HDU 3038 带权并查集

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本文探讨了一种算法，用于解决一系列游戏中的数学谜题，其中玩家需要验证多个子序列和的有效性。通过使用并查集和路径压缩技术，算法能够高效地确定错误的答案数量，帮助玩家快速解决难题。

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

大意：给出一些询问，[l,r]的和为s，问有多少个是错的
我们考虑sum[i]表示i与根节点之间节点的和

也就是sum[i]=sigma(aj) (pre[i]=>j>=i)

另外需要注意的是sum[r]-sum[l-1]表示的是闭区间[l,r]的和

代码如下：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define inf 1000000000
using namespace std;
int par[200010];
int sum[200010];
int ans,s;
void init(int n){
    for(int i=0;i<=n;i++){
        par[i]=i;
        sum[i]=0;
    }
    ans=0;
}
int fi(int x){
    if(x!=par[x])
    {
        int f=par[x];
        par[x]=fi(par[x]);
        sum[x]+=sum[f];
    }
    return par[x];
}
void un(int x,int y){
    int xx=fi(x);
    int yy=fi(y);
    if(xx==yy){
        if(sum[x]-sum[y]!=s)ans++;
    }else{
        par[xx]=yy;
        sum[xx]=sum[y]-sum[x]+s;
    }
}
int main(){
    int n,m,x,y;
    while(scanf("%d %d",&n,&m)!=EOF){
    init(n);
    while(m--){
        scanf("%d %d %d",&x,&y,&s);
        un(x-1,y);
    }
    printf("%d\n",ans);
    }
    return 0;
}