POJ - 2533（LIS）_poj lis-优快云博客

本文链接：https://blog.youkuaiyun.com/AuroraN/article/details/117264629

该博客主要介绍了如何使用动态规划解决寻找给定序列中最长升序子序列的问题。程序输入包含序列长度和序列元素，输出最长升序子序列的长度。示例中，给定序列(1, 7, 3, 5, 9, 4, 8)，最长升序子序列长度为4。

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最大上升子序列（LIS）

A numeric sequence of ai is ordered if a1 < a2 < … < aN. Let the subsequence of the given numeric sequence (a1, a2, …, aN) be any sequence (ai1, ai2, …, aiK), where 1 <= i1 < i2 < … < iK <= 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
4

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<cmath>
using namespace std;
int n;
int s[1005];
int f[1005];
void lis()
{
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<i;j++)
		{
			if(s[j]<s[i])//s数组记录在i位置之前有没有出现比它小的数
			f[i]=max(f[j]+1,f[i]);
		}
	}
	int ans=-1;
	for(int i=1;i<=n;i++)
	{
		ans=max(ans,f[i]);
	}
	printf("%d\n",ans);
}
void init()
{
	for(int i=1;i<=n;i++)
	f[i]=1;//初始可以有它本身一个数则为1
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	scanf("%d",&s[i]);
	init();
	lis();
 	return 0;
}