Longest Ordered Subsequence [dp]

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本文介绍了一种寻找给定数列中最长递增子序列长度的算法实现，并通过一个具体的例子展示了如何使用动态规划的方法解决该问题。该算法能够有效地处理包含数千个元素的数列。

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

题解

#include<stdio.h>
#include<string.h>
#define MAX_N 1002
int a[MAX_N],dp[MAX_N];

int main()
{
    int N;
    while(~scanf("%d",&N)){
        memset(dp,0,sizeof(dp));
        int ans=0;
        for(int i=0;i<N;i++){
            scanf("%d",&a[i]);
            int MAX_cnt=0;
            for(int j=0;j<i;j++)
                if(a[i]>a[j]&&MAX_cnt<dp[j])
                    MAX_cnt=dp[j];
            dp[i]=MAX_cnt+1;
            if(dp[i]>ans) ans=dp[i];
        }
        printf("%d\n",ans);
    }
    return 0;
}