Longest Ordered Subsequence 最长上升子序列

最新推荐文章于 2021-07-30 17:43:41 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，输出为最长递增子序列的长度。

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Longest Ordered Subsequence
Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
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Status
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<string.h>
#include<algorithm>
using namespace std;
int a[1010];
int dp[1010];
int main()
{
    int n;
   // freopen("in.txt","r",stdin);
    while(~scanf("%d",&n))
    {
     for(int i=0;i<n;i++)
         scanf("%d",&a[i]);

     dp[0]=1;
     int ans=1;
     for(int i=1;i<n;i++)
     {
         dp[i]=1;
         for(int j=0;j<i;j++)
         {
           if(a[i]>a[j]) dp[i]=max(dp[i],dp[j]+1);
           else if(a[i]==a[j]) dp[i]=max(dp[i],dp[j]);
          }
         ans=max(ans,dp[i]);
     }
     printf("%d\n",ans);
    }
}