How Many Answers Are Wrong 并查集-优快云博客

本文链接：https://blog.youkuaiyun.com/ctsas/article/details/54708579

本文介绍了一种使用并查集解决区间和查询问题的方法。通过构建区间上下限的集合，并利用并查集的特性来判断区间和的矛盾情况。文章详细解释了算法思路，并提供了完整的C++代码实现。

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题意：提问区间[a,b]的和为sum 是否与之前的信息矛盾。求矛盾提问数量

思路

让所有有关系的区间上下限组成集合。取根的值为0，那么各元素的值为：区间[根,元素]的和。这里给元素取的值没有绝对意义，但是两个元素值的差是有意义的——比如元素a,b : val[a]-val[b] 代表区间[a,b]的和。

如此就跟并查集扯上关系了，如果对并查集不熟悉的可以看看这篇文章（http://blog.youkuaiyun.com/ctsas/article/details/54647581），借那个题理解一下压缩路径的过程。

最后再注意一下边界问题就行了比如S[a,b]=S1,S[b,c]=S2,S[a,c]=S3，然而S1+S2和S3是不一定相等的。//记 S[a,b] 代表区间[a,b]的和
然而S[a,c]==S[a,b]+S[b+1,c]。说明上限元素与下限-1才能等价。及[a,b]与[b+1,c]是有关系的，应该合并到一个集合；而[a,b]与[b,c]没有关系，不能合并集合。
根据这个特性，只要对输入的下限自减1就好了。

题解

#include<stdio.h>
#define MAX_N 200002
int par[MAX_N],val[MAX_N];

void init(int n){
    for(int i=0;i<=n;i++) par[i]=i,val[i]=0;
}

int find(int x){
    if(par[x]==x) return x;

    int t=find(par[x]);
    val[x]+=val[par[x]];
    return par[x]=t;
}
void unite(int x,int y,int s){
    int fx=find(x);
    int fy=find(y);

    par[fy]=fx;
    val[fy]=s-val[y]+val[x];
}
int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m)){
        init(n);
        int x,y,s,ans=0;
        while(m--){
            scanf("%d%d%d",&x,&y,&s);x--;
            if(find(x)!=find(y)) unite(x,y,s);
            else if(val[y]-val[x]!=s) ans++;
        }
        printf("%d\n",ans);
    }
    return 0;
}

Problem

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.