POJ 3298 递推，DP

Antimonotonicity

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 2428 Accepted: 995

Description

I have a sequence Fred of length n comprised of integers between 1 and n inclusive. The elements of Fred are pairwise distinct. I want to find a subsequence Mary of Fred that is as long as possible and has the property that:

Mary₀ > Mary₁ < Mary₂ > Mary₃ < ...

Input

The first line of input will contain a single integer T expressed in decimal with no leading zeroes. T will be at most 50. T test cases will follow.

Each test case is contained on a single line. A line describing a test case is formatted as follows:

n Fred₀ Fred₁ Fred₂ ... Fred_n-1.

where n and each element of Fred is an integer expressed in decimal with no leading zeroes. No line will have leading or trailing whitespace, and two adjacent integers on the same line will be separated by a single space. n will be at most 30000.

Output

For each test case, output a single integer followed by a newline --- the length of the longest subsequence Mary of Fred with the desired properties.

Sample Input

45 1 2 3 4 55 5 4 3 2 15 5 1 4 2 35 2 4 1 3 5

Sample Output

1253

Source

Waterloo Local Contest, 2007.7.14

利用凹凸函数，依次递推。。

Source Code

Problem: 3298 User: bingshenMemory: 252K Time: 172MSLanguage: C++ Result: Accepted

• Source Code

  #include<stdio.h>#include<string.h>int a[30005];int main(){int i,t,n,m,ans;bool s;scanf("%d",&t);while(t--){ans=1;scanf("%d",&n);for(i=1;i<=n;i++)scanf("%d",&a[i]);m=a[1];s=true;for(i=2;i<=n;i++){if(m<a[i]&&s)m=a[i];else if(m>a[i]&&!s)m=a[i];else if(m>a[i]&&s){m=a[i];s=false;ans++;}else if(m<a[i]&&!s){m=a[i];s=true;ans++;}}printf("%d/n",ans);}return 0;}