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=0; char c=getchar();
  x=0;
  while(!isdigit(c)&&c!=EOF)k^=c=='-',c=getchar();
  if (c==EOF)exit(0);
  while(isdigit(c))x=x*10+c-'0',c=getchar();
  x=k?-x:x;
}
int main()
{
  #ifndef ONLINE_JUDGE
  freopen("aa.in","r",stdin);
  #endif
  read(n);
  for(int i=1;i<=n;i++)read(val[i]);
  int r=0,ans=0;
  priority_queue<Node>mx,mi;
  for(int i=1;i<=n;i++)
    {
      while(r+1<=n&&(mi.empty()||
             (fabs(val[r+1]-mx.top().val)<=1&&fabs(val[r+1]+mi.top().val)<=1)))
    {
      mx.push(Node{r+1,val[r+1]});
      mi.push(Node{r+1,-val[r+1]});
      r++;
    }
      ans=max(ans,r-i+1);
      flag[i]=1;
      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][20],mx[N][20];
template<typename T>void read(T&x)
{
  int k=0; char c=getchar();
  x=0;
  while(!isdigit(c)&&c!=EOF)k^=c=='-',c=getchar();
  if (c==EOF)exit(0);
  while(isdigit(c))x=x*10+c-'0',c=getchar();
  x=k?-x:x;
}
void init_ST(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];
    mx[i][0]=mi[i][0]=val[i];
      }
    for(int i=1;i<=20;i++)
      for(int j=1;j+(1<<i)-1<=n;j++)
    {
      mi[j][i]=min(mi[j][i-1],mi[j+(1<<(i-1))][i-1]);
      mx[j][i]=max(mx[j][i-1],mx[j+(1<<(i-1))][i-1]);
    }
  }
int rmq(int x,int y)
  {
    int k=mm[y-x+1];
    int ans=max(mx[x][k],mx[y-(1<<k)+1][k]);
    ans-=min(mi[x][k],mi[y-(1<<k)+1][k]);
    return ans;
  }
int main()
{
  #ifndef ONLINE_JUDGE
  freopen("aa.in","r",stdin);
  #endif
  read(n);
  for(int i=1;i<=n;i++)read(val[i]);
  init_ST(n);
  int r=0,ans=0;
  for(int i=1;i<=n;i++)
    {
      while(r+1<=n&&rmq(i,r+1)<=1)r++;
      ans=max(ans,r-i+1);
    }
  printf("%d\n",ans);
}