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.

Input

The first line of input file contains the length of sequence N. The second line contains the elements of sequence - N integers in the range from 0 to 10000 each, separated by spaces. 1 <= N <= 1000

Output

Output file must contain a single integer - the length of the longest ordered subsequence of the given sequence.

Sample Input

7
1 7 3 5 9 4 8

Sample Output

4
题意:输出最长递增子序列的长度
思路:直接裸LIS,第一次使用,使用两种方法
 
第一种:复杂度n^2
#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std; int a[1005],dp[1005],n; int LIS()
{
int i,j,ans,m;
dp[1] = 1;
ans = 1;
for(i = 2;i<=n;i++)
{
m = 0;
for(j = 1;j<i;j++)
{
if(dp[j]>m && a[j]<a[i])
m = dp[j];
}
dp[i] = m+1;
if(dp[i]>ans)
ans = dp[i];
}
return ans;
} int main()
{
int i;
while(~scanf("%d",&n))
{
for(i = 1;i<=n;i++)
scanf("%d",&a[i]);
printf("%d\n",LIS()); } return 0;
}

第二种:nlogn

#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std; int a[1005],dp[1005],c[1005],n; int bin(int size,int k)
{
int l = 1,r = size;
while(l<=r)
{
int mid = (l+r)/2;
if(k>c[mid] && k<=c[mid+1])
return mid+1;
else if(k<c[mid])
r = mid-1;
else
l = mid+1;
}
} int LIS()
{
int i,j,ans=1;
c[1] = a[1];
dp[1] = 1;
for(i = 2; i<=n; i++)
{
if(a[i]<=c[1])
j = 1;
else if(a[i]>c[ans])
j = ++ans;
else
j = bin(ans,a[i]);
c[j] = a[i];
dp[i] = j;
}
return ans;
} int main()
{
int i;
while(~scanf("%d",&n))
{
for(i = 1; i<=n; i++)
scanf("%d",&a[i]);
printf("%d\n",LIS()); } return 0;
}
												

POJ2533:Longest Ordered Subsequence(LIS)的更多相关文章

  1. POJ2533:Longest Ordered Subsequence

    Longest Ordered Subsequence Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 37454   Acc ...

  2. poj 2533 Longest Ordered Subsequence(LIS)

    Description A numeric sequence of ai is ordered ifa1 <a2 < ... < aN. Let the subsequence of ...

  3. POJ-2533.Longest Ordered Subsequence (LIS模版题)

    本题大意:和LIS一样 本题思路:用dp[ i ]保存前 i 个数中的最长递增序列的长度,则可以得出状态转移方程dp[ i ] = max(dp[ j ] + 1)(j < i) 参考代码: # ...

  4. POJ 2533 Longest Ordered Subsequence LIS O(n*log(n))

    题目链接 最长上升子序列O(n*log(n))的做法,只能用于求长度不能求序列. #include <iostream> #include <algorithm> using ...

  5. 题解报告:poj 2533 Longest Ordered Subsequence(最长上升子序列LIS)

    Description A numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let the subsequence ...

  6. (线性DP LIS)POJ2533 Longest Ordered Subsequence

    Longest Ordered Subsequence Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 66763   Acc ...

  7. POJ2533 Longest Ordered Subsequence —— DP 最长上升子序列(LIS)

    题目链接:http://poj.org/problem?id=2533 Longest Ordered Subsequence Time Limit: 2000MS   Memory Limit: 6 ...

  8. POJ 2533 Longest Ordered Subsequence(LIS模版题)

    Longest Ordered Subsequence Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 47465   Acc ...

  9. POJ2533——Longest Ordered Subsequence(简单的DP)

    Longest Ordered Subsequence DescriptionA numeric sequence of ai is ordered if a1 < a2 < ... &l ...

随机推荐

  1. android中细节效果总结

    android中细节效果总结   andorid取消最上方的标题同时全屏显示 Source code     protected void onCreate(Bundle savedInstanceS ...

  2. angularjs directive 实例 详解

    前面提到了angularjs的factory,service,provider,这个可以理解成php的model,这种model是不带html的,今天所说的directive,也可以理解成php的mo ...

  3. 通过jqueryui实现邮件提示

    //js代码$(function () { var availableTags = ["@qq.com", "@gmail.com", "@126.c ...

  4. 一个超级简单php的留言板

    第一步:配置好测试环境:(详细略了) 第二部:新建一个数据库,命名为guestbook(名字可以随便改),可以直接在phpmyadmin里面操作,在数据库里面新建一张表‘content’,表里面有4个 ...

  5. Flask学习记录之Flask-SQLAlchemy

    Flask-SQLAlchemy库让flask更方便的使用SQLALchemy,是一个强大的关系形数据库框架,既可以使用orm方式操作数据库,也可以使用原始的SQL命令. Flask-Migrate ...

  6. Uva220 Othello

     Othello  Othello is a game played by two people on an 8 x 8 board, using disks that are white on on ...

  7. GestureDetector和SimpleOnGestureListener的使用教程

    1. 当用户触摸屏幕的时候,会产生许多手势,例如down,up,scroll,filing等等,我们知道View类有个View.OnTouchListener内部接口,通过重写他的onTouch(Vi ...

  8. ActionBar Fragment运用最佳实践

    ActionBar Fragment运用最佳实践  

  9. Android和FTP服务器交互,上传下载文件(实例demo)

    今天同学说他备份了联系人的数据放在一个文件里,想把它存到服务器上,以便之后可以进行下载恢复..于是帮他写了个上传,下载文件的demo 主要是 跟FTP服务器打交道-因为这个东东有免费的可以身亲哈 1. ...

  10. 《Algorithms 4th Edition》读书笔记——3.1 符号表(Elementary Symbol Tables)-Ⅱ

    3.1.2 有序的符号表 典型的应用程序中,键都是Comparable的对象,因此可以使用a.compare(b)来比较a和b两个键.许多符号表的实现都利用Comparable接口带来的键的有序性来更 ...