UVA 10534 Wavio Sequence （波浪DP，4级）

C - Wavio Sequence
Crawling in process...Crawling failedTime Limit:3000MS Memory
Limit:0KB 64bit IO Format:%lld & %llu
SubmitStatus

Appoint description:
System Crawler (2013-06-19)

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

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

求最长的子序列从左到中递增，中到有递减，

思路：字符串取反，求个递增序定中间点。求最优

#include<cstring>
#include<cstdio>
#include<iostream>
#define FOR(i,a,b) for(int i=a;i<=b;++i)
#define clr(f,z) memset(f,z,sizeof(f))
using namespace std;
const int mm=10009;
int f[mm],t[mm];
int idp[mm],ddp[mm];
int main()
{
int n;
while(~scanf("%d",&n))
{
FOR(i,1,n)scanf("%d",&f[i]);
clr(t,0x3f);clr(idp,0);
FOR(i,1,n)
{
int x=lower_bound(t+1,t+n+1,f[i])-t;
idp[i]=max(idp[i],x);
t[x]=f[i];
//cout<<i<<" "<<idp[i]<<endl;
}
clr(t,0x3f);clr(ddp,0);
int ans=0;
for(int i=n;i>=1;--i)
{
int x=lower_bound(t+1,t+n+1,f[i])-t;
ddp[i]=max(ddp[i],x);
t[x]=f[i];
//cout<<i<<" "<<ddp[i]<<endl;
ans=max(ans,min(idp[i],ddp[i]));
}
ans=ans+ans-1;
printf("%d\n",ans);
}
return 0;
}