Codeforces Round #FF (Div. 2/C)/Codeforces446A_DZY Loves Sequences(DP)

DZY Loves Sequences

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

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.

解题报告

改变任意一个位置的数（最大只能改变一个），求最大上升子序列长度。

DP1[i]表示以i为终点的最大子序列长度，DP2[i]表示以i为起点的最大子序列长度。

在区间2到n-1里面，如果存在一个区间类似1,4,6,3,8,9/可以改变3（i）的值为7，那么最大上升子序列应该是dp1[i-1]+dp2[i+1]+1。

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

int n,num[100010],dp1[100010],dp2[100010];
int main()
{
int i,j;
while(~scanf("%d",&n))
{
int maxx=1;
for(i=0; i<n; i++)
scanf("%d",&num[i]);
dp1[0]=1;
for(i=1; i<n; i++)
{
if(num[i-1]<num[i])
maxx++;
else
{
dp1[i]=1;
maxx=1;
}
dp1[i]=maxx;
}
maxx=1;
dp2[n-1]=1;
for(i=n-2; i>=0; i--)
{
if(num[i+1]>num[i])maxx++;
else
{
dp2[i]=1;
maxx=1;
}
dp2[i]=maxx;
}
int ans=1;
//        for(i=0;i<n;i++)
//        {
//        cout<<dp1[i]<<" ";
//        }
//        cout<<endl;
//        for(i=0;i<n;i++)
//        cout<<dp2[i]<<" ";
//        cout<<endl;
for(i=0; i<n; i++)
{
if(i==n-1)
ans=max(ans,dp1[i-1]+1);
else if(i==0)
ans=max(ans,dp2[i+1]+1);
else if(num[i+1]-num[i-1]>1)
{
ans=max(ans,dp1[i-1]+dp2[i+1]+1);
}
else ans=max(ans,max(dp1[i-1]+1,dp2[i+1]+1));
}
cout<<ans<<endl;
}
return 0;
}