『妙不可言系列6』CF641D Orac and Medians

$\mathrm{P}\mathrm{r}\mathrm{o}\mathrm{b}\mathrm{l}\mathrm{e}\mathrm{m}$

Slime has a sequence of positive integers $a_1,a_2,\dots ,a_n$ .

In one operation Orac can choose an arbitrary subsegment $[l\dots r]$ of this sequence and replace all values $a_l,a_{l+1},\dots ,a_r$ to the value of median of { a l , a l + 1 , … , a r } \{a_l, a_{l + 1}, \ldots, a_r\} { al?,al+1?,…,ar?} .

In this problem, for the integer multiset $s$ , the median of $s$ is equal to the $?\frac{|s|+1}{2}?$ -th smallest number in it. For example, the median of { 1 , 4 , 4 , 6 , 5 } \{1,4,4,6,5\} { 1,4,4,6,5} is $4$ , and the median of { 1 , 7 , 5 , 8 } \{1,7,5,8\} { 1,7,5,8} is $5$ .

Slime wants Orac to make $a_1=a_2=\dots =a_n=k$ using these operations.

Orac thinks that it is impossible, and he does not want to waste his time, so he decided to ask you if it is possible to satisfy the Slime's requirement, he may ask you these questions several times.

$\mathrm{T}\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{s}\mathrm{l}\mathrm{a}\mathrm{t}\mathrm{e}$

询问$a_1,a_2,?a_n$能否通过若干次将任意区间全部赋值为其中位数这个操作，来使得整个序列全部变为$k$。（中位数指第$?\frac{\mid s\mid +1}{2}?$小的数）

多次询问，每次第一行两个整数，$n$和$k$；第二行$n$个整数，$a_1,a_2,?a_n$。

数据范围:$1\le n\le 10^5,1\le k\le 10^9,1\le a_i\le 10^9$，并保证所有询问中$n$的和不超过$10^5$。

$\mathrm{S}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}$

我们观察一下让所有数字变成$k$的条件：

• 若存在$a_i=k$时满足$a_i\le a_{i?1}$或者$a_i\le a_{i+1}$时，一定存在合法方案。
• 因为每一次可以将$a_{i+1}$或者$a_{i?1}$变成$k$,这样就存在连续的两个$k$，第三个数字加入无论如何答案都是$k$。因此我们便可以将所有数字都变成$k$。

那么如何才能让$a_i$相邻的数字比$k$大呢？

• 若存在连续的三个数中有两个大于等于$k$，那么所有数都可以转变为大于等于$k$的数字。

$\mathrm{C}\mathrm{o}\mathrm{d}\mathrm{e}$

#include <bits/stdc++.h>using namespace std;
const int N = 1e6;int n, k;
int a[N];int read(void)
{
    int s = 0, w = 0; char c = getchar();while (c < '0' || c > '9') w |= c == '-', c = getchar();while (c >= '0' && c <= '9') s = s*10+c-48, c = getchar();return w ? -s : s;
}void work(void)
{
    int tag = 0;n = read(), k = read();for (int i=1;i<=n;++i) {
    a[i] = read();if (a[i] == k) tag = 1;}	if (tag == 0) {
    puts("no");return;}if (n == 1) {
    puts("yes");return;}for (int i=1;i<n;++i) if (a[i] >= k && a[i+1] >= k) {
    puts("yes");return;}for (int i=1;i<n-1;++i)if (a[i] >= k && a[i+2] >= k) {
    puts("yes");return;}puts("no");return;
}int main(void)
{
    int T = read();while (T --) work();
}