【POJ2533】Longest Ordered Subsequence （动态规划&最长上升子序列）

最新推荐文章于 2021-07-25 22:06:01 发布

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本文介绍了一种解决最长有序子序列问题的算法实现。通过动态规划方法，该程序能够找出给定数值序列中最长递增子序列的长度。输入为一串整数，输出则是该序列中最大递增子序列的长度。

这里是题目

Longest Ordered Subsequence
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 54195 Accepted: 24230
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

#include<stdio.h>
#include<math.h>
int s[1010],a[1010];
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
    int m,n;
    while(~scanf("%d",&n))
    {
        int sum=0,temp;
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        s[0]=1;
        for(int i=1;i<n;i++)
        {
            temp=0;
            for(int j=0;j<i;j++)
            {
                if(a[i]>a[j])
                    if(temp<s[j])
                        temp=s[j];
            }
            s[i]=temp+1;
        }
        int max=0;
        for(int i=0;i<n;i++)
        {
            if(max<s[i])
            max=s[i];
        }
        printf("%d\n",max);
    }
return 0;
}