How Many Answers Are Wrong HDU - 3038（带权并查集）-白红宇

How Many Answers Are Wrong HDU - 3038（带权并查集）

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发布时间：2019-05-25

本文共 3219 字，大约阅读时间需要 10 分钟。

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

之前一直没有尝试过带权并查集的题目，虽然大一暑假就学过带权并查集的问题。。

大意：有M个数，不知道它们具体的值，但是知道某两个数之间（包括这两个数）的所有数之和，现在给出N个这样的区间和信息，需要判断有多少个这样的区间和与前边已知的区间和存在矛盾。

一开始想还真不知道怎么联想到并查集上的。后来想想，确实可以用并查集解决这个问题。但是需要注意两点，一个是转化父亲节点之后距离父亲节点的距离怎么算。另一个就是这个是一个闭区间，我们在处理的时候，闭区间不好处理就将它转化成开区间，l–或者r++。其他人说这个题目还有一个坑点就是多组输入，好在单组读入我也是多组读入。。

代码如下：

#include
   
    #define ll long longusing namespace std;const int maxx=2e5+100;const int maxm=4e4+100;int f[maxx],dis[maxx];int n,m;inline int getf(int u){
   	int x;	if(u!=f[u])	{
   		x=getf(f[u]);		dis[u]+=dis[f[u]];//处理转换之后距离父亲节点距离		f[u]=x;	}	return f[u];}inline int merge(int u,int v,int val){
   	int t1=getf(u);	int t2=getf(v);	if(t1==t2)	{
   		if(abs(dis[u]-dis[v])==abs(val)) return 0;		else return 1;	}	else	{
   		f[t1]=t2;		dis[t1]=val+dis[v]-dis[u];		return 0;	}}int main(){
   	int x,y,ans,value;	while(~scanf("%d%d",&n,&m))	{
   		ans=0;		for(int i=0;i<=n;i++) f[i]=i,dis[i]=0;		while(m--)		{
   			scanf("%d%d%d",&x,&y,&value);			x--;			if(merge(x,y,value)) ans++;		}		printf("%d\n",ans);	}	return 0;}

努力加油a啊，(^o)/~

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