<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[随风而行的博客]]></title><description><![CDATA[岁月无声]]></description><link>https://blog.youkuaiyun.com/C_13579</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; C_13579]]></copyright><item><title><![CDATA[Codeforces Round #782 (Div. 2)-D. Reverse Sort Sum]]></title><link>https://blog.youkuaiyun.com/C_13579/article/details/126067010</link><guid>https://blog.youkuaiyun.com/C_13579/article/details/126067010</guid><author>C_13579</author><pubDate>Sat, 30 Jul 2022 03:15:05 +0800</pubDate><description><![CDATA[思路：树状数组]]></description><category></category></item><item><title><![CDATA[Codeforces Round #779 (Div. 2)-D2 - 388535 (Hard Version)]]></title><link>https://blog.youkuaiyun.com/C_13579/article/details/126026240</link><guid>https://blog.youkuaiyun.com/C_13579/article/details/126026240</guid><author>C_13579</author><pubDate>Thu, 28 Jul 2022 01:26:53 +0800</pubDate><description><![CDATA[异或字典树的最大值Max和最小值Min,由于正确答案res^a[i]时会得到[l,r]区间，因此只要。来求得res,但直接暴力遍历会超时，可通过建立字典树，然后求。之中，那么可以通过遍历。首先答案res一定位于。...]]></description><category></category></item><item><title><![CDATA[LeetCode-327. 区间和的个数]]></title><link>https://blog.youkuaiyun.com/C_13579/article/details/125985327</link><guid>https://blog.youkuaiyun.com/C_13579/article/details/125985327</guid><author>C_13579</author><pubDate>Mon, 25 Jul 2022 23:45:50 +0800</pubDate><description><![CDATA[同时是对于前缀和数组进行分析，对于pre[i],当j>i,pre[j]位于区间[pre[i]+lower,pre[i]+upper]之间是合法的，因此可以从左到右遍历pre，对于pre[i],查询线段树中下标pre[i]的个数,然后将区间[pre[i]+lower,pre[i]+upper]加1，需要注意将pre[i]，pre[i]+lower,pre[i]+upper哈希离散化。思路归并排序/树状数组（线段树）2.树状数组（线段树）code2.数组数组。...]]></description><category></category></item><item><title><![CDATA[LeedCode-862. 和至少为 K 的最短子数组]]></title><link>https://blog.youkuaiyun.com/C_13579/article/details/125898338</link><guid>https://blog.youkuaiyun.com/C_13579/article/details/125898338</guid><author>C_13579</author><pubDate>Wed, 20 Jul 2022 18:53:00 +0800</pubDate><description><![CDATA[对于右边窗口，若i=pre[j]，显然可以将pre[i]从窗口中去除，利用j作为左区间显然更好。利用滑动窗口维护一个合法区间，遍历i=0->n,对于当前下标j。而对于左边窗口，pre[i]已经求得其最小区间，故可以将其从窗口去除。先求出数组前缀和，问题就转变成前缀和之差不小于k的最小区间。思路前缀和+滑动窗口。......]]></description><category></category></item><item><title><![CDATA[关于C++的类型转换运算符 static_cast 和 const_cast 的说明]]></title><link>https://blog.youkuaiyun.com/C_13579/article/details/125650992</link><guid>https://blog.youkuaiyun.com/C_13579/article/details/125650992</guid><author>C_13579</author><pubDate>Thu, 07 Jul 2022 18:09:57 +0800</pubDate><description><![CDATA[C++ 对于const_static 和 static_cast 关于去除或添加const/volatile的理解]]></description><category></category></item><item><title><![CDATA[LeetCode-188. 