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

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：

我是先做到它的升级版：此题升级版

我大概讲一下吧：

不难发现：如果(l,r)，那么(l+1,r)必然也合法。

所以我们枚举左端点，右端点是不断递增的，所以是线性的。

可以用ST表做，也可以用优先队列做。

优先队列代码：

 #include<bits/stdc++.h>
 using namespace std;
 #define N 100050
 int n,val[N],flag[N];
 struct Node
 {
   int id,val;
   bool operator <(const Node&b)const
   {return val<b.val;}
 };
 template<typename T>void read(T&x)
 {
   int k=; char c=getchar();
   x=;
   while(!isdigit(c)&&c!=EOF)k^=c=='-',c=getchar();
   if (c==EOF)exit();
   while(isdigit(c))x=x*+c-'',c=getchar();
   x=k?-x:x;
 }
 int main()
 {
   #ifndef ONLINE_JUDGE
   freopen("aa.in","r",stdin);
   #endif
   read(n);
   for(int i=;i<=n;i++)read(val[i]);
   int r=,ans=;
   priority_queue<Node>mx,mi;
   for(int i=;i<=n;i++)
     {
       while(r+<=n&&(mi.empty()||
              (fabs(val[r+]-mx.top().val)<=&&fabs(val[r+]+mi.top().val)<=)))
     {
       mx.push(Node{r+,val[r+]});
       mi.push(Node{r+,-val[r+]});
       r++;
     }
       ans=max(ans,r-i+);
       flag[i]=;
       while(!mi.empty()&&flag[mi.top().id])mi.pop();
       while(!mx.empty()&&flag[mx.top().id])mx.pop();
     }
   printf("%d\n",ans);
 }

ST表代码

 #include<bits/stdc++.h>
 using namespace std;
 #define N 100050
 int n,val[N];
 int mm[N],mi[N][],mx[N][];
 template<typename T>void read(T&x)
 {
   int k=; char c=getchar();
   x=;
   while(!isdigit(c)&&c!=EOF)k^=c=='-',c=getchar();
   if (c==EOF)exit();
   while(isdigit(c))x=x*+c-'',c=getchar();
   x=k?-x:x;
 }
 void init_ST(int n)
   {
     mm[]=-;
     for(int i=;i<=n;i++)
       {
     mm[i]=(i&(i-))==?mm[i-]+:mm[i-];
     mx[i][]=mi[i][]=val[i];
       }
     for(int i=;i<=;i++)
       for(int j=;j+(<<i)-<=n;j++)
     {
       mi[j][i]=min(mi[j][i-],mi[j+(<<(i-))][i-]);
       mx[j][i]=max(mx[j][i-],mx[j+(<<(i-))][i-]);
     }
   }
 int rmq(int x,int y)
   {
     int k=mm[y-x+];
     int ans=max(mx[x][k],mx[y-(<<k)+][k]);
     ans-=min(mi[x][k],mi[y-(<<k)+][k]);
     return ans;
   }
 int main()
 {
   #ifndef ONLINE_JUDGE
   freopen("aa.in","r",stdin);
   #endif
   read(n);
   for(int i=;i<=n;i++)read(val[i]);
   init_ST(n);
   int r=,ans=;
   for(int i=;i<=n;i++)
     {
       while(r+<=n&&rmq(i,r+)<=)r++;
       ans=max(ans,r-i+);
     }
   printf("%d\n",ans);
 }