POJ 2533 Longest Ordered Subsequence（基础DP）

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本文介绍了一种求解最长上升子序列(LIS)的有效算法，并通过C++代码实现了解决方案。该算法利用动态规划思想，定义dp[i]表示以第i个元素结尾的最长上升子序列的长度。

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Longest Ordered Subsequence

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

题目大意：给出一个数字序列，找出它的最长上升子序列LIS。

解题思路：定义dp[i]为以ai为结尾的LIS的长度，则dp[i] = max(dp[i],dp[j] + 1)(j < i且aj < ai)。

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxn = 1005;

int a[maxn],dp[maxn];

int main()
{
    int n;
    scanf("%d",&n);
    for(int i = 0;i < n;i++){
        scanf("%d",&a[i]);
    }
    int ans = 0;
    for(int i = 0;i < n;i++){
        dp[i] = 1;
        for(int j = 0;j < i;j++){
            if(a[j] < a[i]){
                dp[i] = max(dp[i],dp[j] + 1);
            }
        }
        ans = max(ans,dp[i]);
    }
    printf("%d\n",ans);
    return 0;
}