HDU 3038 How Many Answers Are Wrong （并查集）（需复习！）

How Many Answers Are Wrong

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1372    Accepted Submission(s): 553

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 51 10 1007 10 281 3 324 6 416 6 1

 

 

Sample Output

1

 

题意：有n次询问，给出a到b区间的总和，问这n次给出的总和中有几次是和前面已近给出的是矛盾的？

思路：并查集的变形  数组val存储的是前缀和  例如val[i]的值意味，假如他的根为j，则他是区间【i，j】的和

如何求区间[a,b]的和（a，b在统一集合中，根结点为i）相当于区间[i,b]的和减去区间[i,a-1]的和

对于输入的区间，如果在同一个集合即可通过上诉方法直接判断

如果不在同一集合中  则要进行更新：

int t1=find(u);int t2=find(v);if(t1!=t2){    pre[t2]=t1;    val[t2]=val[u]-val[v]+w;}

这个意思是（我理解了很久才明白画个图吧）：

之后还有一个难点就是并查集的查找，他不需要压缩路径！！并且在查找过程中不断对前缀和进行更新，例如将区间【1,3】【4,6】这些连续的区间可以进行更新，一开始val【6】中存储的只是区间【4,6】的和，需要将【1,3】的和也加在val【6】中

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>using namespace std;const int MAXN=200010;int pre[MAXN];int val[MAXN];int find(int x){    if(pre[x]==-1)return x;    int tmp=find(pre[x]);    val[x]+=val[pre[x]];    return pre[x]=tmp;}int main(){    int n,m;    int u,v,w;    while(scanf("%d%d",&n,&m)==2)    {        for(int i=0;i<=n;i++){pre[i]=-1;val[i]=0;}        int ans=0;        while(m--)        {            scanf("%d%d%d",&u,&v,&w);            u=u-1;            int t1=find(u);            int t2=find(v);            if(t1!=t2)            {                pre[t2]=t1;                val[t2]=val[u]-val[v]+w;            }            else            {                if(val[v]-val[u]!=w)ans++;            }        }        printf("%d\n",ans);    }    return 0;}