HOJ 2985 Wavio Sequence（最长递增子序列以及其O(n*logn)算法）

Wavio Sequence

My Tags (Edit)

Source : UVA

Time limit : 1 sec Memory limit : 32 M

Submitted : 296, Accepted : 123

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 multiple 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

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

Sample Output

9

9

1

解法是将数组正着和倒着分别求一下最长递增子序列，然后遍历i，如果i左边的数组和右边的数组的最长子序列相等，就是符合条件的。如果求最长递增子序列用O(N^2)会超时，所以必须用效率高的算法。。关于O(n*logn)算法请参考以下博客

http://blog.csdn.net/dacc123/article/details/50571844

贴上代码

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <algorithm>

using namespace std;
int a[10005];
int b[10005];
int c[10005];
int d[10005];
int dp[10005];
int bp[10005];
int n;
int ans;
int search(int num,int l,int r,int *dp)
{
    int mid;
    while(l<=r)
    {
        mid=(l+r)/2;
        if(num>dp[mid])
            l=mid+1;
        else
            r=mid-1;
    }
    return l;
}
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        ans=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            b[n-i+1]=a[i];
        }
        //memset(dp,0,sizeof(dp));
        dp[1]=a[1];
        c[1]=1;
        int len=1;
        for(int i=2;i<=n;i++)
        {
            if(a[i]>dp[len])
                dp[++len]=a[i];
            else
            {
                int pos=search(a[i],1,len,dp);
                dp[pos]=a[i];
            }
            c[i]=len;
        }
        bp[1]=b[1];
        d[1]=1;
        int len2=1;
        for(int i=2;i<=n;i++)
        {
            if(b[i]>bp[len2])
                bp[++len2]=b[i];
            else
            {
                int pos=search(b[i],1,len2,bp);
                bp[pos]=b[i];
            }
            d[i]=len2;
        }
        for(int i=1;i<=n;i++)
        {
            if(c[i]==d[n-i+1])
                ans=max(ans,c[i]*2-1);
        }
        printf("%d\n",ans);

    }
    return 0;
}