uva 10534 Wavio Sequence (LIS)_wavio sequence uva

本文介绍了一种算法，用于解决寻找给定整数序列中最长的Wavio子序列问题。Wavio序列是一种特殊序列，其长度为奇数且包含严格递增和递减部分。文章详细阐述了解决方案的步骤，并提供了实现该算法的C++代码。

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1、http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1475

2、题目大意：

给定一串数字，求一个子串满足一下要求：子串的长度是L=2*n+1，前n+1个数字是严格的递增序列，后n+1个数字是严格的递减序列，例如123454321就是满足要求的一个子串，输出满足要求的最长的L，

3、正着走一遍LIS，再倒着走一遍LIS，dp[i]表示前i个满足要求的数字的个数，那么dp1[i]和dp2[i]中最小的一个就是L的一半

4、题目：

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

1 2 3 4 5 4 3 2 1 10

1 2 3 2 1 2 3 4 3 2 1 5 4 1 2 3 2 2 1

1 2 3 4 5

4、代码：

#include<stdio.h>
#include<algorithm>
using namespace std;
#define N 10005
int a[N],b[N],dp1[N],dp2[N];
int n;
void LIS(int dp[],int a[])
{
    int stack[N];
    int top=0;
    stack[top]=-99999999;
    for(int i=1; i<=n; i++)
    {
        //如果a[i]>栈顶部元素，则压栈
        if(a[i]>stack[top])
        {
            stack[++top]=a[i];
            dp[i]=top;
        }
        //如果a[i]不大于栈顶部元素，则二分查找第一个比a[i]大的元素
        else
        {
            int l=1,r=top;
            while(l<=r)
            {
                int mid=(l+r)>>1;
                if(a[i]>stack[mid])
                {
                    l=mid+1;
                }
                else
                    r=mid-1;
            }
            //替换a[i]
            stack[l]=a[i];
            dp[i]=l;
        }
    }
}
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&a[i]);
            b[n-i+1]=a[i];
            dp1[i]=0;
            dp2[i]=0;
        }
        LIS(dp1,a);
        LIS(dp2,b);
        int ans=0,maxx=-1;
        for(int i=1; i<=n; i++)
        {
            ans=min(dp1[i],dp2[n-i+1])*2-1;//因为要求递增序列与递减序列的个数相等，则取正反较小的一个
            if(ans>maxx)
                maxx=ans;
        }
        printf("%d\n",maxx);
    }
    return 0;
}