CodeForces 602B_Approximating a Constant Range_DP

Description

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.

本题大意是从一串数字中找出最大值与最小值差不超过1的最长子串。

每个点可设2种情况，以它结束的串的最大值为他本身，以及最大值为点的值+1。

可以很容易得到dp转移方程

#include<iostream>
using namespace std;
int n;
const int MAXN = 1e5 + 10;
int a[MAXN];
int dp[MAXN][2];//0 0,1 +1,//最大值在这2种情况下的以本值结束的最长串
int Abs(int a){ return a > 0 ? a : -a; }
int Max(int a, int b){ return a > b ? a : b; }
int main()
{
while (cin >> n){
dp[0][0] = 0;
dp[0][1] = 0;
a[0] = MAXN;
int ans = 0;
for (int i = 1; i <= n; ++i){
cin >> a[i];
dp[i][0] = 1;
dp[i][1] = 1;
for (int b = 0; b <= 1; ++b){
int k = a[i - 1] + b;
if (k == a[i]){
dp[i][0] = Max(dp[i][0], dp[i - 1][b] + 1);
}
else if (k == a[i] + 1){
dp[i][1] = Max(dp[i][1], dp[i - 1][b] + 1);
}
ans = Max(Max(dp[i][0], dp[i][1]), ans);
}
}
cout << ans << endl;
}
return 0;
}