);

    // Union
    for (i = 0; i < m; i++) {
        x = d[i].x;
        y = d[i].y;
        if (root[y] != y) {
            remap[root[y]] = root[x];
        }
        root[y] = root[x];
    }

    for (i = 0; i < n; i++) {
        p = root[i];
        while(remap[p] != p) {p = remap[p];}
        root[i] = p;
    }

    ans = 0;

    for (i = 0; i < n; i++) {
        if (root[i] == i) {
            ans++;
        }
    }

    printf("%d\n", ans);

    return 0;
}