POJ 2533 Longest Ordered Subsequence（dp LIS）

最新推荐文章于 2021-07-30 17:43:41 发布

原创最新推荐文章于 2021-07-30 17:43:41 发布 · 976 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#dp #poj

本文介绍了一种解决最长递增子序列问题的经典算法，并通过示例解释了该算法的工作原理。文章提供了一个完整的C++代码实现，展示了如何计算给定序列中最长递增子序列的长度。

Language:

Longest Ordered Subsequence

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 33986		Accepted: 14892

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

[Submit] [Go Back] [Status] [Discuss]

求最长递增子序列

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define eps 1e-8
using namespace std;
#define N 1005

int dp[N],n,a[N];

int main()
{
	int i,j;
	while(~scanf("%d",&n))
	{
		for(i=1;i<=n;i++)
			scanf("%d",&a[i]);

		int ans=1;

		dp[1]=1;
		int temp;
		for(i=2;i<=n;i++)
		{
			temp=0;
			for(j=1;j<i;j++)
				if(a[j]<a[i]&&temp<=dp[j])
				  temp=dp[j];

            dp[i]=temp+1;

			if(dp[i]>ans)
				ans=dp[i];
		}
      printf("%d\n",ans);
	}
    return 0;
}