动态规划【Longest Ordered Subsequence】

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

7
1 7 3 5 9 4 8

Sample Output

4

最长不下降子序列

示例分析：

1 7 3 5 9 4 8

1 2 2 3 4 3 4

#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std;
int a[1005],dp[1005],n;
int LIS()
{
    int i,j,ans,m;
    dp[1] = 1;
    ans = 1;
    for(i = 2;i<=n;i++)
    {
        m = 0;
        for(j = 1;j<i;j++)
        {
            if(dp[j]>m && a[j]<a[i])        
            m = dp[j];
        }
        dp[i] = m+1;                      
        if(dp[i]>ans)
        ans = dp[i];
    }
    return ans;
}

int main()
{
    int i;
    while(~scanf("%d",&n))
    {
        for(i = 1;i<=n;i++)
        scanf("%d",&a[i]);
        printf("%d\n",LIS());

    }

    return 0;
}