C. DZY Loves Sequences

最新推荐文章于 2021-10-15 13:15:45 发布

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最新推荐文章于 2021-10-15 13:15:45 发布

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分类专栏：需要注意的 codeforces

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本文介绍了一种算法，用于找到一个整数序列中最长的子段，通过更改该子段中的一个数，使其成为严格递增的序列。算法使用了两个动态规划数组dp1和dp2来分别记录每个位置前后的最长递增子序列长度。

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DZY has a sequence a, consisting of n integers.

We'll call a sequence ai, ai + 1, ..., aj (1 ≤ i ≤ j ≤ n) a subsegment of the sequence a. The value (j - i + 1) denotes the length of the subsegment.

Your task is to find the longest subsegment of a, such that it is possible to change at most one number (change one number to any integer you want) from the subsegment to make the subsegment strictly increasing.

You only need to output the length of the subsegment you find.

Input

The first line contains integer n (1 ≤ n ≤ 105). The next line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109).

Output

In a single line print the answer to the problem — the maximum length of the required subsegment.

        Sample test(s) 
      
          input 
        
6
7 2 3 1 5 6

          output 
        
5

Note

You can choose subsegment a2, a3, a4, a5, a6 and change its 3rd element (that is a4) to 4.

本题坑还算比较多的,我用的递推,dp1[i]表示以i结尾的递增序列的长度,dp2[i]表示已i开始的递增序列的长度,要找出最长的只改变一个数就能构成递增序列的序列长度只需要枚举i就行了,代码如下

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
typedef long long LL;
using namespace std;
int A[100005];
int dp1[100005],dp2[100005];
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        memset(dp1,0,sizeof(dp1));
        memset(dp2,0,sizeof(dp2));
        scanf("%d",A+1);
        dp1[1]=1;
        for(int i=2;i<=n;i++)
        {
            scanf("%d",A+i);
            if(A[i]>A[i-1])dp1[i]=dp1[i-1]+1;
            else dp1[i]=1;
        }
        dp2[n]=1;
        for(int i=n-1;i>=1;i--)
        {
            if(A[i]<A[i+1])dp2[i]=dp2[i+1]+1;
            else dp2[i]=1;
        }
        int m=1;
        A[0]=0;
        A[n+1]=1000000005;
        for(int i=1;i<=n;i++){
                int t=dp1[i-1]+dp2[i+1]+1;
                if(A[i+1]-A[i-1]<2)t=max(dp1[i],dp2[i])+1;
                m=max(m,t);
        }
        cout<<m<<endl;
    }
    return 0;
}