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 Python (3.4.3) (judge verdict: Time Limit Exceeded)

import numpy as np
import sys
from collections import deque

input = sys.stdin.readline
def dfs(start, G):
    stack = deque()
    stack.append(start)
    visited = [0]*len(G)
    while stack:
        label = stack.pop()
        visited[label-1]= 1
        for i in G[label-1]:
          if not visited[i-1]:
            #visited[label-1]= 1
            stack.append(i)
    return visited    
 
N, M =map(int, input().split())
G  = [[] for _ in range(N)]
for i in range(M):
    a,b ,_= map(int, input().split())
    G[a-1].append(b)
    G[b-1].append(a)
 
count = 0
visit = np.array([0]*N)
for i in range(1,N+1):
  if visit[i-1]:
    continue
  if G[i-1]:  
    v_temp = dfs(i,G)
    visit = np.logical_or(visit, v_temp)
  else:
    visit[i-1] = 1
  count += 1
 
print(count)