POJ 2533 Longest Ordered Subsequence

Description
A numeric sequence of ai is ordered if a1 < a2 < … < aN. Let the subsequence of the given numeric sequence (a1, a2, …, aN) be any sequence (ai1, ai2, …, aiK), where 1 <= i1 < i2 < … < iK <= N. For example, sequence (1, 7, 3, 5, 9, 4, 8) has ordered subsequences, e. g., (1, 7), (3, 4, 8) and many others. All longest ordered subsequences are of length 4, e. g., (1, 3, 5, 8).
Your program, when given the numeric sequence, must find the length of its longest ordered subsequence.

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【题目分析】
最长上升子序列，无脑乱写，WA了6次。终于A掉。
不过写nlog

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【代码】
一、n^2

#include <cstdio>#include <cstring>#include <iostream>using namespace std;int dp[1001],a[1001],ans=1;int main(){    int n;    cin>>n;    a[0]=-1;    for (int i=1;i<=n;++i) cin>>a[i],dp[i]=1;    for (int i=1;i<=n;++i)        for (int j=0;j<i;++j)            if (a[j]<a[i]) dp[i]=max(dp[i],dp[j]+1),ans=max(anst<<ans<<endl;}

二、nlogn

#include <iostream>#include <cstdio>#include <cstring>using namespace std;int a[1001],dp[1001];int main(){    int n;    cin>>n;    for (int i=1;i<=n;++i) cin>>a[i];    memset(dp,0x3f,sizeof dp);    for (int i=1;i<=n;++i)    {        int l=0,r=n,ans=0;        while (l<r)        {            int mid=(l+r)/2+1;            if (dp[mid]<a[i]) l=mid;            else r=mid-1;        }        if (dp[l+1]>a[i]) dp[l+1]=a[i];    }    for (int i=n;;--i) if (dp[i]!=0x3f3f3f3f) {cout<<i<<endl;return 0;}}