买卖股票的最佳时机 IV]]></title><link>https://blog.youkuaiyun.com/C_13579/article/details/125645176</link><guid>https://blog.youkuaiyun.com/C_13579/article/details/125645176</guid><author>C_13579</author><pubDate>Wed, 06 Jul 2022 18:22:35 +0800</pubDate><description><![CDATA[Leetcode 188]]></description><category></category></item><item><title><![CDATA[Codeforces Round #791 (Div. 2)-D. Toss a Coin to Your Graph...]]></title><link>https://blog.youkuaiyun.com/C_13579/article/details/124990065</link><guid>https://blog.youkuaiyun.com/C_13579/article/details/124990065</guid><author>C_13579</author><pubDate>Thu, 26 May 2022 18:31:18 +0800</pubDate><description><![CDATA[传送门：https://codeforces.com/contest/1679/problem/D
思路：二分
可以通过二分答案si，对于judge的判断，可通过对于所有不大于si的边重新构图，然后先找图中是否有环，有则true，没有则从入度为0的点开始BFS找最长的路径即可
Code:
#include&lt;iostream&gt;
#include&lt;algorithm&gt;
#include&lt;vector&gt;
#include&lt;queue&gt;
using namespac.]]></description><category></category></item><item><title><![CDATA[Codeforces Round #783 (Div. 2)-D. Optimal Partition]]></title><link>https://blog.youkuaiyun.com/C_13579/article/details/124834099</link><guid>https://blog.youkuaiyun.com/C_13579/article/details/124834099</guid><author>C_13579</author><pubDate>Wed, 18 May 2022 02:13:10 +0800</pubDate><description><![CDATA[传送门：https://codeforces.com/contest/1668/problem/D
思路：dp+树状数组， 参考博客https://www.cnblogs.com/Prgl/p/16169339.html
需要注意dp初始化时需要把0先算上去，即假设数列的左边隐含着一个为0的元素做分割线
Code:
#include&lt;iostream&gt;
#include&lt;algorithm&gt;
#include&lt;set&gt;
#include&lt;unordered_map.]]></description><category></category></item><item><title><![CDATA[Educational Codeforces Round 125 (Rated for Div. 2)-D. For Gamers. By Gamers.]]></title><link>https://blog.youkuaiyun.com/C_13579/article/details/124204508</link><guid>https://blog.youkuaiyun.com/C_13579/article/details/124204508</guid><author>C_13579</author><pubDate>Fri, 15 Apr 2022 21:57:36 +0800</pubDate><description><![CDATA[地址：https://codeforces.com/contest/1657/problem/D
思路：要使得单位 i {di,hi}能够战胜怪兽{Di,Hi},有 Hidi&gt;hiDi\frac{Hi}{di}&gt;\frac{hi}{Di}diHi​&gt;Dihi​，即 hi∗di&gt;Hi∗Dihi*di &gt; Hi*Dihi∗di&gt;Hi∗Di ，而x个单位i战胜怪兽而不是死掉一个单位，即 i∗hi∗di&gt;Hi∗Dii * hi * di &gt; Hi * Dii∗hi∗.]]></description><category></category></item><item><title><![CDATA[LeetCode-1825. 求出 MK 平均值]]></title><link>https://blog.youkuaiyun.com/C_13579/article/details/115746049</link><guid>https://blog.youkuaiyun.com/C_13579/article/details/115746049</guid><author>C_13579</author><pubDate>Fri, 16 Apr 2021 02:58:12 +0800</pubDate><description><![CDATA[地址：https://leetcode-cn.com/problems/finding-mk-average/
思路：https://leetcode-cn.com/problems/finding-mk-average/solution/c-san-ge-multiset-jian-dan-mo-ni-by-newh-y4q9/
维护 3 个 multiset：lower（保存最小的 k 个数）、middle（中间的数）、upper（保存最大的 k 个数）。
插入操作

