UVA-10534-Wavio Sequence（最长单调递增子序列长度NlogN）

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本文介绍了一种算法，用于从给定整数序列中找到满足特定条件的最长Wavio序列。Wavio序列是一种特殊的序列，其长度为奇数，前半部分严格递增而后半部分严格递减，且任意相邻元素不相等。文章详细阐述了求解思路与实现代码。

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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
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

题意：找出满足条件子序列的最大长度。条件要求序列长度为N*2-1，前N个数是严格递增，后N个数是严格递减，相邻的数不能相同。

思路：从左往右求一下递增序列，num_left[i]记录前i个数构成的递增序列最大长度。
再从右往左求一遍递增序列，并记录长度，最后遍历记录的两个序列即可。

代码

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<math.h>
using namespace std;
//最长单调递增子序列
const int maxn=10005;
const int INF=-0x3f3f3f3f;
int num[maxn];
int num_left[maxn];
int num_right[maxn];
int DP[maxn];
int main()
{
    int N;
    while(scanf("%d",&N)!=EOF)
    {
        for(int i=1;i<=N;i++)
            scanf("%d",&num[i]);
        int len=0;
        DP[0]=INF;
        for(int i=1;i<=N;i++)
        {
            if(num[i]>DP[len])
                DP[++len]=num[i];
            else
            {
                int left=1,right=len;
                while(left<=right)
                {
                    int mid=(left+right)>>1;
                    if(DP[mid]>=num[i])
                        right=mid-1;
                    else
                        left=mid+1;
                }
                DP[left]=num[i];
            }
            num_left[i]=len;
        }
        len=0;
        DP[0]=INF;
        for(int i=N;i>0;i--)
        {
            if(num[i]>DP[len])
                DP[++len]=num[i];
            else
            {
                int left=0,right=len;
                while(left<=right)
                {
                    int mid=(left+right)>>1;
                    if(DP[mid]>=num[i])
                        right=mid-1;
                    else
                        left=mid+1;
                }
                DP[left]=num[i];
            }
            num_right[i]=len;
        }
//        for(int i=1;i<=N;i++)
//            printf("%d*",num_left[i]);
//        printf("\n");
//        for(int i=1;i<=N;i++)
//            printf("%d*",num_right[i]);
//        printf("\n");
        int result=1;
        for(int i=1;i<=N;i++)
            if(result<min(num_left[i],num_right[i]))
                result=min(num_left[i],num_right[i]);
        printf("%d\n",result*2-1);
    }
    return 0;
}