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
- 线程池 | Java多线程,彻底搞懂线程池
熟悉Java多线程编程的同学都知道,当我们线程创建过多时,容易引发内存溢出,因此我们就有必要使用线程池的技术了. 最近看了一些相关文章,并亲自研究了一下源码,发现有些文章还是有些问题的,所以我也总结了 ...
- UML部署图介绍
https://www.w3cschool.cn/uml_tutorial/uml_tutorial-mi5w28ur.html
- jquery页面滚动到指定id
//jquery页面滚动到指定id $body = (window.opera) ? (document.compatMode == "CSS1Compat" ? $('html ...
- ElementUI】日期选择器时间选择范围限制,只能选今天之前的时间,或者是只能选今天之后的时间。今天是否可以选。限制结束日期不能大于开始日期
<el-date-picker v-model="value1" type="date" placeholder="选择日期" :pi ...
- SurfaceView双缓冲技术引入
package com.loaderman.customviewdemo; import android.content.Context; import android.graphics.Canvas ...
- requests与urllib.request
requests很明显,在写法上与urllib.request不同,前者多一个 S.导入包时:import requestsimport urllib.requesturllib.request请求模 ...
- nginx+lua 设置跨域
nginx 配置: header_filter_by_lua_file cros.lua; access_by_lua ' if ngx.var.request_method == "OPT ...
- Oracle存在则更新,不存在则插入应用-merge
转: Oracle存在则更新,不存在则插入应用-merge 2017年01月11日 14:15:26 周星猩 阅读数 11354更多 分类专栏: Oracle 版权声明:本文为博主原创文章,遵循C ...
- IDEA同一项目启动多个实例
为了验证负载均衡,服务提供者(EurekaClientServiceProviderApplication)需要启动多个实例,当前已启动了一个实例,端口号8762: -- :: --- [ main] ...
- CentOS7下搭建zabbix监控(三)——Zabbix监控服务配置
CentOS7下搭建zabbix监控(一)——Zabbix监控端配置 CentOS7下搭建zabbix监控(二)——Zabbix被监控端配置 (1).配置Zabbix监控Apache服务 主机名:yo ...