<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[写东西的地方]]></title><description><![CDATA[on the way]]></description><link>https://blog.youkuaiyun.com/HTT_H</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; HTT_H]]></copyright><item><title><![CDATA[hdu 4418 高斯消元+概率dp]]></title><link>https://blog.youkuaiyun.com/HTT_H/article/details/48629409</link><guid>https://blog.youkuaiyun.com/HTT_H/article/details/48629409</guid><author>HTT_H</author><pubDate>Mon, 21 Sep 2015 17:20:07 +0800</pubDate><description><![CDATA[Time travel
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2141    Accepted Submission(s): 489



Problem Description



Agent K is ]]></description><category></category></item><item><title><![CDATA[SPOJ COT2 树上的莫队算法，树上区间查询]]></title><link>https://blog.youkuaiyun.com/HTT_H/article/details/47729479</link><guid>https://blog.youkuaiyun.com/HTT_H/article/details/47729479</guid><author>HTT_H</author><pubDate>Mon, 17 Aug 2015 19:09:00 +0800</pubDate><description><![CDATA[题意：n个节点形成的一棵树。每个节点有一个值。m次查询，求出(u,v)路径上出现了多少个不同的数。
树上的莫队算法，同样将树分成siz=sqrt(n)块，然后离线操作。先对树dfs一遍，每当子树节点个数num>=siz，就将这num个分成一块。读取所有的查询按左端点所在块排序。
重点在于怎么进行区间转移，对路径的lca特殊处理，参考博客http://blog.youkuaiyun.com/kuribohg]]></description><category></category></item><item><title><![CDATA[hdu 5381 莫队算法/gcd]]></title><link>https://blog.youkuaiyun.com/HTT_H/article/details/47729139</link><guid>https://blog.youkuaiyun.com/HTT_H/article/details/47729139</guid><author>HTT_H</author><pubDate>Mon, 17 Aug 2015 18:55:41 +0800</pubDate><description><![CDATA[The sum of gcd
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 645    Accepted Submission(s): 279


Problem Description
You have an array]]></description><category></category></item><item><title><![CDATA[主席树（可持久化线段树）入门专题]]></title><link>https://blog.youkuaiyun.com/HTT_H/article/details/47704209</link><guid>https://blog.youkuaiyun.com/HTT_H/article/details/47704209</guid><author>HTT_H</author><pubDate>Sun, 16 Aug 2015 18:34:52 +0800</pubDate><description><![CDATA[1.poj 2104 查询区间第k小。
主席树其实相当于建立了n棵线段树，第i棵线段树是根据区间【1，i】按值建立的。对于每一棵线段树我们记录它对应的区间每个数出现的次数，所以首先要对所有的数离散化。
先考虑最简单的情况，只查询【1，n】的第k小，对于【1，n】我们按值建立一棵线段树，对于a[i]我们在位置a[i]上加1。查询第k小那么先看左子区间出现了多少个数cnt，假设左区间出现的数cnt]]></description><category></category></item><item><title><![CDATA[zoj 2112 Dynamic Rankings 带修改区间第k大的几种解法]]></title><link>https://blog.youkuaiyun.com/HTT_H/article/details/47701417</link><guid>https://blog.youkuaiyun.com/HTT_H/article/details/47701417</guid><author>HTT_H</author><pubDate>Sun, 16 Aug 2015 12:28:18 +0800</pubDate><description><![CDATA[给出n个数，有两种操作，一种是修改某个数的值，另一种查询指定区间第k大。
比较快的做法是树套树，而分块算法复杂度比较高写起来方便。分块算法可以很简单的处理单独修改某个值的情况。
将n个数分成num块，每块大小siz=n/num。每一块内部排序进行排序，查询[l,r]第k大时，先二分答案，对于完全包含在区间的内块直接二分搜索，而对于区间两端只有部分包含的则直接遍历查找。复杂度是logn*(2*s]]></description><category></category></item><item><title><![CDATA[cf-546E 网络流，状态转移平衡]]></title><link>https://blog.youkuaiyun.com/HTT_H/article/details/47683069</link><guid>https://blog.youkuaiyun.com/HTT_H/article/details/47683069</guid><author>HTT_H</author><pubDate>Sat, 15 Aug 2015 16:44:20 +0800</pubDate><description><![CDATA[E. Soldier and Traveling


