POJ - 2533 Longest Ordered Subsequence

https://vjudge.net/problem/POJ-2533

DP

#include <iostream>
#include <algorithm>
using namespace std;
const int N = 1010;
int n;
int a[N], f[N]; // 
int main()
{
    scanf("%d", &n);
    for (int i = 1; i <= n; i ++ ) scanf("%d", &a[i]);

    for (int i = 1; i <= n; i ++ )
    {
        f[i] = 1; // 初始以a[i]结尾的子序列 只有a[i]一个数
        for (int j = 1; j < i; j ++ )
            if (a[j] < a[i])
                f[i] = max(f[i], f[j] + 1);  //以j结尾的子序列长度 
    }

    int res = 0;
    for (int i = 1; i <= n; i ++ ) res = max(res, f[i]);

    printf("%d\n", res);

    return 0;
}

朴素法

#include<iostream>
#include<algorithm>
#include<cstring>
#define open ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
using namespace std;
const int N=1010;
int n,len;
int d[N],h[N];
int main()
{
	open;
	while(cin>>n)
	{
		for(int i=1;i<=n;i++) cin>>h[i];
		
		len=1;
		d[1]=h[1];
		for(int i=2;i<=n;i++)
			if(h[i]>d[len])
				d[++len]=h[i];
			else
			{
				int j=lower_bound(d+1,d+len+1,h[i])-d;
				d[j]=h[i];
			}
		cout<<len<<endl;
	}
	return 0;
 }