[Kuangbin take you to fly] twelve thematic basis DP1 N - Longest Ordered Subsequence POJ - 2533 (rising longest sequence LIS)

N - Longest Ordered Subsequence POJ - 2533

Topic links: https://vjudge.net/contest/68966#problem/N

topic:

If the longest ordered subsequence a1 <a2 <... <aN, then the sequence of numbers sorted ai. Sequences give the given number sequence (a1, a2, ..., aN ) is any sequence (ai1, ai2, ..., aiK ), where 1 <= i1 <i2 <... <iK <= N For example, the sequence (1,7,3,5,9,4,8) having an ordered sequence, for example. g. , (1,7), (3,4,8) and many others. All longest ordered subsequence of length 4, e. g. , (1,3,5,8).

     When a given sequence of numbers, your program must find its longest length of an ordered sequence.
Input
     The first line of the input file comprises a sequence length of N. The second row contains the elements of a sequence - N integers, ranges, each separated by a space from 0 to 10,000. 1 <= N <= 1000
yields
     an output file must contain an integer - for the longest length of an ordered sequence of a given sequence.
Sample input

    . 7
     . 1. 8. 7. 3 4. 5. 9

Sample Output

    4

Thinking: LIS problems, we must pay attention dp [i] the initial value of 1;

//
// Created by hanyu on 2019/8/8.
//
#include <algorithm>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <set>
#include<math.h>
#include<map>
using namespace std;
typedef long long ll;
const int maxn=1000+7;
#define MAX 0x3f3f3f3f
int main()
{
    int T;
    while(~scanf("%d",&T))
    {
        int dp[maxn],a[maxn];
        memset(dp,0, sizeof(dp));
        for(int i=0;i<T;i++)
        {
            scanf("%d",&a[i]);
        }
        for(int i=0;i<T;i++)
        {
            dp[i]=1;
            for(int j=0;j<=i;j++)
            {
                if(a[j]<a[i])
                    dp[i]=max(dp[i],dp[j]+1);
            }
        }
        int maxx=-1;
        for(int i=0;i<T;i++)
        {
            maxx=max(maxx,dp[i]);
        }
        printf("%d\n",maxx);
    }
    return 0;
}