如果 num≤max⁡(lowe.]]></description><category></category></item><item><title><![CDATA[CentOs7安装apache以及遇到的问题]]></title><link>https://blog.youkuaiyun.com/C_13579/article/details/112597961</link><guid>https://blog.youkuaiyun.com/C_13579/article/details/112597961</guid><author>C_13579</author><pubDate>Thu, 14 Jan 2021 00:35:41 +0800</pubDate><description><![CDATA[转载与https://blog.51cto.com/9083918/2159995
新手在CentOs7装一个Apache时，可以参考本文
1.下载httpd源码包
   [root@localhost utils]# mkdir /usr/local/utils

   [root@localhost utils]# cd /usr/local/utils

   [root@localhost utils]# wget http://mirrors.hust.edu.cn/apache/httpd/ht]]></description><category></category></item><item><title><![CDATA[LeetCode-32. 最长有效括号]]></title><link>https://blog.youkuaiyun.com/C_13579/article/details/112297088</link><guid>https://blog.youkuaiyun.com/C_13579/article/details/112297088</guid><author>C_13579</author><pubDate>Wed, 06 Jan 2021 20:32:36 +0800</pubDate><description><![CDATA[地址：https://leetcode-cn.com/problems/longest-valid-parentheses/
思路：利用栈记录入栈字符的下标位置，从头处理字符串，对于s[i]，若栈头元素与其匹配，则将其从退栈，并求其长度并保存最大长度，若不匹配则入栈。
Code:
class Solution{
public:
	int longestValidParentheses(string s){
		stack&lt;int&gt; sta;
		int res=0;
		for(int i=0.]]></description><category></category></item><item><title><![CDATA[最小生成树-Kruskal算法模板]]></title><link>https://blog.youkuaiyun.com/C_13579/article/details/111601636</link><guid>https://blog.youkuaiyun.com/C_13579/article/details/111601636</guid><author>C_13579</author><pubDate>Wed, 23 Dec 2020 23:40:40 +0800</pubDate><description><![CDATA[最小生成树-Kruskal算法：时间复杂度 O(mlogm) ,m为边数
模板题：P3366 【模板】最小生成树
Code:
#include&lt;iostream&gt;
#include&lt;algorithm&gt;
using namespace std;

struct node{
	int u,v,w;
	bool operator&lt;(const node &amp;A)const{
		return w&lt;A.w;
	}
};
const int MAX_N=5e3+5;
con]]></description><category></category></item><item><title><![CDATA[最小生成树-Prim算法模板]]></title><link>https://blog.youkuaiyun.com/C_13579/article/details/111601595</link><guid>https://blog.youkuaiyun.com/C_13579/article/details/111601595</guid><author>C_13579</author><pubDate>Wed, 23 Dec 2020 23:38:32 +0800</pubDate><description><![CDATA[最小生成树-Prim算法：
未优化：时间复杂度 O(n^2),n为顶点数
堆优化：时间复杂度大概为 O((n+m)logm),m为边数
模板题：P3366 【模板】最小生成树
Code 未优化:
#include&lt;iostream&gt;
#include&lt;vector&gt;
using namespace std;
typedef pair&lt;int,int&gt; pr; 

const int MAX_N=5e3+5;
const int INF=1e9+5;
int n,m;
ve]]></description><category></category></item><item><title><![CDATA[最短路-Dijkstra算法模板]]></title><link>https://blog.youkuaiyun.com/C_13579/article/details/111601100</link><guid>https://blog.youkuaiyun.com/C_13579/article/details/111601100</guid><author>C_13579</author><pubDate>Wed, 23 Dec 2020 23:25:46 +0800</pubDate><description><![CDATA[最短路-Dijkstra算法
不能处理负权边
未优化：时间复杂度为O(n^2) ,n为顶点数
堆优化：时间复杂度大概为O((m+n)logm)，m为边数
模板题（未优化）：P3371 【模板】单源最短路径（弱化版）
模板题（堆优化）：P4779 【模板】单源最短路径（标准版）
Code 未优化：
#include&lt;iostream&gt;
#include&lt;vector&gt;
using namespace std;

