LIS UVA 10534 Wavio Sequence

题目传送门

题意：找对称的，形如：123454321 子序列的最长长度

分析：LIS的nlogn的做法，首先从前扫到尾，记录每个位置的最长上升子序列，从后扫到头同理。因为是对称的，所以取较小值*2-1再取最大值

代码：

/************************************************
* Author        :Running_Time
* Created Time  :2015-8-5 21:38:32
* File Name     :UVA_10534.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int MAXN = 1e4 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int a[MAXN];
int d[MAXN];
int dp[MAXN], dp2[MAXN];

int main(void)    {     //UVA 10534 Wavio Sequence
    int n;
    while (scanf ("%d", &n) == 1)   {
        for (int i=1; i<=n; ++i)    scanf ("%d", &a[i]);
        memset (d, 0, sizeof (d));
        memset (dp, 0, sizeof (dp));
        memset (dp2, 0, sizeof (dp2));
        d[1] = a[1];    int len = 1;    dp[1] = 1;
        for (int i=2; i<=n; ++i)    {
            if (d[len] < a[i])  d[++len] = a[i];
            else    {
                int j = lower_bound (d+1, d+1+len, a[i]) - d;
                d[j] = a[i];
            }
            dp[i] = len;
        }
        d[1] = a[n];    int len2 = 1;   dp2[n] = 1;
        for (int i=n-1; i>=1; --i)    {
            if (d[len2] < a[i]) d[++len2] = a[i];
            else    {
                int j = lower_bound (d+1, d+1+len2, a[i]) - d;
                d[j] = a[i];
            }
            dp2[i] = len2;
        }
        int ans = 0;
        for (int i=1; i<=n; ++i)    {
            ans = max (ans, min (dp[i], dp2[i]) * 2 - 1);
        }
        printf ("%d\n", ans);
    }

    return 0;
}