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
中的最大值。

#include <iostream>
using namespace std;

int main()
{
	int n;
	cin >> n;
	int *a=new int
;
	int *f=new int
;
	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: 65536K
Total Submissions: 14541		Accepted: 6174

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