<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[DJS_K_D的专栏]]></title><description><![CDATA[Maybe fun]]></description><link>https://blog.youkuaiyun.com/DJS_K_D</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; DJS_K_D]]></copyright><item><title><![CDATA[组合数学中的项链计数]]></title><link>https://blog.youkuaiyun.com/DJS_K_D/article/details/48543871</link><guid>https://blog.youkuaiyun.com/DJS_K_D/article/details/48543871</guid><author>DJS_K_D</author><pubDate>Fri, 18 Sep 2015 11:12:57 +0800</pubDate><description><![CDATA[给c种不同颜色宝石能穿成多少种长度为s的宝石项链（本质不同）






Burnside定理的应用：

当n为奇数时，有n种翻转，每种翻转都是以一个顶点和该顶点对边的中点对称。有k^(n/2+1)*n种。

当n为偶数时，有n种翻转，其中一半是以两个对应顶点，另一半是以两条对边对称。有k^(n/2+1)*n/2+k^(n/2)*n/2种。

考虑旋转：枚举旋转角度360]]></description><category></category></item><item><title><![CDATA[最小生成树的3个算法]]></title><link>https://blog.youkuaiyun.com/DJS_K_D/article/details/48131303</link><guid>https://blog.youkuaiyun.com/DJS_K_D/article/details/48131303</guid><author>DJS_K_D</author><pubDate>Mon, 31 Aug 2015 11:16:31 +0800</pubDate><description><![CDATA[最小生成树——Prim、Kruskal、Sollin(Boruvka)

 

本文内容框架：

1.Prim算法及其基于优先队列实现

      2.Kruskal算法

      3.Sollin算法

对于最小生成树，有两种算法可以解决。一种是Prim算法，该算法的时间复杂度为O(n²)，与图中边数无关，该算法适合于稠密图，而另外一种是Kruskal，该算法的时间]]></description><category></category></item><item><title><![CDATA[简单的表达式求值]]></title><link>https://blog.youkuaiyun.com/DJS_K_D/article/details/48130975</link><guid>https://blog.youkuaiyun.com/DJS_K_D/article/details/48130975</guid><author>DJS_K_D</author><pubDate>Mon, 31 Aug 2015 10:55:57 +0800</pubDate><description><![CDATA[#include 
using namespace std;
#define M 100
char ops[7] = {'+', '-', '*', '/', '(', ')', '='};
char cmp[7][7]= {{'>', '>', '', '>'},
    {'>', '>', '', '>'},
    {'>', '>', '>', '>', '', '>'},
    {']]></description><category></category></item><item><title><![CDATA[noip 2006]]></title><link>https://blog.youkuaiyun.com/DJS_K_D/article/details/48108993</link><guid>https://blog.youkuaiyun.com/DJS_K_D/article/details/48108993</guid><author>DJS_K_D</author><pubDate>Sun, 30 Aug 2015 21:08:29 +0800</pubDate><description><![CDATA[能量项链




【问题描述】

    在Mars星球上，每个Mars人都随身佩带着一串能量项链。在项链上有N颗能量珠。能量珠是一颗有头标记与尾标记的珠子，这些标记对应着某个正整数。并且，对于相邻的两颗珠子，前一颗珠子的尾标记一定等于后一颗珠子的头标记。因为只有这样，通过吸盘（吸盘是Mars人吸收能量的一种器官）的作用，这两颗珠子才能聚合成一颗珠子，同时释放出可以被吸盘吸收的能量。]]></description><category></category></item><item><title><![CDATA[noip2015 准备]]></title><link>https://blog.youkuaiyun.com/DJS_K_D/article/details/48108437</link><guid>https://blog.youkuaiyun.com/DJS_K_D/article/details/48108437</guid><author>DJS_K_D</author><pubDate>Sun, 30 Aug 2015 19:56:22 +0800</pubDate><description><![CDATA[从今天开始准备noip2015，先把历年的真题写完，再在各个oj上写些题，打打比赛，写过的题都记录下来


不负初心]]></description><category></category></item><item><title><![CDATA[poj 2828   线段树单点更新]]></title><link>https://blog.youkuaiyun.com/DJS_K_D/article/details/46821263</link><guid>https://blog.youkuaiyun.com/DJS_K_D/article/details/46821263</guid><author>DJS_K_D</author><pubDate>Thu, 09 Jul 2015 20:23:47 +0800</pubDate><description><![CDATA[类似于权值线段树？，就是翻过来插入，单点存当前区间空位置的个数


