<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[HHH_go_的博客]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/HHH_go_</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; HHH_go_]]></copyright><item><title><![CDATA[Codeforces Round #576 (Div. 2) D （线段树+lazy-tag  &  模拟）]]></title><link>https://blog.youkuaiyun.com/HHH_go_/article/details/98174948</link><guid>https://blog.youkuaiyun.com/HHH_go_/article/details/98174948</guid><author>HHH_go_</author><pubDate>Fri, 02 Aug 2019 10:16:09 +0800</pubDate><description><![CDATA[D. Welfare State
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
There is a country with n citizens. The i-th of them initially has ai money. ...]]></description><category></category></item><item><title><![CDATA[ros/pcl 点云分割——分离地面]]></title><link>https://blog.youkuaiyun.com/HHH_go_/article/details/83148472</link><guid>https://blog.youkuaiyun.com/HHH_go_/article/details/83148472</guid><author>HHH_go_</author><pubDate>Thu, 18 Oct 2018 16:25:59 +0800</pubDate><description><![CDATA[代码 


#include &amp;lt;ros/ros.h&amp;gt;
// PCL specific includes
#include &amp;lt;sensor_msgs/PointCloud2.h&amp;gt;
#include &amp;lt;pcl_conversions/pcl_conversions.h&amp;gt;
#include &amp;lt;pcl/ros/conversions.h&amp;gt;
#include ...]]></description><category></category></item><item><title><![CDATA[如何在ROS中使用PCL—数据格式]]></title><link>https://blog.youkuaiyun.com/HHH_go_/article/details/83148297</link><guid>https://blog.youkuaiyun.com/HHH_go_/article/details/83148297</guid><author>HHH_go_</author><pubDate>Thu, 18 Oct 2018 16:06:55 +0800</pubDate><description><![CDATA[原博https://www.cnblogs.com/li-yao7758258/p/6651326.html

 

如何在ROS中使用PCL—数据格式（1）

在ROS中点云的数据类型

在ROS中表示点云的数据结构有： sensor_msgs::PointCloud      sensor_msgs::PointCloud2     pcl::PointCloud&amp;lt;T&amp;gt;

关于PC...]]></description><category></category></item><item><title><![CDATA[L2-012. 关于堆的判断（数据结构）]]></title><link>https://blog.youkuaiyun.com/HHH_go_/article/details/79586787</link><guid>https://blog.youkuaiyun.com/HHH_go_/article/details/79586787</guid><author>HHH_go_</author><pubDate>Fri, 16 Mar 2018 21:02:28 +0800</pubDate><description><![CDATA[L2-012. 关于堆的判断



时间限制

400 ms




内存限制

65536 kB




代码长度限制

8000 B




判题程序

Standard


作者

陈越




将一系列给定数字顺序插入一个初始为空的小顶堆H[]。随后判断一系列相关命题是否为真。命题分下列几种：


“x is ]]></description><category></category></item><item><title><![CDATA[L2-011. 玩转二叉树]]></title><link>https://blog.youkuaiyun.com/HHH_go_/article/details/79572977</link><guid>https://blog.youkuaiyun.com/HHH_go_/article/details/79572977</guid><author>HHH_go_</author><pubDate>Thu, 15 Mar 2018 19:56:22 +0800</pubDate><description><![CDATA[L2-011. 玩转二叉树



时间限制

400 ms




内存限制

65536 kB




代码长度限制

8000 B




判题程序

Standard


作者

陈越




给定一棵二叉树的中序遍历和前序遍历，请你先将树做个镜面反转，再输出反转后的层序遍历的序列。所谓镜面反转，是指将所有非叶结点的左]]></description><category></category></item><item><title><![CDATA[PAT-L2-004. 这是二叉搜索树吗？（数据结构）]]></title><link>https://blog.youkuaiyun.com/HHH_go_/article/details/79518956</link><guid>https://blog.youkuaiyun.com/HHH_go_/article/details/79518956</guid><author>HHH_go_</author><pubDate>Sun, 11 Mar 2018 19:16:25 +0800</pubDate><description><![CDATA[L2-004. 这是二叉搜索树吗？



时间限制

400 ms




内存限制

65536 kB




代码长度限制

8000 B




判题程序

Standard


作者

陈越




一棵二叉搜索树可被递归地定义为具有下列性质的二叉树：对于任一结点，


其左子树中所有结点的键值小于该结点的键值]]></description><category></category></item><item><title><![CDATA[Python语法第3讲：数组]]></title><link>https://blog.youkuaiyun.com/HHH_go_/article/details/79506638</link><guid>https://blog.youkuaiyun.com/HHH_go_/article/details/79506638</guid><author>HHH_go_</author><pubDate>Sat, 10 Mar 2018 12:01:52 +0800</pubDate><description><![CDATA[原博：http://blog.youkuaiyun.com/qzc70919700/article/details/72983513###;

