题目地址：http://codeforces.com/contest/447/problem/C
C. DZY Loves Sequences
time limit per test
memory limit per test
input
output
a, consisting of n
ai, ai + 1, ..., aj (1 ≤ i ≤ j ≤ n) a subsegment of the sequence a. The value (j - i + 1)
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
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
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；
#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;
}