ACM: 最长子序 dp题  poj 2533

                                            Longest Ordered Subsequence

Description

A numeric sequence of a_iis ordered if a₁ <a₂ < ... <a_N. Let the subsequence of the given numericsequence (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 manyothers. 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 thelength of sequence N. The second line contains the elements ofsequence - N integers in the range from 0 to 10000 each, separatedby spaces. 1 <= N <= 1000

Output

Output file must contain a single integer -the length of the longest ordered subsequence of the givensequence.

Sample Input

7 1 7 3 5 9 4 8

Sample Output

4

题意:  最长升序的长度.
解题思路:
              1. 状态转移方程: dp[i] =max(dp[i],dp[j]+1);
                dp[i]表示:当前i位置的最长升序.
代码:
#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
#define MAX 1005

int n;
int a[MAX];
int dp[MAX];

inline int max(int a,int b)
{
    return a> b ? a : b;
}

int main()
{
//   freopen("input.txt","r",stdin);
   while(scanf("%d",&n) !=EOF)
    {
      for(int i = 1; i <= n; ++i)
      {
         scanf("%d",&a[i]);
         dp[i] = 1;
      }
      
      int ans = 0;
      for(int i = 1; i <= n; ++i)
      {
         for(int j = 1; j <= i; ++j)
         {
            if(a[i] > a[j])
            {
               dp[i] = max(dp[j]+1,dp[i]);
            }
         }
         ans = max(ans,dp[i]);
      }
      printf("%d\n",ans);
    }
    return0;
}