<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[ssf_cxdm的博客]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/ssf_cxdm</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; ssf_cxdm]]></copyright><item><title><![CDATA[数据结构总览 李慧琴]]></title><link>https://blog.youkuaiyun.com/ssf_cxdm/article/details/127282741</link><guid>https://blog.youkuaiyun.com/ssf_cxdm/article/details/127282741</guid><author>ssf_cxdm</author><pubDate>Wed, 12 Oct 2022 14:51:35 +0800</pubDate><description><![CDATA[常见的数据结构]]></description><category></category></item><item><title><![CDATA[ROS基础 launch学习]]></title><link>https://blog.youkuaiyun.com/ssf_cxdm/article/details/116701962</link><guid>https://blog.youkuaiyun.com/ssf_cxdm/article/details/116701962</guid><author>ssf_cxdm</author><pubDate>Wed, 12 May 2021 11:56:11 +0800</pubDate><description><![CDATA[ROS基础 launch学习
目的：记录ROS学习过程 ，主要包括结果展示和源码。
实例1 展示

实例2 - 展示
展示视频抽空补上

]]></description><category></category></item><item><title><![CDATA[L2-031 深入虎穴 （25 分)]]></title><link>https://blog.youkuaiyun.com/ssf_cxdm/article/details/89049445</link><guid>https://blog.youkuaiyun.com/ssf_cxdm/article/details/89049445</guid><author>ssf_cxdm</author><pubDate>Sat, 06 Apr 2019 00:13:44 +0800</pubDate><description><![CDATA[题目链接
题目大意与分析：先寻找根结底，然后广度优先搜索，寻找层次最大的结点。
提交代码：
#include&lt;iostream&gt;
#include&lt;vector&gt;
#include&lt;queue&gt;
#include&lt;cstdio&gt;
using namespace std;

