题目链接：
题目
I. Approximating a Constant Range
time limit per test：2 seconds
memory limit per test：256 megabytes
问题描述
When Xellos was doing a practice course in university, he once had to measure the intensity of an effect that slowly approached equilibrium. A good way to determine the equilibrium intensity would be choosing a sufficiently large number of consecutive data points that seems as constant as possible and taking their average. Of course, with the usual sizes of data, it's nothing challenging — but why not make a similar programming contest problem while we're at it?
You're given a sequence of n data points a1, ..., an. There aren't any big jumps between consecutive data points — for each 1 ≤ i < n, it's guaranteed that |ai + 1 - ai| ≤ 1.
A range [l, r] of data points is said to be almost constant if the difference between the largest and the smallest value in that range is at most 1. Formally, let M be the maximum and m the minimum value of ai for l ≤ i ≤ r; the range [l, r] is almost constant if M - m ≤ 1.
Find the length of the longest almost constant range.
输入
The first line of the input contains a single integer n (2 ≤ n ≤ 100 000) — the number of data points.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 100 000).
输出
Print a single number — the maximum length of an almost constant range of the given sequence.
样例
input
5
1 2 3 3 2
output
4
input
11
5 4 5 5 6 7 8 8 8 7 6
output
5
题意
求最大值最小值相差小于2的最大区间长度
题解
这一题由于相邻的数据最多差一，可以直接做，但是这里贴一个单调队列的模板吧，毕竟更通用，
开两个单调队列来维护窗口最大最小值。（这里的实现是有先队列+时间戳）下标其实就相当于时间戳了，对于窗口l，r，小于l的都是过期的。
代码
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#define X first
#define Y second
#define mp make_pair
using namespace std;
typedef __int64 LL;
const int maxn=1e5+10;
int arr[maxn];
priority_queue<pair<int,int> > mi,ma;
int main(){
int n;
scanf("%d",&n);
for(int i=0;i<n;i++) scanf("%d",&arr[i]);
int l=0,r=0;
int ans=-1;
while(r<n){
mi.push(mp(-arr[r],r));
ma.push(mp(arr[r],r));
while(-mi.top().X<ma.top().X-1){
l++;
while(mi.top().Y<l) mi.pop();
while(ma.top().Y<l) ma.pop();
}
ans=max(ans,r-l+1);
r++;
}
printf("%d
",ans);
return 0;
}