POJ 2533 Longest Ordered Subsequence（DP 最长上升子序列）

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本文介绍了一种算法，用于在给定的整数序列中找到最长的有序子序列。通过两个实现版本，一种使用O(n^2)的时间复杂度，另一种优化为O(n log n)，该文详细解释了算法原理及其实现细节。

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Longest Ordered Subsequence

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 38980		Accepted: 17119

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

7
1 7 3 5 9 4 8

Sample Output

Source

Northeastern Europe 2002, Far-Eastern Subregion

n*n算法

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>

using namespace std;

int n;
int a[1001];
int dp[1010];

int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        int maxx = -1;
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
        }
        for(int i=0;i<n;i++)
        {
            dp[i] = 1;
            for(int j=0;j<i;j++)
            {
                if(a[i]>a[j] && dp[j]+1>dp[i])
                {
                    dp[i] = dp[j] + 1;
                }
            }
            if(maxx < dp[i])
            {
                 maxx = dp[i];
            }
        }
        printf("%d\n",maxx);
    }
    return 0;
}

n*logn算法：

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>

#define inf 999999

using namespace std;

int n;
int dp[1010];
int a[1010];

int res(int len,int num)
{
    int l = 0,r = len;
    while(l!=r)
    {
        int mid = (l+r)>>1;
        if(dp[mid] == num)
        {
            return mid;
        }
        else if(dp[mid]<num)
        {
            l = mid + 1;
        }
        else if(dp[mid]>num)
        {
            r = mid;
        }
    }
    return l;
}

int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        int len = 1;
        dp[0] = -1;
        for(int i=1;i<=n;i++)
        {
            dp[i] = inf;
            int k = res(len,a[i]);
            if(k == len)
            {
                len++;
            }
            dp[k] = a[i];
        }
        printf("%d\n",len-1);
    }
    return 0;
}