UVa 10534. Wavio Sequence

这题是要找一个最长（假设长度为2N-1）的子序列,使得前N个元素递增，后N个元素递减。

由于N比较大，直接上n^2的dp会超时……

用另外的方法……贪心+二分……这应该不算dp了……

好吧……也许可用斜率dp解？额……我不会

最长上升子序列问题：

给出一个由n个数组成的序列x[1..n]，找出它的最长单调上升子序列。即求最大的m和a1,
a2……,am，使得a1<a2<……<am且x[a1]<x[a2]<……<x[am]。

动态规划求解思路分析：（O(n^2)）

经典的O(n^2)的动态规划算法，设A[i]表示序列中的第i个数，F[i]表示从1到i这一段中以i结尾的最长上升子序列的长度，初始时设F[i] = 0(i = 1, 2, ..., len(A))。则有动态规划方程：F[i] = max{1, F[j] + 1} (j = 1, 2, ..., i - 1, 且A[j] < A[i])。

贪心+二分查找：（O(nlogn)）   
开辟一个栈，每次取栈顶元素s和读到的元素a做比较，如果a>s，  则加入栈；如果a<s，则二分查找栈中的比a大的第1个数，并替换。  最后序列长度为栈的长度。  
这也是很好理解的，对x和y，如果x<y且E[y]<E[x],用E[x]替换  E[y],此时的最长序列长度没有改变但序列Q的''潜力''增大。  
举例：原序列为1，5，8，3，6，7  
栈为1，5，8，此时读到3，则用3替换5，得到栈中元素为1，3，8，  再读6，用6替换8，得到1，3，6，再读7，得到最终栈为1，3，6，7  ，最长递增子序列为长度4。

#include<iostream>#include<map>#include<string>#include<cstring>#include<cstdio>#include<cstdlib>#include<cmath>#include<queue>#include<vector>#include<algorithm>using namespace std;int a[10010];int dp1[10010];int dp2[10010];int main(){int i,n,t;vector<int>box;while(cin>>n){for(i=0;i<n;i++)cin>>a[i];box.clear();for(i=0;i<n;i++){if(box.empty()||a[i]>box.back()){box.push_back(a[i]);dp1[i]=box.size();}else{t=lower_bound(box.begin(),box.end(),a[i])-box.begin();box[t]=a[i];dp1[i]=t+1;}}box.clear();for(i=n-1;i>-1;i--){if(box.empty()||a[i]>box.back()){box.push_back(a[i]);dp2[i]=box.size();}else{t=lower_bound(box.begin(),box.end(),a[i])-box.begin();box[t]=a[i];dp2[i]=t+1;}}t=0;for(i=0;i<n;i++)t=max(t,min(dp1[i],dp2[i]));cout<<t*2-1<<endl;}return 0;}

 

Wavio Sequence

Time Limit: 3000MS Memory Limit: Unknown 64bit IO Format: %lld & %llu

SubmitStatus

Description

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 length9. 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 be9.

 

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 beN 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

 

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Problemsetter: Md. Kamruzzaman, Member of Elite Problemsetters' Panel

Source

Root :: Competitive Programming 2: This increases the lower bound of Programming Contests. Again (Steven & Felix Halim) :: Problem Solving Paradigms :: Dynamic Programming ::Longest Increasing Subsequence (LIS)
Root :: AOAPC I: Beginning Algorithm Contests (Rujia Liu) :: Volume 5. Dynamic Programming
Root :: AOAPC I: Beginning Algorithm Contests -- Training Guide (Rujia Liu) :: Chapter 1. Algorithm Design :: Dynamic Programming ::Exercises: Beginner
Root :: Prominent Problemsetters :: Md. Kamruzzaman (KZaman)
Root :: Competitive Programming 3: The New Lower Bound of Programming Contests (Steven & Felix Halim) :: Problem Solving Paradigms :: Dynamic Programming ::Longest Increasing Subsequence (LIS)

Root :: Competitive Programming: Increasing the Lower Bound of Programming Contests (Steven & Felix Halim) :: Chapter 3. Problem Solving Paradigms :: Dynamic Programming ::Longest Increasing Subsequence (LIS) - Classical