F - Longest Ordered Subsequence

最新推荐文章于 2021-03-23 11:19:13 发布

superxtong

最新推荐文章于 2021-03-23 11:19:13 发布

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分类专栏： dp 文章标签： dp

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本文介绍了一种求解最长递增子序列问题的有效算法，通过使用C++实现，详细展示了如何利用动态规划思想和标准库函数来高效解决此类问题。

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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
4

题意：求最长递增子序列，套模板

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
const int INF=0x7fffffff;
const int MAXN=1005;
using namespace std;
int dp[MAXN];
long a[MAXN];
int main (void)
{
    int n;
    cin>>n;
    fill(dp,dp+n,INF);
    for(int i=0;i<n;i++)
    {
        scanf("%ld",&a[i]);
    }
    for(int i=0;i<n;i++)
    {
        *lower_bound(dp,dp+n,a[i])=a[i];
    }
    printf("%ld\n",lower_bound(dp,dp+n,INF)-dp);
    return 0;
}