int main(){
    int n,root,door;
    sc...]]></description><category></category></item><item><title><![CDATA[L2-032 彩虹瓶 （25 分)]]></title><link>https://blog.youkuaiyun.com/ssf_cxdm/article/details/89049240</link><guid>https://blog.youkuaiyun.com/ssf_cxdm/article/details/89049240</guid><author>ssf_cxdm</author><pubDate>Fri, 05 Apr 2019 23:54:52 +0800</pubDate><description><![CDATA[题目链接
题目大意：彩虹瓶有n层，必须按照1，2，3…n的顺序堆放小球。假设现在需要编号为m的小球，此时送来的恰好是m，则继续判断m+1层；若不是，如果此时货架上有小球，并且顶层就是m号，则继续判断m+1层。若不是，如果货架上还能放小球，则放小球，否则无法成功制作彩虹瓶。
思路：利用堆栈进行模拟。
提交代码：
#include&lt;bits/stdc++.h&gt;
using namespac...]]></description><category></category></item><item><title><![CDATA[L2-030 冰岛人 （25 分)]]></title><link>https://blog.youkuaiyun.com/ssf_cxdm/article/details/89049088</link><guid>https://blog.youkuaiyun.com/ssf_cxdm/article/details/89049088</guid><author>ssf_cxdm</author><pubDate>Fri, 05 Apr 2019 23:32:49 +0800</pubDate><description><![CDATA[题目链接
题目大意：找两结点的最近公共祖先结点。
思路：
1、将名字用数字来编号，并记录性别。
2、寻找并记录各个结点的父结点的编号。
3、判断两人是否可以交往。这里难点是如何寻找公共结点 。可以这样，a到根结点有一条路径，记录下a到路径中各个结点的距离。同样对b进行同样的操作，在b到根结点的过程中，若两条路径相交，则是公共祖先，若距离都大于等于4，则是五代以外，满足要求，可以交往。
注意，“五代...]]></description><category></category></item><item><title><![CDATA[L2-029 特立独行的幸福 （25 分)]]></title><link>https://blog.youkuaiyun.com/ssf_cxdm/article/details/89048754</link><guid>https://blog.youkuaiyun.com/ssf_cxdm/article/details/89048754</guid><author>ssf_cxdm</author><pubDate>Fri, 05 Apr 2019 22:53:02 +0800</pubDate><description><![CDATA[题目链接
题目大意：注意这几个概念，幸福数，幸福依附于初始数字的幸福数，特立独行的幸福数，独立性，不幸福的数。给定一区间，列出区间内所有的特立独行的幸福数和它的独立性。如果区间内没有幸福数，则输出SAD。
思路：逐个遍历区间内的数字，判断是否是幸福数，若是幸福数则保存在一数组内；并且将该过程中经历的数字标记下来，用来下一步判断是否是特立独行的幸福数。
若数组为空，说明没有幸福数；否则，遍历数组，判...]]></description><category></category></item><item><title><![CDATA[L1-064 估值一亿的AI核心代码 （20 分)]]></title><link>https://blog.youkuaiyun.com/ssf_cxdm/article/details/89045701</link><guid>https://blog.youkuaiyun.com/ssf_cxdm/article/details/89045701</guid><author>ssf_cxdm</author><pubDate>Fri, 05 Apr 2019 17:47:47 +0800</pubDate><description><![CDATA[题目链接
题目大意：按照要求，改变字符串，有些复杂。
思路：方法对了，就很简单。“在每个空格和标点之前加一个空格，将字符串化成一块一块，逐个判断输出”，另一位博主写到。
提交代码（参考别人）：
#include&lt;bits/stdc++.h&gt;
using namespace std;
int main(){
    int n;
    scanf("%d",&amp;n);
    g...]]></description><category></category></item><item><title><![CDATA[L1-058 6翻了 （15 分)]]></title><link>https://blog.youkuaiyun.com/ssf_cxdm/article/details/89045500</link><guid>https://blog.youkuaiyun.com/ssf_cxdm/article/details/89045500</guid><author>ssf_cxdm</author><pubDate>Fri, 05 Apr 2019 17:30:48 +0800</pubDate><description><![CDATA[题目链接
题目大意：简单，不说了。
思路：从左到右扫描，分两种情况：不是字符’6’，直接输出；是字符’6’，继续往后扫描，统计’6’的个数，然后再输出。
提交代码：
#include&lt;bits/stdc++.h&gt;
using namespace std;
int main(){
    string s;
    int cnt=0;
    getline(cin,s);
    f...]]></description><category></category></item><item><title><![CDATA[L1-059 敲笨钟]]></title><link>https://blog.youkuaiyun.com/ssf_cxdm/article/details/89045211</link><guid>https://blog.youkuaiyun.com/ssf_cxdm/article/details/89045211</guid><author>ssf_cxdm</author><pubDate>Fri, 05 Apr 2019 17:08:01 +0800</pubDate><description><![CDATA[题目链接
题目大意：题意很简单，直接讲思路。
思路：分两步完成
1、判断是否押韵。分别判断上句和下午，看最后三个字母是否是"ing"，放心的判断，不用担心越界；也不用担心最后一个汉字的拼音只有两个字母（当时在考场，这点浪费了将近半小时的时间，吐血啊！），只有两个字母一定不符合"ing"。
2、输出。找到倒数第三个汉字的位置。
提交代码（参考了别人）：
#include&lt;bits/stdc++...]]></description><category></category></item><item><title><![CDATA[2019第四届团体程序设计天梯赛 CCCC经验、 总结！]]></title><link>https://blog.youkuaiyun.com/ssf_cxdm/article/details/89044659</link><guid>https://blog.youkuaiyun.com/ssf_cxdm/article/details/89044659</guid><author>ssf_cxdm</author><pubDate>Fri, 05 Apr 2019 16:29:08 +0800</pubDate><description><![CDATA[第四届C4愉快地结束了！现在来写写经验！！！

]]></description><category></category></item><item><title><![CDATA[CodeBlocks的调试技巧]]></title><link>https://blog.youkuaiyun.com/ssf_cxdm/article/details/88847402</link><guid>https://blog.youkuaiyun.com/ssf_cxdm/article/details/88847402</guid><author>ssf_cxdm</author><pubDate>Wed, 27 Mar 2019 15:58:19 +0800</pubDate><description><![CDATA[设置断点、运行到光标、单步调试
添加链接描述
软件调试工具设置
添加链接描述

