<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[Windrunner]]></title><description><![CDATA[                            ——最可怕的事情不是别人比你聪明，而是比你聪明的人比你更努力。 ]]></description><link>https://blog.youkuaiyun.com/E_Windrunner</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; E_Windrunner]]></copyright><item><title><![CDATA[UVa 11624 Fire!]]></title><link>https://blog.youkuaiyun.com/E_Windrunner/article/details/8998871</link><guid>https://blog.youkuaiyun.com/E_Windrunner/article/details/8998871</guid><author>E_Windrunner</author><pubDate>Fri, 31 May 2013 12:16:13 +0800</pubDate><description><![CDATA[谁能帮我看看为什么第一份代码是AC，而第二份是WA呢- -！，就改了起火时间bfs那个判断条件而已
AC

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#d]]></description><category></category></item><item><title><![CDATA[HDU 1403 Longest Common Substring]]></title><link>https://blog.youkuaiyun.com/E_Windrunner/article/details/8988141</link><guid>https://blog.youkuaiyun.com/E_Windrunner/article/details/8988141</guid><author>E_Windrunner</author><pubDate>Wed, 29 May 2013 10:26:19 +0800</pubDate><description><![CDATA[好久没学新算法了，从现在开始努力吧。
花了3天时间看了后缀数组的论文，差不多看明白了，贴上一份应用的代码吧。
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#d]]></description><category></category></item><item><title><![CDATA[HDU 2586 How far away?]]></title><link>https://blog.youkuaiyun.com/E_Windrunner/article/details/8619863</link><guid>https://blog.youkuaiyun.com/E_Windrunner/article/details/8619863</guid><author>E_Windrunner</author><pubDate>Wed, 27 Feb 2013 23:10:38 +0800</pubDate><description><![CDATA[用dfs和lca均可，贴上两份代码
dfs
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#define ll long long
#define MP make_pair
#define PB push_back
#define SZ(x)]]></description><category></category></item><item><title><![CDATA[HDU 1541 Stars]]></title><link>https://blog.youkuaiyun.com/E_Windrunner/article/details/8615895</link><guid>https://blog.youkuaiyun.com/E_Windrunner/article/details/8615895</guid><author>E_Windrunner</author><pubDate>Tue, 26 Feb 2013 23:37:58 +0800</pubDate><description><![CDATA[看起来是二维，其实用一维数组就可以解决了。
#include 
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#define ll long long
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#define SZ(x)]]></description><category></category></item><item><title><![CDATA[HDU 2852 KiKi's K-Number]]></title><link>https://blog.youkuaiyun.com/E_Windrunner/article/details/8611434</link><guid>https://blog.youkuaiyun.com/E_Windrunner/article/details/8611434</guid><author>E_Windrunner</author><pubDate>Mon, 25 Feb 2013 22:03:15 +0800</pubDate><description><![CDATA[因为一个输出超时N次。

