HDU3038 How Many Answers Are Wrong[带权并查集]

How Many Answers Are Wrong

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6336    Accepted Submission(s): 2391

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

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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

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又是并查集看不出来系列

维护sum[i]为节点i到根(在左面)的距离

一个区间和a,b,s 可以变成b到a-1的距离是s

a,b在同一个集合(即sum到同一个节点)，就判断是否成立sum[b]-sum[a]=s

否则就合并这两个集合，把后面的合并到前面的，距离的话画一下图就好了

//
//  main.cpp
//  hdu3038
//
//  Created by Candy on 31/10/2016.
//  Copyright © 2016 Candy. All rights reserved.
//

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int N=2e5+5;
inline int read(){
    char c=getchar();int x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}
int n,m,a,b,s,ans=0;
int fa[N],sum[N];
inline int find(int x){
    if(fa[x]==x) return x;
    int root=find(fa[x]);
    sum[x]+=sum[fa[x]];
    return fa[x]=root;
}
int main(int argc, const char * argv[]){
    while(scanf("%d%d",&n,&m)!=EOF){
        int ans=0;
        for(int i=1;i<=n;i++) fa[i]=i,sum[i]=0;
        for(int i=1;i<=m;i++){
            a=read()-1;b=read();s=read();
            int f1=find(a),f2=find(b);
            if(f1==f2){
                if(sum[b]-sum[a]!=s) ans++;
            }else{
                if(f1<f2){
                    fa[f2]=f1;
                    sum[f2]=sum[a]+s-sum[b];
                }else{
                    fa[f1]=f2;
                    sum[f1]=sum[b]-sum[a]-s;
                }
            }
        }
        printf("%d
",ans);
    }
    return 0;
}