贴代码：
#include
#include
#include
#define M 200010
#define ls t*2
#define rs t*2+1


using namespace std;
struct Node
{
    int s,l,r;
} a[M*4];
int id,pos[M],val[M]]></description><category></category></item><item><title><![CDATA[Codeforces  Round 307 Div 2]]></title><link>https://blog.youkuaiyun.com/DJS_K_D/article/details/46503753</link><guid>https://blog.youkuaiyun.com/DJS_K_D/article/details/46503753</guid><author>DJS_K_D</author><pubDate>Mon, 15 Jun 2015 14:51:50 +0800</pubDate><description><![CDATA[A. GukiZ and Contest


time limit per test
 2 seconds


memory limit per test
 256 megabytes


input
 standard input


output
 standard output



Professor GukiZ likes programming ]]></description><category></category></item><item><title><![CDATA[主席树]]></title><link>https://blog.youkuaiyun.com/DJS_K_D/article/details/46390711</link><guid>https://blog.youkuaiyun.com/DJS_K_D/article/details/46390711</guid><author>DJS_K_D</author><pubDate>Sat, 06 Jun 2015 19:06:46 +0800</pubDate><description><![CDATA[主席树


好神的数据结构，不过挺好写的。。


POJ 2104 K-th number


题目大意：
     对一静态序列，每次询问[l,r]中第k大d


主席树模板题


贴代码：
#include
#include
#include
#define M 100010
using namespace std;


struct Node{
    in]]></description><category></category></item><item><title><![CDATA[bzoj 1034 n [ZJOI2008]泡泡堂BNB 贪心 田忌赛马]]></title><link>https://blog.youkuaiyun.com/DJS_K_D/article/details/46010129</link><guid>https://blog.youkuaiyun.com/DJS_K_D/article/details/46010129</guid><author>DJS_K_D</author><pubDate>Tue, 26 May 2015 16:57:56 +0800</pubDate><description><![CDATA[bzoj   1034
http://www.lydsy.com/JudgeOnline/problem.php?id=1034





最强的比能赢就赢，最弱的比能赢就赢
其他的用最弱的跟别人最强的比


#include
#include
#define MAXN 100010

using namespace std;

int a[MAXN],b[MAXN],n;
]]></description><category></category></item><item><title><![CDATA[bzoj 1047 [HAOI2007]理想的正方形 单调队列]]></title><link>https://blog.youkuaiyun.com/DJS_K_D/article/details/46009081</link><guid>https://blog.youkuaiyun.com/DJS_K_D/article/details/46009081</guid><author>DJS_K_D</author><pubDate>Tue, 26 May 2015 15:58:53 +0800</pubDate><description><![CDATA[bzoj  1047
http://www.lydsy.com/JudgeOnline/problem.php?id=1047






Description

有一个a*b的整数组成的矩阵，现请你从中找出一个n*n的正方形区域，使得该区域所有数中的最大值和最小值的差最小。

Input

第一行为3个整数，分别表示a,b,n的值第二行至第a+1行每行为b个非负整数]]></description><category></category></item><item><title><![CDATA[bzoj 3240  [Noi2013]矩阵游戏]]></title><link>https://blog.youkuaiyun.com/DJS_K_D/article/details/45875387</link><guid>https://blog.youkuaiyun.com/DJS_K_D/article/details/45875387</guid><author>DJS_K_D</author><pubDate>Wed, 20 May 2015 21:06:27 +0800</pubDate><description><![CDATA[bzoj   3240
传送门：
http://www.lydsy.com/JudgeOnline/problem.php?id=3240




Description

婷婷是个喜欢矩阵的小朋友,有一天她想用电脑生成一个巨大的n行m列的矩阵(你不用担心她如何存储)。她生成的这个矩阵满足一个神奇的性质：若用F[i][j]来表示矩阵中第i行第j列的元素，则F[i][j]满足下面的]]></description><category></category></item><item><title><![CDATA[bzoj 1207 [HNOI2004]打鼹鼠]]></title><link>https://blog.youkuaiyun.com/DJS_K_D/article/details/45848795</link><guid>https://blog.youkuaiyun.com/DJS_K_D/article/details/45848795</guid><author>DJS_K_D</author><pubDate>Tue, 19 May 2015 19:36:11 +0800</pubDate><description><![CDATA[bzoj 1207
http://www.lydsy.com/JudgeOnline/problem.php?id=1207






Description

鼹鼠是一种很喜欢挖洞的动物，但每过一定的时间，它还是喜欢把头探出到地面上来透透气的。根据这个特点阿Q编写了一个打鼹鼠的游戏：在一个n*n的网格中，在某些时刻鼹鼠会在某一个网格探出头来透透气。你可以控制一个机器人来打鼹]]></description><category></category></item><item><title><![CDATA[bzoj 1071 [SCOI2007]组队  单调性]]></title><link>https://blog.youkuaiyun.com/DJS_K_D/article/details/45848217</link><guid>https://blog.youkuaiyun.com/DJS_K_D/article/details/45848217</guid><author>DJS_K_D</author><pubDate>Tue, 19 May 2015 18:37:28 +0800</pubDate><description><![CDATA[bzoj  1071
http://www.lydsy.com/JudgeOnline/problem.php?id=1071






