CodeForces - 602B

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题目链接：https://vjudge.net/problem/284704/origin

Approximating a Constant Range 

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, ..., a_n. There aren‘t any big jumps between consecutive data points — for each 1 ≤ i < n, it‘s guaranteed that |a_i + 1 - a_i| ≤ 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 a_i 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, ..., a_n (1 ≤ a_i ≤ 100 000).

Output

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

Examples

5
1 2 3 3 2

4

11
5 4 5 5 6 7 8 8 8 7 6

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].

题目意思：就是给你一个数组，保证数组中前后两个元素的差<=1，要你找出连续的并且任意两个数相差不超过1的最长串的长度。

题目分析：可以有以下做法：

1.dp

2.优化扩展串的长度

3.rmq做法

#include <bits/stdc++.h>
using namespace std;
const int maxn=1e5+50;
int a[maxn],dp[maxn];
int main()
{
        int n;
        scanf("%d",&n);
        int ans=0;
        for(  int i=1; i<=n; i++ )
        {
              int x;
              scanf( "%d",&x );
              if(  dp[x-1]>dp[x+1] ) ans=max(  ans,i-max(  dp[x-2],dp[x+1] )  );
              else ans=max(  ans, i-max(dp[x+2],dp[x-1] ) );
              dp[x]=i;
        }
        printf( "%d\n",ans );
        return 0;
}

DP

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=1e6+10;
int n,cnt,ans,a[maxn];
int main()
{
    cin>>n;
    for(int i=1;i<=n;i++)cin>>a[i];
    ans=0;
    int l=1,r=2;
    int one=a[1],two=a[2];//两个元素前后的差<=1 
    while(r<=n){
        while(a[r]==one||a[r]==two) r++;//固定两个元素l不变r往右拓展 
        ans=max(ans,r-l);//确定长度
        one=a[r-1];two=a[r];//更新两个元素
        l=r-1;//l更新 
        while(a[l]==one) l--;//l回溯 
        l++;
    }
    cout<<ans<<endl;
    return 0;
} 

边更新边扩展

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=1e5+10;
int n,cnt,ans,a[maxn],dp[maxn][20],fp[maxn][20];
void init_rmq(){
    for(int i=1;i<=n;i++)dp[i][0]=fp[i][0]=a[i];
    for(int i=1;(1<<i)<=n;i++){
        for(int j=1;j+(1<<i)-1<=n;j++){
            dp[j][i]=max(dp[j][i-1],dp[j+(1<<i-1)][i-1]);
            fp[j][i]=min(fp[j][i-1],fp[j+(1<<i-1)][i-1]);
        }
    }
}
int query_max(int l,int r){
    int k=log2(r-l+1);
    return max(dp[l][k],dp[r-(1<<k)+1][k]);
}
int query_min(int l,int r){
    int k=log2(r-l+1);
    return min(fp[l][k],fp[r-(1<<k)+1][k]);
}
int main()
{
    cin>>n;
    for(int i=1;i<=n;i++)cin>>a[i];
    ans=0;
    init_rmq();
    int j=1;
    for(int i=1;i<=n;i++)
    {
        while(query_max(j,i)-query_min(j,i)>1&&j<=i)
        {
            j++;
        }
        ans=max(ans,i-j+1);
    }
    cout<<ans<<endl;
    return 0;
} 

RMQ

 

CodeForces - 602B

标签：连续   contains   mil   app   main   was   个数   single   rmi   

原文地址：https://www.cnblogs.com/Mingusu/p/12001295.html