uva 10534 - Wavio Sequence （dp）

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. But1, 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

 

经典的最长上升子序列的变形题

必须用nlogn的解法，这个类似单调队列的做法严格上来说已经不是dp了

容易想到枚举个位置作为“浪尖”，从前往后扫一遍，求最大的

需要预处理出起始位置到 i 的最长上升序列，和 i 到末尾位置的最长下降序列

 

#include <cstdio>#include <iostream>#include <algorithm>#include <vector>using namespace std;int num[10005],increase[10005],decrease[10005];vector <int> vi(10000),vd(10000);vector <int>::iterator it;int main(){    int n;    while(scanf("%d",&n) != EOF){        vi.clear();vd.clear();        int top;        for(int i = 0;i < n;i++)            cin >> num[i];        //先算一遍最长上升序列        vi.push_back(num[0]);        increase[0] = 1;        top = 0;        for(int i = 1;i < n;i++){            if(num[i] > vi[top]){                top++;                vi.push_back(num[i]);            }            else{                it = lower_bound(vi.begin(),vi.end(),num[i]);                *it = num[i];            }            increase[i] = top+1;        }        //再倒着算一遍上升序列，即正着的下降序列        vd.push_back(num[n-1]);        decrease[n-1] = 1;        top = 0;        for(int i = n-2;i >= 0;i--){            if(num[i] > vd[top]){                top++;                vd.push_back(num[i]);            }            else{                it = lower_bound(vd.begin(),vd.end(),num[i]);                *it = num[i];            }            decrease[i] = top+1;        }        //然后从左到右扫一遍，看以哪个位置为“浪尖”，浪波序列最长        int ans = 1,tem;        for(int i = 0;i < n;i++){            tem = min(increase[i],decrease[i]);            ans = max(tem*2-1,ans);        }        cout << ans << endl;    }    return 0;}