POJ2533——Longest Ordered Subsequence(简单的DP)

Longest Ordered Subsequence

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

题目大意：

　　　　给定一个数列，输出其最长递增子序列的长度。

解题思路：

　　　　水题，不解释。

Code：

 1 /*************************************************************************
 2     > File Name: poj2533.cpp
 3     > Author: Enumz
 4     > Mail: 369372123@qq.com
 5     > Created Time: 2014年10月27日 星期一 20时07分22秒
 6  ************************************************************************/
 7 
 8 #include<iostream>
 9 #include<cstdio>
10 #include<cstdlib>
11 #include<string>
12 #include<cstring>
13 #include<list>
14 #include<queue>
15 #include<stack>
16 #include<map>
17 #include<set>
18 #include<algorithm>
19 #include<cmath>
20 #include<bitset>
21 #include<climits>
22 #define MAXN 1001
23 using namespace std;
24 int date[MAXN];
25 int dp[MAXN];
26 int main()
27 {
28     int N;
29     while (cin>>N)
30     {
31         for (int i=1;i<=N;i++)
32             cin>>date[i];
33         for (int i=1;i<=MAXN-1;i++)
34             dp[i]=1;
35         int Max=1;
36         for (int i=2;i<=N;i++)
37         {
38             for (int j=1;j<=i-1;j++)
39                 if (date[j]<date[i])
40                     dp[i]=max(dp[i],dp[j]+1);
41             Max=max(Max,dp[i]);
42         }
43         cout<<Max<<endl;
44     }
45     return 0;
46 }