HDU3038 How Many Answers Are Wrong 解题报告【带权并查集】

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Problem Description
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)
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
解题报告
题意是，给我们n个数以及m个区间的数字和，然而有一些区间信息是矛盾的，让我们输出他们的个数。
这道题我们采用带权并查集的套路。每给我们一个区间，我们就将区间的两个端点合并在一起，如果已经合并过了，那么就检验w[v]-w[u]（每个并查集所带的权值之差）是否等于其给的值，如果没有合并过，就把这个值用于更改w数组。
代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=200000;
int father[N+5],w[N+5];
int n,m,ans;
int find(int a)
{if(a==father[a])return a;int root=find(father[a]);w[a]+=w[father[a]];return father[a]=root;
}
int main()
{while(~scanf("%d%d",&n,&m)){ans=0;for(int i=1;i<=n;i++)father[i]=i,w[i]=0;while(m--){int u,v,wi;scanf("%d%d%d",&u,&v,&wi);u--;int u1=find(u),v1=find(v);if(u1!=v1)father[v1]=u1,w[v1]=w[u]-w[v]+wi;//将u、v两个并查集合并到v所在的那个并查集里面 else if(w[v]-w[u]!=wi)ans++;}printf("%d\n",ans);}return 0;
}