How Many Answers Are Wrong HDU

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.

Boring~~Boring~~a 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) 19:47:1019:47:17

InputLine 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.
OutputA single line with a integer denotes how many answers are wrong.

不清楚带权并查集的点这里 ----> 传送门

思路：根据带权并查集的套路，我们可以直接考虑两种情况:

①fax == fay 则我们直接判断fa[x].v + d 与 fa[y].v是否相同就可以了

②fax != fay 则直接合并，fa[fay].rt = fay, fa[fay].v = fa[x].v + d - fa[y].v。因为两者未合并过，则他们两个区间的单位数值可以是任意的，合并了不过是确定了区间的数值，单位数值可以是负数。

这里有个细节问题，因为我们求得是 A[R] - A[L-1] = D,则输入了L,R,我们应该让L-1再进行操作。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <queue>
 5 #include <vector>
 6 #include <cmath>
 7 
 8 using namespace std;
 9 
10 #define ll long long
11 #define pb push_back
12 #define fi first
13 #define se second
14 
15 const int N = 2e5 + 10;
16 struct node
17 {
18     int rt, v;
19 }fa[N];
20 int n, m;
21 
22 int Find(int x)
23 {
24     if(fa[x].rt == x){
25         return x;
26     }else{
27         int tmp = fa[x].rt;
28         fa[x].rt = Find(fa[x].rt);
29         fa[x].v += fa[tmp].v;
30         return fa[x].rt;
31     }
32 }
33 
34 bool Union(int x, int y, int d){
35     int fax = Find(x);
36     int fay = Find(y);
37     //if(fax > fay) swap(fax, fay);
38     if(fax == fay){
39         if(0 == fa[x].v + d - fa[y].v) return true;
40         else return false;
41     }
42     else if(fax != fay){
43         fa[fay].rt = fax;
44         fa[fay].v = fa[x].v + d - fa[y].v;
45         return true;
46     }
47 }
48 
49 void solve()
50 {   
51     while(~scanf("%d%d", &n, &m)){
52         for(int i = 0; i <= n; ++i) fa[i].rt = i, fa[i].v = 0;
53         
54         int ans = 0;
55         for(int i = 1; i <= m; ++i){
56             int x, y, d;
57             scanf("%d%d%d", &x, &y, &d);
58             --x;
59             if(!Union(x, y, d)) ++ans;
60         }   
61 
62         //printf("ans = %d
", ans);
63         printf("%d
", ans);
64     }
65 }
66 
67 int main()
68 {
69 
70     solve();
71 
72     return 0;
73 }