Description

NBA每年都有球员选秀环节。通常用速度和身高两项数据来衡量一个篮球运动员的基本素质。假如一支球队里速度最慢的球员速度为minV，身高最矮的球员高度为minH，那么这支球队的所有队员都应该满足: A * ( heig]]></description><category></category></item><item><title><![CDATA[bzoj  3675 [Apio2014]序列分割   斜率优化]]></title><link>https://blog.youkuaiyun.com/DJS_K_D/article/details/45827803</link><guid>https://blog.youkuaiyun.com/DJS_K_D/article/details/45827803</guid><author>DJS_K_D</author><pubDate>Mon, 18 May 2015 21:26:17 +0800</pubDate><description><![CDATA[bzoj 3675
传送门：
http://www.lydsy.com/JudgeOnline/problem.php?id=3675






Description

小H最近迷上了一个分割序列的游戏。在这个游戏里，小H需要将一个长 
度为N的非负整数序列分割成k+l个非空的子序列。为了得到k+l个子序列， 
小H将重复进行七次以下的步骤： 
1．小H首先选择一个]]></description><category></category></item><item><title><![CDATA[bzoj 1858  序列操作  线段树]]></title><link>https://blog.youkuaiyun.com/DJS_K_D/article/details/45049295</link><guid>https://blog.youkuaiyun.com/DJS_K_D/article/details/45049295</guid><author>DJS_K_D</author><pubDate>Tue, 14 Apr 2015 22:44:20 +0800</pubDate><description><![CDATA[传送门：
http://www.lydsy.com/JudgeOnline/problem.php?id=1858


Description

lxhgww最近收到了一个01序列，序列里面包含了n个数，这些数要么是0，要么是1，现在对于这个序列有五种变换操作和询问操作： 0 a b 把[a, b]区间内的所有数全变成0 1 a b 把[a, b]区间内的所有数全变成1 2 a b 把]]></description><category></category></item><item><title><![CDATA[网络流复习]]></title><link>https://blog.youkuaiyun.com/DJS_K_D/article/details/45024255</link><guid>https://blog.youkuaiyun.com/DJS_K_D/article/details/45024255</guid><author>DJS_K_D</author><pubDate>Mon, 13 Apr 2015 12:44:54 +0800</pubDate><description><![CDATA[这几天做了些网络流的题，在此复习一下：
bzoj 1066 蜥蜴
http://www.lydsy.com/JudgeOnline/problem.php?id=1066
        最大流
先拆点，对于每个石柱i，拆成in(i),ou(i)两点，连一条容量height[i]的边，能跳出去的向T连容量INF的边，然后对于每条蜥蜴，向能够到的石柱连一条容量INF的边表示跳跃，S向每条蜥蜴]]></description><category></category></item><item><title><![CDATA[凸包]]></title><link>https://blog.youkuaiyun.com/DJS_K_D/article/details/44678339</link><guid>https://blog.youkuaiyun.com/DJS_K_D/article/details/44678339</guid><author>DJS_K_D</author><pubDate>Fri, 27 Mar 2015 16:14:03 +0800</pubDate><description><![CDATA[自己凭着机房神犇的讲解写的求凸包～～～～到现在才写。。。太弱了


                   仅是自己的纪念～～～不要炒啊～～～（复杂度方面不保证优）

#include
#include
#include
#include
#include
#include
#define MAXN 200000
using namespace std;


int n;
int Tup[]]></description><category></category></item><item><title><![CDATA[网络流]]></title><link>https://blog.youkuaiyun.com/DJS_K_D/article/details/44023905</link><guid>https://blog.youkuaiyun.com/DJS_K_D/article/details/44023905</guid><author>DJS_K_D</author><pubDate>Mon, 02 Mar 2015 20:56:45 +0800</pubDate><description><![CDATA[网络流
     要细致学网络流了。。。虽然狼抓兔子自己写出来了 ，，可是那个建模太明显了~~~~~~~
       
http://blog.sina.com.cn/s/blog_68629c7701010r91.html
      ~~~~讲的好细致~~~~
    唉，图复制不过来，就去那看吧........
﻿﻿]]></description><category></category></item><item><title><![CDATA[Splay]]></title><link>https://blog.youkuaiyun.com/DJS_K_D/article/details/44005343</link><guid>https://blog.youkuaiyun.com/DJS_K_D/article/details/44005343</guid><author>DJS_K_D</author><pubDate>Sun, 01 Mar 2015 20:42:12 +0800</pubDate><description><![CDATA[Splay
     写这玩意好累~~~~
     这里贴一份用指针写的（为什么周围的神犇不喜欢用]]></description><category></category></item><item><title><![CDATA[矩阵乘法]]></title><link>https://blog.youkuaiyun.com/DJS_K_D/article/details/44004771</link><guid>https://blog.youkuaiyun.com/DJS_K_D/article/details/44004771</guid><author>DJS_K_D</author><pubDate>Sun, 01 Mar 2015 19:54:15 +0800</pubDate><description><![CDATA[矩阵乘法     本来在看bzoj1009GT考试题解~~~不会做~~~，看到有矩阵乘法这个知识点，就去补知识点了       基础可以看这个http://zh.wikipedia.org/wiki/%E7%9F%A9%E9%99%A3%E4%B9%98%E6%B3%95       进一步了解（代码）的看http://baike.baidu.com/link?url=E_X4GHOFjNlD5ar]]></description><category></category></item></channel></rss>