POJ 2533-Longest Ordered Subsequence 动态规划

题目来源：http://124.205.79.250/JudgeOnline/problem?id=2533

 

解题报告：

 

这道题要求最长递增子序列。

 

由于之前做的POJ 1836中用到的就是求最长递增子序列的思路，所以这道题相对而言就很简单了。

设f[i]表示包含a[i]在内的最长递增子序列的长度，则

 

f[i]=max(1,f[k1]+1, f[k2]+1,..., f[kn]+1)，其中,k1,k2,...,kn满足条件：k1,k2,...,kn 都小于i,且
a[k1],a[k2],...a[kn]都小于a[i]。

 

最后要求的是f[1],...,f[n]中的最大值。

#include <iostream>using namespace std;int main(){int n;cin >> n;int *a=new int[n];int *f=new int[n];for(int i=0;i<n;i++)cin >> a[i];int m=0;for(int i=0;i<n;i++){f[i]=1;for(int j=0;j<i;j++){if(a[j]<a[i])f[i]=max(f[i],f[j]+1);}if(m < f[i])m = f[i];}cout << m << endl;}

 

附录：

Longest Ordered Subsequence

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 14541 Accepted: 6174

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