Longest Ordered Subsequence

最新推荐文章于 2022-08-30 19:25:01 发布

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最新推荐文章于 2022-08-30 19:25:01 发布

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本文介绍了一种求解最长递增子序列问题的经典算法实现，通过动态规划的方法找到给定序列中最长递增子序列的长度。示例中使用了一个包含7个元素的序列进行说明，并提供了一份完整的C++代码实现。

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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

题目要求最长递增子序列，需要注意不能相等，然后直接用模板就行了。

代码：

#include<stdio.h>
#include<iostream>
#include<string.h>
using namespace std;
const int N = 1000 + 10;
int a[N], dp[N];
int main()
{
    int n;
    while(scanf("%d", &n) != EOF)
    {
        int ans = -1;
        for(int i = 1; i <= n; ++ i)
        {
            scanf("%d", &a[i]);
        }
        memset(dp, 0, sizeof(dp));
        for(int i = 1; i <= n; ++ i)
        {
            dp[i] = 1;
            for(int j = 1; 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;
}