Wavio Sequence - UVa 10534 dp

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

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

 

题意：找到最长的n*2+1的序列使得前n+1是严格递增的，后n+1是严格递减的。

思路：前后各扫一遍最长递增子序列。

AC代码如下：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int num[10010],dp[10010],ans1[10010],ans2[10010],len,ans;int solve(int k){ int l=1,r=len,mi;  while(l<r)  { mi=(l+r)/2;    if(dp[mi]<k)     l=mi+1;    else     r=mi;  }  return l;}int main(){ int n,i,j,k;  dp[0]=-1000000000;  while(~scanf("%d",&n))  { for(i=1;i<=n;i++)     scanf("%d",&num[i]);    len=0;    for(i=1;i<=n;i++)    { if(num[i]>dp[len])      { dp[++len]=num[i];        ans1[i]=len;      }      else      { k=solve(num[i]);        dp[k]=num[i];        ans1[i]=k;      }    }    len=0;    for(i=n;i>=1;i--)    { if(num[i]>dp[len])      { dp[++len]=num[i];        ans2[i]=len;      }      else      { k=solve(num[i]);        dp[k]=num[i];        ans2[i]=k;      }    }    ans=1;    for(i=1;i<=n;i++)      ans=max(ans,min(ans1[i],ans2[i])*2-1);    printf("%d\n",ans);  }}