struct node{
	int v;
	int w;
};
const]]></description><category></category></item><item><title><![CDATA[数据结构-平衡二叉树（C++代码实现）]]></title><link>https://blog.youkuaiyun.com/C_13579/article/details/111570474</link><guid>https://blog.youkuaiyun.com/C_13579/article/details/111570474</guid><author>C_13579</author><pubDate>Wed, 23 Dec 2020 00:39:17 +0800</pubDate><description><![CDATA[数据结构中平衡二叉树的C++代码实现
有关参考博客：https://blog.youkuaiyun.com/qq_25940921/article/details/82183093
https://www.cnblogs.com/zhangbaochong/p/5164994.html
代码及其测试结果如下
Code :
/*
二叉平衡搜索树(AVL):对于二叉搜索树的所有结点，其左右节点的高度之差不超过 1 
*/ 
#include&lt;iostream&gt;
#include&lt;queue&gt;
usi]]></description><category></category></item><item><title><![CDATA[数据结构-二叉排序树（C++代码实现）]]></title><link>https://blog.youkuaiyun.com/C_13579/article/details/111569695</link><guid>https://blog.youkuaiyun.com/C_13579/article/details/111569695</guid><author>C_13579</author><pubDate>Wed, 23 Dec 2020 00:33:43 +0800</pubDate><description><![CDATA[数据结构中二叉排序树的C++代码实现
有关参考博客：https://blog.youkuaiyun.com/kang___xi/article/details/80392565
Code :
/*
二叉排序树（二叉搜索树）：树上所有结点的左子树的值均小于该节点的值，右子树的值均大于该节点的值
https://blog.youkuaiyun.com/kang___xi/article/details/80392565
*/ 
#include&lt;iostream&gt;
#include&lt;queue&gt;
using n]]></description><category></category></item><item><title><![CDATA[LeetCode-102. 二叉树的层序遍历]]></title><link>https://blog.youkuaiyun.com/C_13579/article/details/111360278</link><guid>https://blog.youkuaiyun.com/C_13579/article/details/111360278</guid><author>C_13579</author><pubDate>Fri, 18 Dec 2020 14:23:26 +0800</pubDate><description><![CDATA[地址：https://leetcode-cn.com/problems/binary-tree-level-order-traversal/
思路：BFS/DFS
思路一、BFS：利用广度优先搜索逐层搜索，每次搜索一层，将下一层的加入队列中即可
思路二、DFS：利用深度优先搜索，则为从左向右搜索，对于层数可用变量进行记录即可
Code BFS:
class Solution {
    vector&lt;vector&lt;int&gt;&gt; res;
    queue&lt;TreeNode *.]]></description><category></category></item><item><title><![CDATA[HDU-4192-Guess the Numbers]]></title><link>https://blog.youkuaiyun.com/C_13579/article/details/111357212</link><guid>https://blog.youkuaiyun.com/C_13579/article/details/111357212</guid><author>C_13579</author><pubDate>Fri, 18 Dec 2020 12:53:40 +0800</pubDate><description><![CDATA[地址：http://acm.hdu.edu.cn/showproblem.php?pid=4192
思路：首先将中缀表达式转为后缀表达式，然后将数组全排列取计算每一个排列的后缀表达式的值即可
Code：
#include&lt;iostream&gt;
#include&lt;algorithm&gt;
#include&lt;stack&gt;
using namespace std;

int n;
int d[55];
stack&lt;string&gt; sta;

string transfo.]]></description><category></category></item><item><title><![CDATA[LeetCode-714.买卖股票的最佳时机含手续费]]></title><link>https://blog.youkuaiyun.com/C_13579/article/details/110602380</link><guid>https://blog.youkuaiyun.com/C_13579/article/details/110602380</guid><author>C_13579</author><pubDate>Fri, 04 Dec 2020 00:43:35 +0800</pubDate><description><![CDATA[地址：https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock-with-transaction-fee/
思路：贪心 || DP
思路一、贪心：利用贪心思路，当出现峰值（买卖间有正差值）时即将股票卖掉，同时寻找下一个峰值。对于卖出时需要手续费fee,其峰值相当于卖出价格prices[i]-fee,记录当前最小值Min,当prices[i]-fee&gt;Min时，则答案加上其差值res+=prices[i]-fee-Min;同时令 .]]></description><category></category></item></channel></rss>