href="https://www.cnblogs.com/ljbguanli/p/19623084"

title="发布于

xmlns="http://www.w3.org/2000/svg">

d="M5,0

style="-webkit-tap-highlight-color:

rgba(0,

href="https://www.luogu.com.cn/problem/P14978"

rel="noopener

id="_1">思路

将限制建双向边，设一个联通块的根为

class="katex">

class="katex-mathml">rtrt

class="base">

mathnormal">t，则每个

class="katex-mathml">aia_i

class="base">

mtight">i

class="vlist-s">​

class="vlist"

class="">一定允许表示为

class="katex">

class="katex-mathml">ai=k×art+ba_i=k\times

a_{rt}+b

mtight">i

class="vlist-s">​

class="vlist"

class="">

class="mspace"

mtight">t

class="vlist-s">​

class="vlist"

class="">

class="mspace"

mathnormal">b，其中

class="katex-mathml">k∈{−1,1}k\in

\left

0">}。

如果有环，那么每个

class="katex">

class="katex-mathml">aia_i

class="base">

mtight">i

class="vlist-s">​

class="vlist"

class="">的值就确定了，检查一下有多少

class="katex">

class="katex-mathml">aia_i

class="base">

mtight">i

class="vlist-s">​

class="vlist"

class="">产生动力即可。

否则我们解出满足

class="katex">

class="katex-mathml">li≤ai≤ril_i\le

a_i

mtight">i

class="vlist-s">​

class="vlist"

class="">

class="mspace"

mtight">i

class="vlist-s">​

class="vlist"

class="">

class="mspace"

mtight">i

class="vlist-s">​

class="vlist"

class="">

class="katex-mathml">arta_{rt}

class="base">

mtight">t

class="vlist-s">​

class="vlist"

class="">取值范围。

现在问题转换为：有若干线段，求一个点使包含这个点的线段数量最多。

扫描线即可，时间复杂度

class="katex">

class="katex-mathml">O(nlog⁡n)O(n\log_{}{n})

class="base">

mtight">

class="vlist-s">​

class="vlist"

class="">

class="mspace"

mathnormal">n

class="mclose">)，瓶颈在于排序。

id="_3">代码

压行导致代码有点猎奇。

string"><bits/stdc++.h>

class="token

punctuation">{queue

class="token

number">1000000000000000

class="token

operator">&&b

class="token

operator">&&

class="token

operator">&&val

class="token

number">1000000000000000

class="token

operator">&&val

class="token

punctuation">}vector

class="token

function">push_back

class="token

function">push_back

class="token

operator">&&v

class="token

function">sync_with_stdio

class="token

operator">>>t

class="token

operator">>>n

class="token

operator">>>m

class="token

operator">>>l

class="token

operator">>>r

class="token

operator">>>x

class="token

operator">>>y

class="token

operator">>>z

class="token

number">1000000000000000

class="token

punctuation">.second

class="token

punctuation">}