[Poj 2533] Longest Ordered Subsequence 최장 상승 하위 시퀀스

Longest Ordered Subsequence Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 43410 Accepted: 19139
Description 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
제목 링크:http://poj.org/problem?id=2533
제목:
Longest Ordered Subsequence
아이디어: Longest Ordered Subsequence
코드:

#include<iostream>
#include<stdio.h>
#include<algorithm>
using namespace std;
int n;
int dp[1005];
int a[1005];


int maxx;
int main()
{
    scanf("%d",&n);

    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        dp[i]=1;
    }

    maxx=0;

    for(int i=1;i<=n;i++)
    for(int j=1;j<i;j++)
    if(a[i]>a[j])
    {
        dp[i]=max(dp[i],dp[j]+1);
    }
    for(int i=1;i<=n;i++)
    maxx=max(maxx,dp[i]);
    printf("%d
",maxx);



}