The Heaviest Non-decreasing Subsequence Problem
最长非递减子序列变形题,把大于等于10000的copy五次放回去就可以了
ac代码:
#include <cstdio>
#include <cstring>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
#define LL long long
const LL N=;
int a[N];
int dp[N];
int LIS(int a[],int len)
{
int dlen=;
memset(dp,,sizeof(dp));
dp[]=a[];//
for(int i=;i<len;i++)
{
if(a[i] < dp[]) // 最开始的位置
{
dp[]=a[i];
continue;
}
if(a[i] >= dp[dlen-])
{
dp[dlen++]=a[i];// new insert
}
else
{
int pos=upper_bound(dp,dp+dlen,a[i])-dp;//
dp[pos]=a[i];
}
}
return dlen;
}
int main()
{
int len=;
int x;
while(~scanf("%d",&x))
{
if(x<) continue;
if(x>=)
{
x-=;
for(int i=;i<=;i++) a[len++]=x;
}
else a[len++]=x;
}
/*
for(int i=1;i<=14;i++)
{
scanf("%d",&x);
if(x<0) continue;
if(x>=10000)
{
x-=10000;
for(int i=1;i<=5;i++) a[len++]=x;
}
else a[len++]=x;
}
*/
// for(int i=1;i<=len;i++) cout<<a[i]<<' ';
// cout<<endl;
cout<<LIS(a,len)<<endl;
return ;
}
The Heaviest Non-decreasing Subsequence Problem的更多相关文章
- SPOJ LIS2 Another Longest Increasing Subsequence Problem 三维偏序最长链 CDQ分治
Another Longest Increasing Subsequence Problem Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://a ...
- SPOJ Another Longest Increasing Subsequence Problem 三维最长链
SPOJ Another Longest Increasing Subsequence Problem 传送门:https://www.spoj.com/problems/LIS2/en/ 题意: 给 ...
- 2017ICPC南宁赛区网络赛 The Heaviest Non-decreasing Subsequence Problem (最长不下降子序列)
Let SSS be a sequence of integers s1s_{1}s1, s2s_{2}s2, ........., sns_{n}sn Each integer i ...
- 2017 ACM-ICPC 亚洲区(南宁赛区)网络赛 The Heaviest Non-decreasing Subsequence Problem
Let SS be a sequence of integers s_{1}s1, s_{2}s2, ......, s_{n}snEach integer is is associ ...
- 2017 ACM/ICPC Asia 南宁区 L The Heaviest Non-decreasing Subsequence Problem
2017-09-24 20:15:22 writer:pprp 题目链接:https://nanti.jisuanke.com/t/17319 题意:给你一串数,给你一个处理方法,确定出这串数的权值, ...
- [Algorithms] Using Dynamic Programming to Solve longest common subsequence problem
Let's say we have two strings: str1 = 'ACDEB' str2 = 'AEBC' We need to find the longest common subse ...
- SPOJ:Another Longest Increasing Subsequence Problem(CDQ分治求三维偏序)
Given a sequence of N pairs of integers, find the length of the longest increasing subsequence of it ...
- SPOJ - LIS2 Another Longest Increasing Subsequence Problem
cdq分治,dp(i)表示以i为结尾的最长LIS,那么dp的递推是依赖于左边的. 因此在分治的时候需要利用左边的子问题来递推右边. (345ms? 区间树TLE /****************** ...
- SPOJ LIS2 - Another Longest Increasing Subsequence Problem(CDQ分治优化DP)
题目链接 LIS2 经典的三维偏序问题. 考虑$cdq$分治. 不过这题的顺序应该是 $cdq(l, mid)$ $solve(l, r)$ $cdq(mid+1, r)$ 因为有个$DP$. #i ...
随机推荐
- 监控zabbix 3.4.11异常通过邮件报警步骤
监控的目的一个是可以查看历史状态,可以对比零晨和工作区间数据的对比,以便后期进行优化指导.还有一个是报警,总不能等到服务器出现异常了才去从头查是什么问题吧.所以这篇主要介绍报警中最基础的一个 配置邮件 ...
- php手记之05-tp5软删除
01-需要在设置软删除的模型里设置
- 2019-8-15C#MDI窗体实现多窗口效果
C#MDI窗体实现多窗口效果 Visual C#是微软公司推出的下一代主流程序开发语言,他也是一种功能十分强大的程序设计语言,正在受到越来越多的编程人员的喜欢.在Visual C#中,提供了为实现M ...
- 监控、日志、APM整个监控体系思考 我为峰2014 简书作者 4.6092018-11-19 11:39打开App 序言
监控.日志.APM整个监控体系思考 我为峰2014 简书作者 4.6092018-11-19 11:39打开App 序言
- git如何获取获取子模块的代码?
答: 步骤如下: 1. git submodule init 2. git submodule update
- VMware与宿主机同一网段
将VMware做为一个物理的虚拟机,设置网段与宿主机在同一子网.
- Android 摇一摇监听实现
package com.loaderman.androiddemo; import android.content.Context; import android.hardware.Sensor; i ...
- std::wstring std::string w2m m2w
static std::wstring m2w(std::string ch, unsigned int CodePage = CP_ACP) { if (ch.empty())return L&qu ...
- java获取properties配置文件中某个属性最简单方法
假如我想获取src目录下sysConfig.properties中的uploadpath属性的值 方法如下所示: private static final ResourceBundle bundle ...
- 123457123456#0#-----com.twoapp.mathGame13--前拼后广--13种数学方法jiemei
com.twoapp.mathGame13--前拼后广--13种数学方法jiemei