1、数组的创建 
[python] view plain copy

intArr = [1, 4, 2, 5, 3, 7, 9, 0] #整型数组  
strArr = [&quot;北京&quot;, &quot;上海&quot;, &quot;杭州&quot;, &quot;深圳&quot;] #字符串数组  
mixArr = [1, 3...]]></description><category></category></item><item><title><![CDATA[codeforces 893E Counting Arrays (组合数学)]]></title><link>https://blog.youkuaiyun.com/HHH_go_/article/details/78862003</link><guid>https://blog.youkuaiyun.com/HHH_go_/article/details/78862003</guid><author>HHH_go_</author><pubDate>Thu, 21 Dec 2017 11:43:17 +0800</pubDate><description><![CDATA[E. Counting Arrays


time limit per test
3 seconds


memory limit per test
256 megabytes


input
standard input


output
standard output



You are given two positive integer numbe]]></description><category></category></item><item><title><![CDATA[codeforces 900D Unusual Sequences （数论）]]></title><link>https://blog.youkuaiyun.com/HHH_go_/article/details/78861868</link><guid>https://blog.youkuaiyun.com/HHH_go_/article/details/78861868</guid><author>HHH_go_</author><pubDate>Thu, 21 Dec 2017 11:35:25 +0800</pubDate><description><![CDATA[D. Unusual Sequences


time limit per test
1 second


memory limit per test
256 megabytes


input
standard input


output
standard output



Count the number of distinct sequences ]]></description><category></category></item><item><title><![CDATA[codeforces 901B GCD of Polynomials （数论+构造）]]></title><link>https://blog.youkuaiyun.com/HHH_go_/article/details/78861712</link><guid>https://blog.youkuaiyun.com/HHH_go_/article/details/78861712</guid><author>HHH_go_</author><pubDate>Thu, 21 Dec 2017 11:27:37 +0800</pubDate><description><![CDATA[B. GCD of Polynomials


time limit per test
2 seconds


memory limit per test
256 megabytes


input
standard input


output
standard output



Suppose you have two polynomials  and]]></description><category></category></item><item><title><![CDATA[磁盘调度算法（FCFS,SSTF,SCAN）]]></title><link>https://blog.youkuaiyun.com/HHH_go_/article/details/78861669</link><guid>https://blog.youkuaiyun.com/HHH_go_/article/details/78861669</guid><author>HHH_go_</author><pubDate>Thu, 21 Dec 2017 11:25:24 +0800</pubDate><description><![CDATA[#include
using namespace std;
int a[2000]; //磁道位置 
int n; //磁道数量 
int begin;  //初始位置 
void init()
{
	srand(time(0));
	n=(rand()%20)+1; // 0-1000
	for(int i=0;i<n;i++)
	a[i]=(rand()%20)+1; //0-1000
	be]]></description><category></category></item><item><title><![CDATA[最佳淘汰算法（OPT） 先进先出的算法（FIFO） 最近最久未使用算法（LRU）]]></title><link>https://blog.youkuaiyun.com/HHH_go_/article/details/78808926</link><guid>https://blog.youkuaiyun.com/HHH_go_/article/details/78808926</guid><author>HHH_go_</author><pubDate>Fri, 15 Dec 2017 02:24:50 +0800</pubDate><description><![CDATA[#include
#define Del(a,b) memset(a,b,sizeof(a))
using namespace std;
int a[4000];  //随机生成的页 
int c;//用户页面容量 
void init() //生成页 
{
	srand(time(0));
	for(int i=1;i<=320;i++)
	{
		int q=rand()%317;
		//a]]></description><category></category></item><item><title><![CDATA[银行家算法]]></title><link>https://blog.youkuaiyun.com/HHH_go_/article/details/78808924</link><guid>https://blog.youkuaiyun.com/HHH_go_/article/details/78808924</guid><author>HHH_go_</author><pubDate>Fri, 15 Dec 2017 02:18:26 +0800</pubDate><description><![CDATA[银行家算法是用于解决死锁问题。
相当于动态的模拟一个资源分配的过程，判断哪些可以运行，而后回收资源 重复上两个过程
直到所有进程都能完成。


这里用队列模拟，先将所有进程放进队列，当满足条件出队同时回收资源
当不满足，继续入队
如果在资源未更新的情况下扫到一个进程两次，那么不存在安全序列。


#include
#define Del(a,b) memset(a,b,size]]></description><category></category></item><item><title><![CDATA[例析Alpha-Beta剪枝]]></title><link>https://blog.youkuaiyun.com/HHH_go_/article/details/78731461</link><guid>https://blog.youkuaiyun.com/HHH_go_/article/details/78731461</guid><author>HHH_go_</author><pubDate>Wed, 06 Dec 2017 15:46:58 +0800</pubDate><description><![CDATA[本文使用三子棋问题简单描述Alpha-Beta剪枝的原理。顺序是：先描述三子棋问题，接着描述三子棋问题的极小极大算法，最后描述三子棋问题的Alpha-Beta剪枝算法。