time limit per test
1 second


memory limit per test
256 megabytes


input
standard input


output
standard output



In the country there are n cities a]]></description><category></category></item><item><title><![CDATA[hdu 3998 最长上升子序列个数+网络流 (最多不相交合法路径数）]]></title><link>https://blog.youkuaiyun.com/HTT_H/article/details/47453111</link><guid>https://blog.youkuaiyun.com/HTT_H/article/details/47453111</guid><author>HTT_H</author><pubDate>Wed, 12 Aug 2015 22:37:40 +0800</pubDate><description><![CDATA[Sequence
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1886    Accepted Submission(s): 687



Problem Description

There is a sequenc]]></description><category></category></item><item><title><![CDATA[hdu 3416 最短路+网络流（不重叠最短路径计数）]]></title><link>https://blog.youkuaiyun.com/HTT_H/article/details/47452295</link><guid>https://blog.youkuaiyun.com/HTT_H/article/details/47452295</guid><author>HTT_H</author><pubDate>Wed, 12 Aug 2015 22:11:57 +0800</pubDate><description><![CDATA[Marriage Match IV
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2594    Accepted Submission(s): 772



Problem Description

Do not si]]></description><category></category></item><item><title><![CDATA[hdu 3338 网络流填数字]]></title><link>https://blog.youkuaiyun.com/HTT_H/article/details/47398473</link><guid>https://blog.youkuaiyun.com/HTT_H/article/details/47398473</guid><author>HTT_H</author><pubDate>Mon, 10 Aug 2015 14:55:28 +0800</pubDate><description><![CDATA[Kakuro Extension
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1268    Accepted Submission(s): 439
Special Judge

Problem Description
I]]></description><category></category></item><item><title><![CDATA[poj 1741 树分治入门]]></title><link>https://blog.youkuaiyun.com/HTT_H/article/details/47265201</link><guid>https://blog.youkuaiyun.com/HTT_H/article/details/47265201</guid><author>HTT_H</author><pubDate>Mon, 03 Aug 2015 23:20:45 +0800</pubDate><description><![CDATA[题意:统计距离
这道题用的树的点分治方法。点分治基于树的重心。
树的重心，定义是删除某个点后得到的最大（节点数）子树的节点数最小。性质是，可以证明删除掉重心后，每个子树的大小]]></description><category></category></item><item><title><![CDATA[hdu 5324 树套树、cdq分治]]></title><link>https://blog.youkuaiyun.com/HTT_H/article/details/47205241</link><guid>https://blog.youkuaiyun.com/HTT_H/article/details/47205241</guid><author>HTT_H</author><pubDate>Sun, 02 Aug 2015 11:33:53 +0800</pubDate><description><![CDATA[一、线段树/树状数组套平衡树

题目大意是给你n个(L[i], R[i])的点，找出最长的子序列，满足L[i]递减，R[i]递增。为了看起来顺眼点，我们可以把其中一维取相反数，使得两个都找递减，或者都找递增的。
我们把L[i]取相反数，找两个都递增的。假设不考虑R[i]，只考虑L[i],那么就相当于LIS问题了。


dp[i]记录以L[i]为结尾的最大长度，dp[i]=max{dp[]]></description><category></category></item><item><title><![CDATA[hdu 4630 线段树+离线处理]]></title><link>https://blog.youkuaiyun.com/HTT_H/article/details/47124709</link><guid>https://blog.youkuaiyun.com/HTT_H/article/details/47124709</guid><author>HTT_H</author><pubDate>Wed, 29 Jul 2015 11:46:02 +0800</pubDate><description><![CDATA[No Pain No Game
Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1790    Accepted Submission(s): 763



Problem Description

Life is a g]]></description><category></category></item><item><title><![CDATA[poj 3415 后缀数组+单调队列]]></title><link>https://blog.youkuaiyun.com/HTT_H/article/details/45825199</link><guid>https://blog.youkuaiyun.com/HTT_H/article/details/45825199</guid><author>HTT_H</author><pubDate>Mon, 18 May 2015 17:47:46 +0800</pubDate><description><![CDATA[Common Substrings




