Wavio Sequence

Input: Standard Input

Output: Standard Output

Time Limit: 2 Seconds

Wavio is a sequence of integers. It has some interesting properties.

·  Wavio is of odd length i.e. L = 2*n + 1.

·  The first (n+1) integers of Wavio sequence makes a strictly increasing sequence.

·  The last (n+1) integers of Wavio sequence makes a strictly decreasing sequence.

·  No two adjacent integers are same in a Wavio sequence.

For example 1, 2, 3, 4, 5, 4, 3, 2, 0 is an Wavio sequence of length9. But
1, 2, 3, 4, 5, 4, 3, 2, 2 is not a valid wavio sequence. In this problem, you will be given a sequence of integers. You have to find out the length of the longest Wavio sequence which is a subsequence of the given sequence. Consider, the given
sequence as :

1 2 3 2 1 2 3 4 3 2 1 5 4 1 2 3 2 2 1.

Here the longest Wavio sequence is : 1 2 3 4 5 4 3 2 1. So, the output will be9.

Input

The input file contains less than 75 test cases. The description of each test case is given below: Input is terminated by end of file.

Each set starts with a postive integer, N(1<=N<=10000). In next few lines there will beN integers.

Output

For each set of input print the length of longest wavio sequence in a line.

Sample Input                   Output for Sample Input

10  

1 2 3 4 5 4 3 2 1 10 

19 

1 2 3 2 1 2 3 4 3 2 1 5 4 1 2 3 2 2 1 

5 

1 2 3 4 5

     9       

     9     

     1

题意：给定一串数字，求一个子串满足一下要求：子串的长度是L=2*n+1，前n+1个数字是严格的递增序列，后n+1个数字是严格的递减序列，例如123454321就是满足要求的一个子串，输出满足要求的最长的L，

思路：正着走一遍LIS，再倒着走一遍LIS，分别用 pre 数组和 suf 数组保存。pre[i]表示表示已a[i]为位数的最长递增子序列的长度，suf 亦然。然后用 ans = max(ans, min(pre[i], suf[n-1-i])*2-1) 遍历即可。

<span style="font-size:24px;">#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <string>
#include <algorithm>
#include <queue>
#include <stack>
using namespace std;

const int INF = 1<<29;
const double PI = acos(-1.0);
const double e = 2.718281828459;
const double eps = 1e-8;
const int MAXN  = 10010;
int t[MAXN];
int n;

void LIS(int *dp, int *a)
{
    fill(t, t+n, INF);
    for(int i = 0; i < n; i++)
    {
        *lower_bound(t, t+n, a[i]) = a[i];
        dp[i] = lower_bound(t, t+n, a[i])-t+1;
    }
}

int main()
{
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    int a[MAXN], b[MAXN];
    int pre[MAXN], suf[MAXN];
    while(cin>>n)
    {
        for(int i = 0; i < n; i++)
        {
            scanf("%d", &a[i]);
            b[n-1-i] = a[i];
        }
        LIS(pre, a);
        LIS(suf, b);
        int ans = 0;
        for(int i = 0; i < n; i++)
        {
            ans = max(ans, min(pre[i], suf[n-1-i])*2-1);
        }
        cout<<ans<<endl;
    }
    return 0;
}</span>