zoukankan      html  css  js  c++  java
  • POJ 2533 Longest Ordered Subsequence

    A numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let the subsequence of the given numeric sequence ( a1a2, ..., aN) be any sequence ( ai1ai2, ..., 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

    分析
    最长上升子序列。定义d[i]为以第i个元素为起始的LIS,那么d[i]=max(d[j])+1,(j>i)。初始是d[n]=1。
    #include<iostream>
    #include<cstdio>
    #include<cmath>
    #include<cstdlib>
    #include<algorithm>
    #include<cstring>
    #include <queue>
    #include <vector>
    #include<bitset>
    #include<map>
    #include<deque>
    using namespace std;
    typedef long long LL;
    const int maxn = 1e4+5;
    const int mod = 77200211+233;
    typedef pair<int,int> pii;
    #define X first
    #define Y second
    #define pb push_back
    //#define mp make_pair
    #define ms(a,b) memset(a,b,sizeof(a))
    const int inf = 0x3f3f3f3f;
    #define lson l,m,2*rt
    #define rson m+1,r,2*rt+1
    typedef long long ll;
    #define N 100010
    
    int a[maxn];
    int dp[maxn];
    int main(){
        int n;
        scanf("%d",&n);
    
        for(int i=1;i<=n;i++) scanf("%d",&a[i]);
        ms(dp,-1);
        for(int i=n;i>=1;i--){
            dp[i]=1;
            int res=0;
            for(int j=i+1;j<=n;j++){
                if(a[i]<a[j]){
                    res=max(res,dp[j]);
                }
            }
            dp[i]+=res;
        }
        int ans=-1;
        for(int i=1;i<=n;i++){
            ans=max(ans,dp[i]);
        }
        cout<<ans<<endl;
        return 0;
    }
     
  • 相关阅读:
    接口的幂等性原则
    SpringBoot热部署-解决方案
    @Resource 与 @Service注解的区别
    软件概要设计做什么,怎么做
    First Show
    Glide源码解析一,初始化
    android使用giflib加载gif
    android的APT技术
    RxJava的concat操作符
    RxJava基本使用
  • 原文地址:https://www.cnblogs.com/fht-litost/p/8855310.html
Copyright © 2011-2022 走看看