zoukankan      html  css  js  c++  java
  • POJ 2533 动态规划入门 (LIS)

    Longest Ordered Subsequence
    Time Limit: 2000MS Memory Limit: 65536K
    Total Submissions: 42914 Accepted: 18914
    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

    一个裸的 LIS
    直接给状态转移方程: if (a[j]>a[i]) f[j]=max(f[j],f[i]+1);

    #include <cstdio>
    #include <algorithm>
    using namespace std;
    int n,a[1005],f[1005];
    int main()
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]); 
        for(int i=1;i<=n;i++)
            for(int j=i;j<=n;j++)
                if(a[j]>a[i])   f[j]=max(f[i]+1,f[j]);
        for(int i=1;i<=n;i++)
            f[0]=max(f[0],f[i]);
        printf("%d",f[0]+1);//因为是从0开始的 所以要+1
    }

    还是0msAC的

  • 相关阅读:
    玩转xss
    Anonim小白成长计划
    mssql注入与绕过
    了解mssql数据库
    2020年度学习规划
    access 注入
    bypasswaf 之报错注入
    bypasswaf之盲注
    sql注入常用函数与bypasswaf
    一篇关于数据库的另类操作
  • 原文地址:https://www.cnblogs.com/SiriusRen/p/5831199.html
Copyright © 2011-2022 走看看