题意：

TT写一个数列，现在TT会选择一个区间，然后让FF计算这个区间里面所有数的和，FF准备捉弄一下TT，有时候她会故意计算出来一个错的答案，当然TT也比较聪明，他会发现这个答案跟以前的答案会有冲突，问有多少话是假的？

题目：

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

分析：

1.对于A~B区间的和是S，根据前缀和思维，其实可以理解成B比A-1大S；记录前缀和，每一条信息转化成S(A)-S(B)=SUM。
2.并查集的权值为差值，如果r和l在一个并查集里面，就看看差值是否为sum，判矛盾。
3.对于权值，每次迭代都加上根节点的值（初始值），求区间差则都减去，不影响。

AC代码：

/**记录前缀和，每一条信息转化成S(R)-S(L)=SUM
并查集的权值为差值，如果r和l在一个并查集里面，就看看差值是否为sum，判矛盾。*/
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int M=2e5+10;
int n,m,ans,a,b,c;
int f[M],dp[M];
int getf(int x)
{if(x!=f[x]){int t=f[x];f[x]=getf(f[x]);dp[x]+=dp[t];}return f[x];
}
int main()
{while(~scanf("%d%d",&n,&m)){ans=0;for(int i=0;i<=n;i++){f[i]=i;dp[i]=0;}for(int i=0;i<m;i++){scanf("%d%d%d",&a,&b,&c);a--;int x=getf(a);int y=getf(b);if(x==y){if(dp[a]-dp[b]!=c)ans++;}else{f[x]=y;dp[x]=dp[b]-dp[a]+c;}}printf("%d\n",ans);}return 0;
}