POJ 2533 Longest Ordered Subsequence

最新推荐文章于 2018-08-05 20:33:09 发布

原创最新推荐文章于 2018-08-05 20:33:09 发布 · 264 阅读

1 ·

CC 4.0 BY-SA版权

文章标签：

#poj

动态规划专栏收录该内容

107 篇文章

订阅专栏

本文介绍了两种求解最长上升子序列问题的算法实现：一种是时间复杂度为O(n^2)的动态规划方法；另一种是更为高效的O(n log n)算法。通过具体示例展示了如何寻找给定数列中最长有序子序列的长度。

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.

【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

【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;}
}