#include 
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#define ll long long
#define MP make_pair
#define PB push_back
#define SZ(x) ((int)(x]]></description><category></category></item><item><title><![CDATA[HDU 3038 How Many Answers Are Wrong]]></title><link>https://blog.youkuaiyun.com/E_Windrunner/article/details/8575839</link><guid>https://blog.youkuaiyun.com/E_Windrunner/article/details/8575839</guid><author>E_Windrunner</author><pubDate>Thu, 07 Feb 2013 17:29:29 +0800</pubDate><description><![CDATA[这题也是看题解的。
A到B之间的和可用k到B的和减去k到A-1的和代替（k
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#define ll long long
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#defin]]></description><category></category></item><item><title><![CDATA[HDU 3635 Dragon Balls]]></title><link>https://blog.youkuaiyun.com/E_Windrunner/article/details/8571600</link><guid>https://blog.youkuaiyun.com/E_Windrunner/article/details/8571600</guid><author>E_Windrunner</author><pubDate>Tue, 05 Feb 2013 14:12:15 +0800</pubDate><description><![CDATA[这题的并查集是看题解的，想不出怎么算转移次数，题解好像用的是并入的次序来计算的，这真没想到。。
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#define MP make_pai]]></description><category></category></item><item><title><![CDATA[HDU 1856 More is better]]></title><link>https://blog.youkuaiyun.com/E_Windrunner/article/details/8565690</link><guid>https://blog.youkuaiyun.com/E_Windrunner/article/details/8565690</guid><author>E_Windrunner</author><pubDate>Sat, 02 Feb 2013 17:56:14 +0800</pubDate><description><![CDATA[别人用的并查集。。我这又钻牛角尖了，硬是要用dfs，TLE了好几次才AC。。已经越来越依赖STL了。。
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#define MP make]]></description><category></category></item><item><title><![CDATA[HDU 1829 A Bug's Life]]></title><link>https://blog.youkuaiyun.com/E_Windrunner/article/details/8561000</link><guid>https://blog.youkuaiyun.com/E_Windrunner/article/details/8561000</guid><author>E_Windrunner</author><pubDate>Thu, 31 Jan 2013 20:49:22 +0800</pubDate><description><![CDATA[看别人都用并查集。。我用的是bfs
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#defi]]></description><category></category></item><item><title><![CDATA[HDU 2616 Kill the monster]]></title><link>https://blog.youkuaiyun.com/E_Windrunner/article/details/8560918</link><guid>https://blog.youkuaiyun.com/E_Windrunner/article/details/8560918</guid><author>E_Windrunner</author><pubDate>Thu, 31 Jan 2013 20:10:27 +0800</pubDate><description><![CDATA[数据太弱，dfs水过。。
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#define SZ]]></description><category></category></item><item><title><![CDATA[HDU 2821 Pusher]]></title><link>https://blog.youkuaiyun.com/E_Windrunner/article/details/8560871</link><guid>https://blog.youkuaiyun.com/E_Windrunner/article/details/8560871</guid><author>E_Windrunner</author><pubDate>Thu, 31 Jan 2013 19:50:38 +0800</pubDate><description><![CDATA[dfs
#include 
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#define SZ(x) ((int]]></description><category></category></item><item><title><![CDATA[HDU 2610 Sequence one]]></title><link>https://blog.youkuaiyun.com/E_Windrunner/article/details/8557599</link><guid>https://blog.youkuaiyun.com/E_Windrunner/article/details/8557599</guid><author>E_Windrunner</author><pubDate>Wed, 30 Jan 2013 20:52:34 +0800</pubDate><description><![CDATA[我用的是bfs。1046ms，这。。都倒数了。我偷懒用了set。而且c++的string里面居然没有逆置函数，我勒个去，还得自己写，最新的标准才支持从整型转成字符串(to_string)。。这。。
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#]]></description><category></category></item><item><title><![CDATA[HDU 1426 Sudoku Killer]]></title><link>https://blog.youkuaiyun.com/E_Windrunner/article/details/8555853</link><guid>https://blog.youkuaiyun.com/E_Windrunner/article/details/8555853</guid><author>E_Windrunner</author><pubDate>Wed, 30 Jan 2013 13:24:11 +0800</pubDate><description><![CDATA[dfs,每次在？处尝试每个数字，若符合条件，则进行下一层dfs，知道找到足够的数字
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#de]]></description><category></category></item><item><title><![CDATA[HDU 1104 Remainder]]></title><link>https://blog.youkuaiyun.com/E_Windrunner/article/details/8552935</link><guid>https://blog.youkuaiyun.com/E_Windrunner/article/details/8552935</guid><author>E_Windrunner</author><pubDate>Tue, 29 Jan 2013 15:35:53 +0800</pubDate><description><![CDATA[156ms,比其他人慢好多。。
bfs暴搜
#include
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using namespace std;