Time Limit: 5000MS
 
Memory Limit: 65536K


Total Submissions: 8106
 
Accepted: 2688





Description


A substring of a string T is defined as:
T(i, k)=TiT]]></description><category></category></item><item><title><![CDATA[poj 1325 最小顶点覆盖]]></title><link>https://blog.youkuaiyun.com/HTT_H/article/details/45787547</link><guid>https://blog.youkuaiyun.com/HTT_H/article/details/45787547</guid><author>HTT_H</author><pubDate>Sun, 17 May 2015 11:15:30 +0800</pubDate><description><![CDATA[Machine Schedule




Time Limit: 1000MS
 
Memory Limit: 10000K


Total Submissions: 12559
 
Accepted: 5368





Description
As we all know, machine scheduling is a very classical pr]]></description><category></category></item><item><title><![CDATA[poj 2446 二分图最大匹配 匈牙利算法]]></title><link>https://blog.youkuaiyun.com/HTT_H/article/details/45787489</link><guid>https://blog.youkuaiyun.com/HTT_H/article/details/45787489</guid><author>HTT_H</author><pubDate>Sun, 17 May 2015 11:10:02 +0800</pubDate><description><![CDATA[Chessboard




Time Limit: 2000MS
 
Memory Limit: 65536K


Total Submissions: 14525
 
Accepted: 4513





Description
Alice and Bob often play games on chessboard. One day, Alice dr]]></description><category></category></item><item><title><![CDATA[poj 1422 最小路径覆盖]]></title><link>https://blog.youkuaiyun.com/HTT_H/article/details/45787429</link><guid>https://blog.youkuaiyun.com/HTT_H/article/details/45787429</guid><author>HTT_H</author><pubDate>Sun, 17 May 2015 11:06:17 +0800</pubDate><description><![CDATA[Air Raid




Time Limit: 1000MS
 
Memory Limit: 10000K


Total Submissions: 7172
 
Accepted: 4261





Description
Consider a town where all the streets are one-way and each street ]]></description><category></category></item><item><title><![CDATA[poj 3352 双连通分量]]></title><link>https://blog.youkuaiyun.com/HTT_H/article/details/45673933</link><guid>https://blog.youkuaiyun.com/HTT_H/article/details/45673933</guid><author>HTT_H</author><pubDate>Tue, 12 May 2015 19:43:58 +0800</pubDate><description><![CDATA[Road Construction




Time Limit: 2000MS
 
Memory Limit: 65536K


Total Submissions: 9542
 
Accepted: 4739





Description


It's almost summer time, and that means that it's alm]]></description><category></category></item><item><title><![CDATA[hdu 4738 边双连通&桥，有重边情况]]></title><link>https://blog.youkuaiyun.com/HTT_H/article/details/45673899</link><guid>https://blog.youkuaiyun.com/HTT_H/article/details/45673899</guid><author>HTT_H</author><pubDate>Tue, 12 May 2015 19:41:39 +0800</pubDate><description><![CDATA[Caocao's Bridges
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1955    Accepted Submission(s): 699


Problem Description
Caocao was def]]></description><category></category></item><item><title><![CDATA[zoj 3649 树上的倍增法]]></title><link>https://blog.youkuaiyun.com/HTT_H/article/details/45601477</link><guid>https://blog.youkuaiyun.com/HTT_H/article/details/45601477</guid><author>HTT_H</author><pubDate>Sat, 09 May 2015 15:03:24 +0800</pubDate><description><![CDATA[Social Net

Time Limit: 5 Seconds      Memory Limit: 65536 KB


There are n individuals(2 n <= 30000). Everyone has one or more friends. And everyone can contact all people by friend-relation. If]]></description><category></category></item><item><title><![CDATA[hdu 2255 二分图最大权匹配 km算法模板]]></title><link>https://blog.youkuaiyun.com/HTT_H/article/details/45601437</link><guid>https://blog.youkuaiyun.com/HTT_H/article/details/45601437</guid><author>HTT_H</author><pubDate>Sat, 09 May 2015 15:02:28 +0800</pubDate><description><![CDATA[代码是打的kuangbin的模板


#include 
#include 
#include 
using namespace std;

const int maxn= 305;
const int inf= 0x3f3f3f3f;
int g[maxn][maxn];
int nx,ny;//二分图两类点的个数
int linker[maxn],lx[maxn],ly[maxn];
i]]></description><category></category></item></channel></rss>