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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                                   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

对每个点前后LIS：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<limits.h>
typedef long long LL;
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
const int maxn=10000+100;
int a[maxn],s[maxn];
int t1[maxn],t2[maxn];
int n;
int main()
{
     std::ios::sync_with_stdio(false);
     while(~scanf("%d",&n))
     {
         REPF(i,1,n)   scanf("%d",&a[i]);
         REPF(i,1,n)
         {
             s[i]=INT_MAX;
             int tt=lower_bound(s+1,s+1+i,a[i])-s;
             t1[i]=tt;
             s[tt]=a[i];
         }
         for(int i=n;i>=1;i--)
         {
             s[n-i+1]=INT_MAX;
             int tt=lower_bound(s+1,s+n-i+2,a[i])-s;
             t2[i]=tt;
             s[tt]=a[i];
         }
         int ans=1;
         REPF(i,1,n)
         {
             int temp=min(t1[i],t2[i])*2-1;//合并的时候一定要注意前后长度一样所以取min(t1[i],t2[i])
             if(temp>ans)   ans=temp;
         }
         printf("%d\n",ans);
     }
    return 0;
}

UVA 10534 Wavio Sequence（LIS）

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原文地址：http://blog.csdn.net/u013582254/article/details/41392651