poj 2533 Longest Ordered Subsequence

本文介绍了一种求解最长递增子序列长度的算法实现。通过对给定数值序列进行处理,该算法能够找到序列中最长递增子序列的长度。示例中使用了一个包含7个元素的序列进行说明,并提供了完整的C++代码实现。

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/*Longest Ordered Subsequence
Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 26720		Accepted: 11648

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

Source
Northeastern Europe 2002, Far-Eastern Subregion*/


#include <cstdio>
using namespace std;

int main()
{
    int n = 0;
    short num[1005] = {0};
    short len[1005] = {0};
    scanf("%d", &n);
    for(int i = 0; i < n; ++i)
    scanf("%hd", &num[i]);
    len[0] = 1;
    if(n > 1)
    {
        for(int i = 1; i < n; ++i)
        {
            int max = 0;
            for(int j = 0; j < i; ++j)
            {
                if(num[j] < num[i] && len[j] > max) max = len[j];
            }
            len[i] = max + 1;
        }
    }
    int max = 0;
    for(int i = 0; i < n; ++i)
    {
        if(max < len[i]) max = len[i];
    }
    printf("%d\n", max);
}

转载于:https://my.oschina.net/locusxt/blog/133401

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