D - How Many Answers Are Wrong

Portal ：How Many Answers Are Wrong

TT and FF are … friends. Uh… very very good friends -________-b

FF is a bad boy, he is always wooing TT to play the following game with him. This is a very humdrum game. To begin with, TT should write down a sequence of integers-_-!!(bored).

Then, FF can choose a continuous subsequence from it(for example the subsequence from the third to the fifth integer inclusively). After that, FF will ask TT what the sum of the subsequence he chose is. The next, TT will answer FF’s question. Then, FF can redo this process. In the end, FF must work out the entire sequence of integers.

BoringBoringa very very boring game!!! TT doesn’t want to play with FF at all. To punish FF, she often tells FF the wrong answers on purpose.

The bad boy is not a fool man. FF detects some answers are incompatible. Of course, these contradictions make it difficult to calculate the sequence.

However, TT is a nice and lovely girl. She doesn’t have the heart to be hard on FF. To save time, she guarantees that the answers are all right if there is no logical mistakes indeed.

What’s more, if FF finds an answer to be wrong, he will ignore it when judging next answers.

But there will be so many questions that poor FF can’t make sure whether the current answer is right or wrong in a moment. So he decides to write a program to help him with this matter. The program will receive a series of questions from FF together with the answers FF has received from TT. The aim of this program is to find how many answers are wrong. Only by ignoring the wrong answers can FF work out the entire sequence of integers. Poor FF has no time to do this job. And now he is asking for your help~(Why asking trouble for himself~~Bad boy)
Input
Line 1: Two integers, N and M (1 <= N <= 200000, 1 <= M <= 40000). Means TT wrote N integers and FF asked her M questions.

Line 2…M+1: Line i+1 contains three integer: Ai, Bi and Si. Means TT answered FF that the sum from Ai to Bi is Si. It’s guaranteed that 0 < Ai <= Bi <= N.

You can assume that any sum of subsequence is fit in 32-bit integer.
Output
A single line with a integer denotes how many answers are wrong.
Sample Input
10 5
1 10 100
7 10 28
1 3 32
4 6 41
6 6 1
Sample Output
1

The question ： Tell you m Weights of intervals , Find out the number of weights of the wrong interval . Be careful , Before finding out the weight between obvious errors , They all think that everything from the previous one to the previous error interval is right .
Ideas ： It is a classic topic of interval and search set , First contact with this type of topic , mentally . After reading some people's blogs, I understand .

First sum Array is the core of this problem , You can understand it as the weight from this node to its ancestor node , If it is understood as an interval , It's the left closed and right closed interval , The left end point of the interval is the node , The right endpoint is its ancestor node .
Here is a schematic diagram ：

among B yes A The ancestor node of .

Next, back to the topic itself , When the ancestor nodes of two nodes are the same , It means they are in a range , We assume that b Below ,a On the top , Then it's judgment sum[b]-sum[a] Is it equal to s 了 , It's not equal to ans++;
Here is a schematic diagram ：

because sum It is a left closed and right closed interval , therefore a One less , such sum[b]-sum[a] Namely [a,b] The weight of this interval .

When the ancestor nodes of two nodes are different , The schematic diagram is as follows ：

We first make pa Become pb The ancestor node of , because sum[a],sum[b] All in find It is found in the function ,s It is known that ,sum[pb] You can find out ,sum[pb]=s+sum[a]-sum[b];
See code for details .
Code ：

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
const int N=2e5+5;
int f[N],sum[N];//f[i] Express i Father ,sum[i] Express i This point to f[i] The sum between this point 
int find(int x)
{
	if(x!=f[x])
	{
		int pre=f[x];
		f[x]=find(f[x]);// Path compression , Make all nodes point to the root node ,
		sum[x]+=sum[pre];// The reason for adding here is that after all the parts are updated, it's good to update here at one time 
	}
	return f[x];
}
int main()
{
	int n,m;
	scanf("%d%d",&n,&m);
	
	memset(sum,0,sizeof(sum));
	int ans=0;
	for(int i=1;i<=n;i++)
	f[i]=i;
	while(m--)
	{
		int a,b,s;
		scanf("%d%d%d",&a,&b,&s);
		a--;
		int pa=find(a);
		int pb=find(b);
		if(pa==pb)
		{
			if(sum[b]-sum[a]!=s)
			ans++;
		}
		else{
			f[pb]=pa;
			sum[pb]=s+sum[a]-sum[b];
		}
	}
	printf("%d\n",ans);	
	return 0;
}