POJ2533-Longest Ordered Subsequence(dp)

POJ2533 原题链接:http://poj.org/problem?id=2533

Longest Ordered Subsequence

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 55259 Accepted: 24781

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_iK), 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

71 7 3 5 9 4 8

Sample Output

4

题目大意:最长上升子序列,给出一个序列,求出最长的升序子序列;

思路:dp。很简单,一个大循环i从1到n,一个从小循环j从1到i,大循环找出最差解,j循环里面优化结果

代码如下:

#include<iostream>#include<cstdio>#include<cstdlib>using namespace std;int main(){    int a[1001];    int dp[1001];    int n;    cin>>n;    for(int i=1;i<=n;i++)    {        cin>>a[i];    }    for(int i=1;i<=n;i++)    {        dp[i]=1;//最坏的情况        for(int j=1;j<i;j++)        {            if(a[j]<a[i])            {                dp[i]=max(dp[i],dp[j]+1);//优化结果            }        }    }    int Max=1;    for(int i=1;i<=n;i++)    {        if (dp[i]>Max)            Max=dp[i];    }    cout<<Max<<endl;}