POJ 2533 LIS N2

2024-08-28 07:38

本文主要是介绍POJ 2533 LIS N2，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

状态转移方程 dp[i]=max(dp[i],dp[j]+1); dp[i]为以第i个数结尾的能得到的最长的上升子序列。1<=j<i。这个题目注意a[i]可能为0，因此要注意初始化a[0]=-1;

http://poj.org/problem?id=2533

Longest Ordered Subsequence

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 34360		Accepted: 15083

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

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<vector>
#include<cmath>
#include<stdlib.h>
#include<iomanip>
#include<list>
#include<deque>
#include<map>
#include <stdio.h>
#include <queue>

#define maxn 10000+5
#define ull unsigned long long
#define ll long long
#define reP(i,n) for(i=1;i<=n;i++)
#define rep(i,n) for(i=0;i<n;i++)
#define cle(a) memset(a,0,sizeof(a))
#define mod 90001
#define PI 3.141592657
#define INF 1<<30
const ull inf = 1LL << 61;
const double eps=1e-5;

using namespace std;

bool cmp(int a,int b){
    return a>b;
}
int a[1001],dp[1001];
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int n;
    while(cin>>n)
	{
		for(int i=1;i<=n;i++)
		cin>>a[i];
		a[0]=-1;
		cle(dp);
		int ans=-1;
		for(int i=1;i<=n;i++)//4    0 1 2 3
		{
			for(int j=0;j<i;j++)
			if(a[i]>a[j])dp[i]=max(dp[i],dp[j]+1);
			ans=max(dp[i],ans);
		}
		cout<<ans<<endl;
	}
    return 0;
}