codeforces#FF DIV2C题DZY Loves Sequences（DP）

C. 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.

可以先进行预处理，将每一个数字可向左延伸的数目记录下来，再将可向右延伸的记录下来，最后遍历一遍，如果这个数可以变成大于左边小于右边的数，即右边-左边>1。这是就可以为左边延伸的+右边延伸的+1。如果不行，则只能满足某一边，这是就要为左边或右边较大的延伸数+1.

代码如下；

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#include <iostream>

#include <stdio.h>

#include <string.h>

#include <stdlib.h>

#include <math.h>

#include <ctype.h>

#include <queue>

#include <map>

#include<algorithm>

using namespace std;

int a[200000], b[200000], c[200000];

int main()

{

int n, i, j, max1=-1;

scanf("%d",&n);

for(i=0;i<n;i++)

scanf("%d",&a[i]);

b[0]=1;

for(i=1;i<n;i++)

{

if(a[i]>a[i-1])

{

b[i]=b[i-1]+1;

}

else

b[i]=1;

}

c[n-1]=1;

for(i=n-2;i>=0;i--)

{

if(a[i]<a[i+1])

c[i]=c[i+1]+1;

else

c[i]=1;

}

if(n==1)

printf("1\n");

else

{max1=2;

if(max1<b[n-2]+1)

max1=b[n-2]+1;

if(max1<c[1]+1)

max1=c[1]+1;

for(i=1;i<=n-2;i++)

{

if(a[i+1]-a[i-1]>1)

{

if(max1<b[i-1]+c[i+1]+1)

max1=b[i-1]+c[i+1]+1;

}

else

{

if(max1<b[i-1]+1)

max1=b[i-1]+1;

if(max1<c[i+1]+1)

max1=c[i+1]+1;

}

}

printf("%d\n",max1);}

return 0;

}