UVA 10534 Wavio Sequence

 Wavio Sequence

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

 

Wavio is of odd length i.e.

L= 2n+ 1.

 

The rst (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 length 9. 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 be `9'.

Input

The input le 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 be N integers.

Output

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

SampleInput

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

SampleOutput

9

9

1

这是一个最长上升子序列和最长下降序列的合体版， 寻找以a[i]为中心， 以 n 为半径， 左侧为升序， 右侧为降序， 求n的最大值。

 普通的方法做这个会超时的， 这有最长上升子序列(LIS)长度的O(nlogn)算法

https://www.slyar.com/blog/longest-ordered-subsequence.html

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <climits>
using namespace std;

const int maxn = 10010;
const int INF = 0x3f3f3f3f;
int a[maxn], b[maxn], dp1[maxn], dp2[maxn], s[maxn];
int n;
void LIS(int dp[], int tmp[]) {
    int top = 0;
    s[top] = -INF;
    for(int i = 1; i <= n; i++) {
        if(tmp[i] > s[top]) {
            s[++top] = tmp[i];
            dp[i] = top;
        } else {
            int l = 1, r = top;
            while(l <= r) {
                int mid = (r+l)>>1;
                if ( tmp[i] > s[mid]) l = mid + 1;
                else    r = mid - 1;
            }
            s[l] = tmp[i];
            dp[i] = l;
        }
    }
}

int main() {
    while(~scanf("%d", &n)) {
        for(int i = 1; i <= n; i++) {
            scanf("%d", &a[i]);
            b[n-i+1] = a[i];
            dp1[i] =  dp2[i] = 0;
        }
        LIS(dp1, a);
        LIS(dp2, b);
        int ans = -1;
        for(int i = 1; i <= n; i++) {
            ans = max(min(dp1[i], dp2[n-i+1]), ans);
        }
        printf("%d
", ans*2 -  1);
    }
    return 0;
}