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

最新推荐文章于 2024-08-27 16:42:59 发布

Rank92

最新推荐文章于 2024-08-27 16:42:59 发布

阅读量150

点赞数

文章标签：数据结构与算法

本文介绍了一种使用动态规划解决最长上升子序列问题的方法。通过实例演示了如何寻找给定序列中最长的按数值递增排列的子序列。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Longest Ordered Subsequence

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 19643		Accepted: 8501

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

http://poj.org/problem?id=2533

/*
POJ2533Longest Ordered Subsequence
最长上升子序列
描述：给一队排列整齐的数列ai，找到数列ai中最长上升子序列。
输入：第一行是数列的元素个数N，第二行是N个整数，范围从0到10,000。（1<=N<=1000）
输出：包括一个整数，表示最长上升子序列的长度。

Sample Input
7
1 7 3 5 9 4 8

Sample Output
4 
*/
/*
用动态规划做。
用dp[k]表示以a[k]作为终点的最大上升子序列
则：
dp[1] = 1;
dp[k] = Max (dp[i]:1 <= i < k 且 a[i ]< a[k] 且 k != 1) + 1.
*/ 
#include<stdio.h>
#define MAXN 1000
int dp[MAXN+10],a[MAXN+10];//a数组记录输入的序列
int main()
{
int n,i,j;
    scanf("%d",&n);
for(i=1;i<=n;i++)
       scanf("%d",&a[i]);
    dp[1]=1;
for(i=2;i<=n;i++)
    {
int temp=0;
for(j=1;j<i;j++)
if(a[i]>a[j])
if(temp<dp[j])
                temp=dp[j];
        dp[i]=temp+1;
    }  
int maxlen=0;
for(i=1;i<=n;i++)
if(maxlen<dp[i])
           maxlen=dp[i];
    printf("%d\n",maxlen);
return 0;  
}