并查集-HDU-3038-How Many Answers Are Wrong

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How Many Answers Are Wrong

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4535 Accepted Submission(s): 1741

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

Source
2009 Multi-University Training Contest 13 - Host by HIT

题意：
给出一个序列，并给出一系列a b c这样的问答，其意思是序列中第a个到第b个元素的和为c，问题是这一系列问答中，答案有明显逻辑问题的有多少个。

题解：
其实一开始并不知道该怎么做，但是想到这道题是在并查集专题里（真的是难度大减，毕竟水平不够随机刷题），于是想着怎么和并查集联系在一起。最后的方案是，保存其前驱以及和。
举个栗子就是，如果有个问答是a b c，我们这里都先将a–，原因是这里表示的是sum[a,b]=c，而不是sum(a,b]=c，例如6 6 6指的是第6个是6，相信大家都能明白这一点。那么father[a]保存的是能够找到的a的前驱，father[b]同理。在这个问答里，a相当于是b的前驱。于是我们要来看这个问答是否合法，那么当father[a]==father[b]时，sum[a]保存的是father[a]到a的和，sum[b]同理，所以我们只需要比较c和sum[b]-sum[a]就行了。你怎么当father[a]!=father[b]时，我们只需要推一下公式就知道，其实让fa=father[a]，fb=father[b]，那么令father[fb]=fa，sum[fb]=c+sum[a]-sum[b]。推理过程很简单，c表示(a,b](因为a已经减一，所以左边是不包含），sum[a]表示[fa,a]，sum[b]表示[fb,b]，那么[fa,a]+(a,b]=[fa,b]，然后[fa,b]-[fb,b]=[fa,fb],即是sum[fb]了。
这样一来，就能判断不合法的问答了，需要注意的是，在并查集操作路径压缩的时候，不要忘了sum值的刷新。

//
//  main.cpp
//  并查集-D-How Many Answers Are Wrong
//
//  Created by 袁子涵 on 16/2/18.
//  Copyright © 2016年 袁子涵. All rights reserved.
//
//  78ms    4348KB#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#define MAXN 200005
using namespace::std;
long long int n,m,a,b,sum,total=0;
long long int father[MAXN],num[MAXN];long long int find(long long int p)
{long long int tmp;if (p!=father[p]){tmp=father[p];father[p]=find(father[p]);num[p]+=num[tmp];}return father[p];
}int main(int argc, const char * argv[]) {while (cin >> n >> m) {total=0;memset(num, 0, sizeof(num));for (long long int i=0; i<=n; i++)father[i]=i;for (long long int i=1; i<=m; i++) {scanf("%lld %lld %lld",&a,&b,&sum);a--;long long int fa=find(a),fb=find(b);if (fa==fb) {if (num[b]-num[a]!=sum)total++;}else{father[fb]=fa;num[fb]=sum+num[a]-num[b];}}cout << total << endl;}return 0;
}