UVa 10534 Wavio Sequence LIS

Problem D
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 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 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 be N 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

--------------------------------------------

f(i)表示从1到i的最长递增子序列，g(i)表示从n到i的最长递增子序列。

ans=min( min(g(i),g(i))*2-1 ) 1<=i<=n

求LIS复杂度为O(nlogn) 枚举i复杂度为O(n) 总时间复杂度O(nlogn)

-------------------------------------------

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int OO=1e9;

int a[11111];
int b[11111];
int f[11111];
int g[11111];
int d[11111];
int n;

int main()
{
    while (~scanf("%d",&n))
    {
        memset(f,0,sizeof(f));
        memset(g,0,sizeof(g));
        memset(d,0,sizeof(d));
        for (int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            b[n-i+1]=a[i];
        }
        for (int i=1;i<=n;i++) d[i]=OO;
        for (int i=1;i<=n;i++)
        {
            int k=lower_bound(d+1,d+n+1,a[i])-d;
            f[i]=k;
            d[k]=a[i];
        }
        for (int i=1;i<=n;i++) d[i]=OO;
        for (int i=1;i<=n;i++)
        {
            int k=lower_bound(d+1,d+n+1,b[i])-d;
            g[n-i+1]=k;
            d[k]=b[i];
        }
        //for (int i=1;i<=n;i++) cerr<<f[i]<<" ";cerr<<endl;
        //for (int i=1;i<=n;i++) cerr<<g[i]<<" ";cerr<<endl;
        int ans=0;
        for (int i=1;i<=n;i++)
        {
            int tmp=min( f[i], g[i] );
            tmp=(tmp-1)*2+1;
            if (tmp>ans) ans=tmp;
        }
        printf("%d\n",ans);
    }
    return 0;
}