Codeforces 602B - Approximating a Constant Range(DP)

B. Approximating a Constant Range
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
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.

Input
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).

Output
Print a single number — the maximum length of an almost constant range of the given sequence.

Examples
input
5
1 2 3 3 2
output
4
input
11
5 4 5 5 6 7 8 8 8 7 6
output
5
Note
In the first sample, the longest almost constant range is [2, 5]; its length (the number of data points in it) is 4.

In the second sample, there are three almost constant ranges of length 4: [1, 4], [6, 9] and [7, 10]; the only almost constant range of the maximum length 5 is [6, 10].

题意:
给出一串数字,每个数字和他相邻的数相差不超过1,求最长子串长度,子串的最大值和最小值差值不超过1.

解题思路:
因为每个数他相邻的都不超过1,所以对于当前的数x来说.
只要讨论x-1,x-2,x+1,x+2出现的最后位置即可.

对于x+1 > x-1(x-2出现在x-1后面),那么就看x-1和x+2最后出现的位置(x和x-2不符合,x-1和x+1不符合)

同理,对于x+1 < x-1,就看x-1的位置和x+2的位置.

AC代码:

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+5;
int p[maxn];
int main()
{
    ios::sync_with_stdio(false);
    int n;
    cin>>n;
    int ans = 2;
    for(int i = 1;i <= n;i++)
    {
        int x;
        cin>>x;
        if(p[x-1] > p[x+1]) ans = max(ans,i-max(p[x+1],p[x-2]));
        else                ans = max(ans,i-max(p[x+2],p[x-1]));
        p[x] = i;
    }
    cout<<ans;
    return 0;
}
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