对三子棋问题描述如下：

下棋的一方是计算机（记为“MAX"，执棋子“X”），另一方是人（记为”MIN“，执棋子“O”）；

机器先手，双方轮流在九宫格中落子；

先取得三子一线的一方胜，可以是横向、]]></description><category></category></item><item><title><![CDATA[hdu 1824 Let's go home (2-sate)]]></title><link>https://blog.youkuaiyun.com/HHH_go_/article/details/78604388</link><guid>https://blog.youkuaiyun.com/HHH_go_/article/details/78604388</guid><author>HHH_go_</author><pubDate>Wed, 22 Nov 2017 15:42:21 +0800</pubDate><description><![CDATA[Let's go home
Time Limit: 10000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2141    Accepted Submission(s): 925



Problem Description

小时候，乡愁是一枚小小的]]></description><category></category></item><item><title><![CDATA[poj 1741 Tree (点分治)]]></title><link>https://blog.youkuaiyun.com/HHH_go_/article/details/78183913</link><guid>https://blog.youkuaiyun.com/HHH_go_/article/details/78183913</guid><author>HHH_go_</author><pubDate>Mon, 09 Oct 2017 16:26:45 +0800</pubDate><description><![CDATA[Tree




Time Limit: 1000MS
 
Memory Limit: 30000K


Total Submissions: 24493
 
Accepted: 8161




Description

Give a tree with n vertices,each edge has a length(positive integer l]]></description><category></category></item><item><title><![CDATA[开关灯类问题  （高斯消元）]]></title><link>https://blog.youkuaiyun.com/HHH_go_/article/details/78178219</link><guid>https://blog.youkuaiyun.com/HHH_go_/article/details/78178219</guid><author>HHH_go_</author><pubDate>Sun, 08 Oct 2017 22:38:53 +0800</pubDate><description><![CDATA[POJ 1222

EXTENDED LIGHTS OUT



#include
#include
#include
using namespace std;
int a[35][35];
int x[35];
void gauss(int n,int m)
{
	int r=0,c=0;
	while(r<n&&c<m)
	{
		int t=r;
		for(int i=r+1;]]></description><category></category></item><item><title><![CDATA[lca 离线模板]]></title><link>https://blog.youkuaiyun.com/HHH_go_/article/details/78178181</link><guid>https://blog.youkuaiyun.com/HHH_go_/article/details/78178181</guid><author>HHH_go_</author><pubDate>Sun, 08 Oct 2017 22:33:54 +0800</pubDate><description><![CDATA[struct node
{
	int v,w;
	node(){}
	node(int vv,int ww)
	{
		v=vv;w=ww;
	}
};
struct edge
{
	int u,v,lca;
	edge(){}
	edge(int uu,int vv,int ll)
	{
		u=uu,v=vv,lca=ll;
	}
};
vectorG;
vectorv[50000];
vec]]></description><category></category></item><item><title><![CDATA[强联通 tarjan （模板）]]></title><link>https://blog.youkuaiyun.com/HHH_go_/article/details/78178162</link><guid>https://blog.youkuaiyun.com/HHH_go_/article/details/78178162</guid><author>HHH_go_</author><pubDate>Sun, 08 Oct 2017 22:31:58 +0800</pubDate><description><![CDATA[struct tarjan
{
	const int maxn=50050;
	vectorv[maxn];
	stackS;
	int dfn[maxn],low[maxn],viss[maxn];
	int index,cnt;
	void init(int n)
	{
		for(int i=0;i<=n;i++)
		v[i].clear();
		memset(dfn,0,sizeof(]]></description><category></category></item><item><title><![CDATA[树状数组 （模板）]]></title><link>https://blog.youkuaiyun.com/HHH_go_/article/details/78178142</link><guid>https://blog.youkuaiyun.com/HHH_go_/article/details/78178142</guid><author>HHH_go_</author><pubDate>Sun, 08 Oct 2017 22:29:33 +0800</pubDate><description><![CDATA[struct szsz
{
	int c[maxn];
	int lowbit(int x)
	{
		return x&(-x);
	} 
	int getadd(int x)
	{
		int ans=0;
		while(x>0)
		{
			ans+=c[x];
			x-=lowbit(x);
		}
		return ans;
	}
	void update(int x,int n,]]></description><category></category></item></channel></rss>