acm pku 2533 Longest Ordered Subsequence 动态规划

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本文介绍了一种使用动态规划解决最长递增子序列问题的方法。通过一个具体的例子展示了如何找到给定序列中最长递增子序列的长度。该算法的时间复杂度为O(n^2)，适用于解决此类问题。

Longest Ordered Subsequence
Time Limit:2000MS Memory Limit:65536K
Total Submit:2685 Accepted:964

Description
A numeric sequence of a_i is ordered if a₁ < a₂ < ... < a_N. Let the subsequence of the given numeric sequence (a₁, a₂, ..., a_N) be any sequence (a_i₁, a_i₂, ..., a_{i_K}), where 1 <= i₁ < i₂ < ... < i_K <= 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

Sample Output

Source
Northeastern Europe 2002, Far-Eastern Subregion

#include <iostream>
using namespace std;

int main()
{
    int i,j,n,max;
    int a[1000];
    int f[1000];
    cin>>n;
    for(i=0;i<n;i++)cin>>a[i];
    //for(i=0;i<n;i++)cout<<a[i]<<endl;
    f[0]=1;
    for(i=1;i<n;i++)
    {
                    f[i]=1;
                    for(j=0;j<i;j++)
                    {
                                    if(a[i]>a[j] && f[i]<=f[j])
                                                 f[i]=f[j]+1;
                    }
    }

    max=f[0];
    for(i=1;i<n;i++)
                    if(max<f[i])max=f[i];
    cout<<max<<endl;


}

动态规划经典题