POJ 2533（最大递增子序列）

最新推荐文章于 2021-03-18 22:38:22 发布

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最新推荐文章于 2021-03-18 22:38:22 发布

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分类专栏： POJ 文章标签：最大递增子序列

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本文介绍了一种求解最长递增子序列的算法实现，通过动态规划的方法找到给定序列中最长递增子序列的长度。算法首先初始化dp数组，然后遍历整个序列，比较每个元素与之前的所有元素，更新dp数组以存储到当前位置为止的最长递增子序列长度。最终，通过遍历dp数组找出最大值，即为所求的最长递增子序列的长度。

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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

代码+题解

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int N;
int arr[1005],dp[1005];
int mmp()
{
    int ans=1;
    for(int i=1;i<=N;i++)
    {
        dp[i]=1;
        for(int j=1;j<i;j++)
        {
            if(arr[j]<arr[i])            //每进行一次循环，dp[i]都会变为当前最大的递增子序列。
                dp[i]=max(dp[j]+1,dp[i]);
        }
        ans=max(ans,dp[i]);          //找出N组递增子序列中最大的即可。
    }
    return ans;
}
int main()
{
    scanf("%d",&N);
    for(int i=1;i<=N;i++)
        scanf("%d",&arr[i]);
    int ans=mmp();
    printf("%d\n",ans);
    return 0;
}