Codeforces Round #333 (Div. 2) B. Approximating a Constant Range st 二分

B. Approximating a Constant Range

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/602/problem/B

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.

Sample Input

5
1 2 3 3 2

Sample Output

4

HINT

题意

给你n个数，要求你找到最长的区间，使得这个区间的最大值减去最小值之差的绝对值小于等于1

题解：

枚举每一个数，以这个数为这个区间的最小值，能够往左边延伸多少，往右边延伸多少

再枚举每一个数，以这个数为区间的最大值，能够往左边延伸多少，往右边延伸多少就好了

可以O(n) 也可以 像我一样 用倍增然后二分去找

代码：

#include<iostream>
#include<stdio.h>
#include<algorithm>
using namespace std;
#define maxn 100005
int n;
int dp[maxn][20];
int dp1[maxn][20];
int a[maxn];
int L[maxn],R[maxn];
int mm[maxn];
void initrmp(int n)
{
    mm[0]=-1;
    for(int i=1;i<=n;i++)
    {
        mm[i]=((i&(i-1))==0)?mm[i-1]+1:mm[i-1];
        dp[i][0]=a[i];
        dp1[i][0]=a[i];
    }
    for(int j = 1;j<=mm[n];j++)
        for(int i=1;i+(1<<j)-1<=n;i++)
            dp[i][j]=max(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
    for(int j = 1;j<=mm[n];j++)
        for(int i=1;i+(1<<j)-1<=n;i++)
            dp1[i][j]=min(dp1[i][j-1],dp1[i+(1<<(j-1))][j-1]);
}
int queryMax(int l,int r)
{
    int k = mm[r-l+1];
    return max(dp[l][k],dp[r-(1<<k)+1][k]);
}
int queryMin(int l,int r)
{
    int k = mm[r-l+1];
    return min(dp1[l][k],dp1[r-(1<<k)+1][k]);
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    int Ans = 0;
    initrmp(n);
    for(int i=1;i<=n;i++)
    {
        int l=1,r=i;
        while(l<=r)
        {
            int mid = (l+r)/2;
            if(abs(a[i]-queryMin(mid,i))<=0 && abs(a[i]-queryMax(mid,i))<=1)r=mid-1;
            else l=mid+1;
        }
        L[i]=l;
        l=i,r=n;
        while(l<=r)
        {
            int mid = (l+r)/2;
            if(abs(a[i]-queryMin(i,mid))<=0 && abs(a[i]-queryMax(i,mid))<=1)l=mid+1;
            else r=mid-1;
        }
        R[i]=l-1;
    }
    for(int i=1;i<=n;i++)
        Ans = max(Ans,R[i]-L[i]+1);
    for(int i=1;i<=n;i++)
    {
        int l=1,r=i;
        while(l<=r)
        {
            int mid = (l+r)/2;
            if(abs(a[i]-queryMin(mid,i))<=1 && abs(a[i]-queryMax(mid,i))<=0)r=mid-1;
            else l=mid+1;
        }
        L[i]=l;
        l=i,r=n;
        while(l<=r)
        {
            int mid = (l+r)/2;
            if(abs(a[i]-queryMin(i,mid))<=1 && abs(a[i]-queryMax(i,mid))<=0)l=mid+1;
            else r=mid-1;
        }
        R[i]=l-1;
    }
    for(int i=1;i<=n;i++)
        Ans = max(Ans,R[i]-L[i]+1);
    cout<<Ans<<endl;
}