]]></description><category></category></item><item><title><![CDATA[1131 Subway Map]]></title><link>https://blog.youkuaiyun.com/ssf_cxdm/article/details/84308867</link><guid>https://blog.youkuaiyun.com/ssf_cxdm/article/details/84308867</guid><author>ssf_cxdm</author><pubDate>Tue, 20 Nov 2018 20:04:18 +0800</pubDate><description><![CDATA[Subway Map
In the big cities, the subway systems always look so complex to the visitors. To give you some sense, the following figure shows the map of Beijing subway. Now you are supposed to help peop...]]></description><category></category></item><item><title><![CDATA[1087 All Roads Lead to Rome]]></title><link>https://blog.youkuaiyun.com/ssf_cxdm/article/details/84145386</link><guid>https://blog.youkuaiyun.com/ssf_cxdm/article/details/84145386</guid><author>ssf_cxdm</author><pubDate>Fri, 16 Nov 2018 20:32:55 +0800</pubDate><description><![CDATA[1087 All Roads Lead to Rome
Indeed there are many different tourist routes from our city to Rome. You are supposed to find your clients the route with the least cost while gaining the most happiness.
...]]></description><category></category></item><item><title><![CDATA[1053 Path of Equal Weight]]></title><link>https://blog.youkuaiyun.com/ssf_cxdm/article/details/83097301</link><guid>https://blog.youkuaiyun.com/ssf_cxdm/article/details/83097301</guid><author>ssf_cxdm</author><pubDate>Tue, 16 Oct 2018 20:51:14 +0800</pubDate><description><![CDATA[1053 Path of Equal Weight
题目链接
Given a non-empty tree with root R, and with weight W​i
​​  assigned to each tree node T​i
​​ . The weight of a path from R to L is defined to be the sum of the weights ...]]></description><category></category></item><item><title><![CDATA[1026 Table Tennis]]></title><link>https://blog.youkuaiyun.com/ssf_cxdm/article/details/83012340</link><guid>https://blog.youkuaiyun.com/ssf_cxdm/article/details/83012340</guid><author>ssf_cxdm</author><pubDate>Thu, 11 Oct 2018 14:47:53 +0800</pubDate><description><![CDATA[1026 Table Tennis
A table tennis club has N tables available to the public. The tables are numbered from 1 to N. For any pair of players, if there are some tables open when they arrive, they will be a...]]></description><category></category></item><item><title><![CDATA[1030 Travel Plan]]></title><link>https://blog.youkuaiyun.com/ssf_cxdm/article/details/83012289</link><guid>https://blog.youkuaiyun.com/ssf_cxdm/article/details/83012289</guid><author>ssf_cxdm</author><pubDate>Thu, 11 Oct 2018 14:44:59 +0800</pubDate><description><![CDATA[1030 Travel Plan
A traveler’s map gives the distances between cities along the highways, together with the cost of each highway. Now you are supposed to write a program to help a traveler to decide th...]]></description><category></category></item><item><title><![CDATA[1018 Public Bike Management]]></title><link>https://blog.youkuaiyun.com/ssf_cxdm/article/details/82987495</link><guid>https://blog.youkuaiyun.com/ssf_cxdm/article/details/82987495</guid><author>ssf_cxdm</author><pubDate>Tue, 09 Oct 2018 19:41:09 +0800</pubDate><description><![CDATA[1018 Public Bike Management
There is a public bike service in Hangzhou City which provides great convenience to the tourists from all over the world. One may rent a bike at any station and return it t...]]></description><category></category></item><item><title><![CDATA[1022 Digital Library]]></title><link>https://blog.youkuaiyun.com/ssf_cxdm/article/details/82987377</link><guid>https://blog.youkuaiyun.com/ssf_cxdm/article/details/82987377</guid><author>ssf_cxdm</author><pubDate>Tue, 09 Oct 2018 19:32:09 +0800</pubDate><description><![CDATA[1022 Digital Library
A Digital Library contains millions of books, stored according to their titles, authors, key words of their abstracts, publishers, and published years. Each book is assigned an un...]]></description><category></category></item><item><title><![CDATA[1148 Werewolf - Simple Version]]></title><link>https://blog.youkuaiyun.com/ssf_cxdm/article/details/82951794</link><guid>https://blog.youkuaiyun.com/ssf_cxdm/article/details/82951794</guid><author>ssf_cxdm</author><pubDate>Sat, 06 Oct 2018 16:11:59 +0800</pubDate><description><![CDATA[1148 Werewolf - Simple Version
题目链接
Werewolf（狼人杀） is a game in which the players are partitioned into two parties: the werewolves and the human beings. Suppose that in a game,
player #1 said: “Player ...]]></description><category></category></item><item><title><![CDATA[1144 The Missing Number]]></title><link>https://blog.youkuaiyun.com/ssf_cxdm/article/details/82951359</link><guid>https://blog.youkuaiyun.com/ssf_cxdm/article/details/82951359</guid><author>ssf_cxdm</author><pubDate>Sat, 06 Oct 2018 15:27:56 +0800</pubDate><description><![CDATA[1144 The Missing Number
Given N integers, you are supposed to find the smallest positive integer that is NOT in the given list.
Input Specification:
Each input file contains one test case. For each ca...]]></description><category></category></item></channel></rss>