Longest Ordered Subsequence

          Longest Ordered Subsequence

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

 

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

解析： 最长上升子序列

 状态转移方程：

    dp[i]表示以i为结尾的最长递增子序列的长度

    dp[i] = max{dp[j]+1}, 1<=j<i,a[j]<a[i].

 

# include<stdio.h>
int max(int a,int b)
{
    return a>b?a:b;
} 
int main()
{
    int nNum,i,j;
    int num[10002];
    int dp[10002];
    int Max=1;
    scanf("%d",&nNum);
    for(i=1;i<=nNum;i++)
    {
        scanf("%d",&num[i]);
        dp[i]=1;
    }
    for(i=2;i<=nNum;i++)//这是以每个点为一次起点，得到最大上升子序列，时间复杂度O(n^2);后面要写一种时间复杂度O(nlogn)的代码
    {
        for(j=1;j<i;j++)
        {
            if(num[i]>num[j])
            {
                dp[i]=max(dp[i],dp[j]+1);
            }
        }
        if(Max<dp[i])Max=dp[i];
    }
    printf("%d\n",Max);
    return 0;
}