const int maxN=2000010;
int n,k,m,step;
bool vis[maxN];
string anstr;
struct Node{
    int n,k;
    string]]></description><category></category></item><item><title><![CDATA[HDU 1796 How many integers can you find]]></title><link>https://blog.youkuaiyun.com/E_Windrunner/article/details/8331331</link><guid>https://blog.youkuaiyun.com/E_Windrunner/article/details/8331331</guid><author>E_Windrunner</author><pubDate>Wed, 19 Dec 2012 13:24:29 +0800</pubDate><description><![CDATA[一个月没做题了。。。
容斥原理，注意处理非互素的数，应用最小公倍数，注意0不用处理，若集合里有1，则直接输出n-1，同时要注意最小公倍数不要溢出，适当的时候应该跳出dfs。
#include 
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#define SZ]]></description><category></category></item><item><title><![CDATA[POJ 1149 PIGS]]></title><link>https://blog.youkuaiyun.com/E_Windrunner/article/details/8196034</link><guid>https://blog.youkuaiyun.com/E_Windrunner/article/details/8196034</guid><author>E_Windrunner</author><pubDate>Sun, 18 Nov 2012 12:52:20 +0800</pubDate><description><![CDATA[最大流。关键在于建图。注意：每次打开猪笼，重新分配猪的数目后，所有已打开的猪笼要重新关闭。把每位顾客当成一个结点，若笼子k曾经被之前某个顾客a打开过，当前顾客b又能打开笼子k，则容量cap[a][b]=inf，若当前顾客是第一个打开笼子k的人，则cap[0][a]=pigNum[k]，其中以0为源点，n+1为汇点，易知每个顾客到汇点的边的容量为顾客想要购买猪的数目。至此，构图完成。
#inclu]]></description><category></category></item><item><title><![CDATA[POJ 1166 The Clocks]]></title><link>https://blog.youkuaiyun.com/E_Windrunner/article/details/8192642</link><guid>https://blog.youkuaiyun.com/E_Windrunner/article/details/8192642</guid><author>E_Windrunner</author><pubDate>Fri, 16 Nov 2012 23:01:24 +0800</pubDate><description><![CDATA[代码略丑。。已经懒得去改了，华丽丽的563ms。。。
bfs暴搜，记录一下状态就行了。
#include 
#include 
#include 
#include 
#define SZ(x) ((int)(x).size())
#define FOR(it,c) for ( __typeof((c).begin()) it=(c).begin(); it!=(c).end(); it++]]></description><category></category></item><item><title><![CDATA[POJ 1144 Network]]></title><link>https://blog.youkuaiyun.com/E_Windrunner/article/details/8171878</link><guid>https://blog.youkuaiyun.com/E_Windrunner/article/details/8171878</guid><author>E_Windrunner</author><pubDate>Sun, 11 Nov 2012 20:33:33 +0800</pubDate><description><![CDATA[假若在删去顶点v以及和v相关联的各边之后，将图的一个连通分量分割成两个或两个以上的连通分量，则称顶点v为该图的一个割点。
生成树为从根结点出发深度优先搜索遍历图，未遍历到的边为虚边。
割点判定：
（1）若生成树的根有两棵或两棵以上子树，则此根结点必为割点
（2）若生成树中某个非叶子结点v，其某棵子树的根和子树中的其他结点均没有指向v的祖先的回边，则v为割点
具体请看数据结构课本。
#i]]></description><category></category></item><item><title><![CDATA[POJ 1087 A Plug for UNIX]]></title><link>https://blog.youkuaiyun.com/E_Windrunner/article/details/8170939</link><guid>https://blog.youkuaiyun.com/E_Windrunner/article/details/8170939</guid><author>E_Windrunner</author><pubDate>Sun, 11 Nov 2012 12:45:58 +0800</pubDate><description><![CDATA[二分图匹配，主要在于建图，貌似建得有点丑。。把能插座X上能插入的插头（包括通过转化器的）通过dfs全找出来，图就建好了，我是用map将字符串映射成数组下标的。
二分图一边是插座，一边是插头。

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using namespace std;

const i]]></description><category></category></item><item><title><![CDATA[POJ 1064 Cable master]]></title><link>https://blog.youkuaiyun.com/E_Windrunner/article/details/8170626</link><guid>https://blog.youkuaiyun.com/E_Windrunner/article/details/8170626</guid><author>E_Windrunner</author><pubDate>Sun, 11 Nov 2012 10:02:37 +0800</pubDate><description><![CDATA[二分答案

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using namespace std;

int a[10010];

int main(){
#ifndef ONLINE_JUDGE	          
	freopen("a.in","r",stdin);
    freopen("a.out","w]]></description><category></category></item></channel></rss>