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<math.h>

int main(){
 int N,M;
 
  scanf("%d %d",&N,&M);
  
  int x[M],y[M],z[M],ju[N];
  
  for(int i=0;i<M;i++){
  scanf("%d %d %d",&x[i],&y[i],&z[i]);
  }
  
    for(int i=0;i<N;i++){
   ju[i]=0;
  }
  
  int gr=0,mi=0;
  
  for(int i=0;i<M;i++){
    
  if((ju[x[i]-1]==0)&&(ju[y[i]-1]==0)){
    gr++;
    ju[x[i]-1]=gr;
    ju[y[i]-1]=gr;
  }
    
  if((ju[x[i]-1]==0)||(ju[y[i]-1]==0)){
    if(ju[x[i]-1]==0){
      ju[x[i]-1]=ju[y[i]-1];
    }
    else
      ju[y[i]-1]=ju[x[i]-1];  
  }
    
   if(((ju[x[i]-1]!=0)&&(ju[y[i]-1]!=0))&&(ju[y[i]-1]!=ju[x[i]-1])){
	mi++;
}
  }
   int count=0;      
    for(int i=0;i<N;i++){
      if(ju[i]==0)count++;
    }
    count=count-mi+gr;
         printf("%d\n",count);
}