codeforce 447C

life4711發表於2014-07-15

http://codeforces.com/contest/447/problem/C

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.

題目大意：這道題是最長公共上升子序列的變形，給定的條件是在可以任意改變其中一個數的大小，當然也可以不改動。

大體思路：我的思路是先找出以每個a[i]為開頭和結尾的最長公共子序列的長度，然後依次列舉每個a[i],具體見程式碼。

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <string.h>
using namespace std;
int a[100006];
int dp1[100006],dp2[100006];
int n;
int main()
{
    while(~scanf("%d",&n))
    {
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        int maxn=1;
        dp1[1]=1;
        for(int i=2;i<=n;i++)
        {
            if(a[i]>a[i-1])
                maxn++;
            else
                maxn=1;
            dp1[i]=maxn;
        }
        maxn=1;
        dp2[n]=1;
        for(int i=n-1;i>=1;i--)
        {
            if(a[i]<a[i+1])
                maxn++;
            else
                maxn=1;
            dp2[i]=maxn;
        }
        int ans=0;
        for(int i=1;i<=n;i++)
        {
            if(i==1)
                ans=max(ans,dp2[i]);
            else if(i==n)
                ans=max(ans,dp1[i]);
            else if(a[i+1]-a[i-1]>1)
                ans=max(ans,dp1[i-1]+dp2[i+1]);
            else
                ans=max(ans,max(dp1[i],dp2[i]));
        }
        if(ans<n)
            ans++;
        printf("%d\n",ans);
    }
    return 0;
}


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