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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <stdio.h>

int main(){
  int n,m,ans=0;
  scanf("%d %d",&n,&m);
  
  int x[m+1],y[m+1],z[m+1];
  for(int i=1;i<=m;i++){
    scanf("%d %d %d",&x[i],&y[i],&z[i]);
  }
  
  int a[n+1],b[n+1];
  for(int i=1;i<=n;i++){
    a[i]=1;
  }
  
  for(int i=1;i<=m;i++){
    if(x[i]>y[i] && a[x[i]]==0){
      a[b[x[i]]]+=a[y[i]];
      a[y[i]]=0;
      b[y[i]]=b[x[i]];
    }
    
    if(x[i]>y[i] && a[x[i]]>0){
      a[x[i]]+=a[y[i]];
      a[y[i]]=0;
      b[y[i]]=x[i];
    }
    
    if(y[i]>x[i] && a[y[i]]==0){
      a[b[y[i]]]+=a[x[i]];
      a[x[i]]=0;
      b[x[i]]=b[y[i]];
    }
    
    if(y[i]>x[i] && a[y[i]]>0){
      a[y[i]]+=a[x[i]];
      a[x[i]]=0;
      b[x[i]]=y[i];
    }
    
  }
  
  for(int i=1;i<=n;i++){
    if(a[i]>0){
      ans+=1;
    }
  }
  for(int i=1;i<=n;i++){
    printf("%d\n",a[i]);
  }
  printf("%d",ans);
  return 0;
}