<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[yc5-yc的博客]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/yc5_yc</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; yc5_yc]]></copyright><item><title><![CDATA[AVL树代码]]></title><link>https://blog.youkuaiyun.com/yc5_yc/article/details/8510614</link><guid>https://blog.youkuaiyun.com/yc5_yc/article/details/8510614</guid><author>yc5_yc</author><pubDate>Wed, 16 Jan 2013 18:30:52 +0800</pubDate><description><![CDATA[这个代码输出了map与我的AVL效率的差异，明显AVL快。
#include 
#include 
#include 
using namespace std;
const int NMax=1000000;
template
struct node{
	KEY k;DATA d;
	int h;
	node *left,*right;
};
#define HIGH(_)]]></description><category></category></item><item><title><![CDATA[stoer wagner代码]]></title><link>https://blog.youkuaiyun.com/yc5_yc/article/details/8497935</link><guid>https://blog.youkuaiyun.com/yc5_yc/article/details/8497935</guid><author>yc5_yc</author><pubDate>Sun, 13 Jan 2013 13:40:37 +0800</pubDate><description><![CDATA[合并的代码很容易写错！
#include 
#include 
#include 
#include 
using namespace std;
const int NMax=550;
int N,M,G[NMax][NMax],dist[NMax],map[NMax];
bool inA[NMax];
int Dinic() {
//1.min=MAXINT，固定一个顶点P]]></description><category></category></item><item><title><![CDATA[Haskell读入一个字符串，把它转换成数字]]></title><link>https://blog.youkuaiyun.com/yc5_yc/article/details/8245830</link><guid>https://blog.youkuaiyun.com/yc5_yc/article/details/8245830</guid><author>yc5_yc</author><pubDate>Sat, 01 Dec 2012 10:37:01 +0800</pubDate><description><![CDATA[toint :: String->Int
toint' :: Char->Int
toint' s
	|s=='1' =1
	|s=='2' =2
	|s=='3' =3
	|s=='4' =4
	|s=='5' =5
	|s=='6' =6
	|s=='7' =7
	|s=='8' =8
	|s=='9' =9
	|s=='0' =0
	|otherwise = 0]]></description><category></category></item><item><title><![CDATA[Haskell线段树建树]]></title><link>https://blog.youkuaiyun.com/yc5_yc/article/details/8245250</link><guid>https://blog.youkuaiyun.com/yc5_yc/article/details/8245250</guid><author>yc5_yc</author><pubDate>Fri, 30 Nov 2012 22:23:47 +0800</pubDate><description><![CDATA[调了一晚上……
data Tree=Tree Int Int [Tree] [Tree]|Empty deriving (Show)
build :: (Int,Int)->Tree
build (x,y)=if x<y then Tree x y [build (x,(div (x+y) 2))] [build ((div (x+y) 2)+1,y)] else Tree x y [Emp]]></description><category></category></item><item><title><![CDATA[完整版线段树求最小值，支持区间减]]></title><link>https://blog.youkuaiyun.com/yc5_yc/article/details/8227702</link><guid>https://blog.youkuaiyun.com/yc5_yc/article/details/8227702</guid><author>yc5_yc</author><pubDate>Mon, 26 Nov 2012 21:45:27 +0800</pubDate><description><![CDATA[def Min(a,b):
    if a
        return a
    else:
        return b
class node:
    def __init__(self,x,y):
        self.l=x
        self.r=y
        self.key=0
        self.lazy=0
        i]]></description><category></category></item><item><title><![CDATA[用Python实现的基本版的线段树]]></title><link>https://blog.youkuaiyun.com/yc5_yc/article/details/8227523</link><guid>https://blog.youkuaiyun.com/yc5_yc/article/details/8227523</guid><author>yc5_yc</author><pubDate>Mon, 26 Nov 2012 20:57:04 +0800</pubDate><description><![CDATA[class node:
    def __init__(self,x,y):
        self.l=x
        self.r=y
        self.key=0
        if x
            mid=(x+y)/2
            self.left=node(x,mid)
            self.right=node(]]></description><category></category></item><item><title><![CDATA[用 Python实现C的读入方式]]></title><link>https://blog.youkuaiyun.com/yc5_yc/article/details/8226949</link><guid>https://blog.youkuaiyun.com/yc5_yc/article/details/8226949</guid><author>yc5_yc</author><pubDate>Mon, 26 Nov 2012 17:59:43 +0800</pubDate><description><![CDATA[A=[]
N=int(input("N:"))
x=""
tot=0
while True:
    flag=0
    y=x.split(" ")
    for j in y:
        if j!='':
            j=int(j)
            A.append(j)
            tot=tot+1]]></description><category></category></item><item><title><![CDATA[用Linux配置pacman]]></title><link>https://blog.youkuaiyun.com/yc5_yc/article/details/8192435</link><guid>https://blog.youkuaiyun.com/yc5_yc/article/details/8192435</guid><author>yc5_yc</author><pubDate>Fri, 16 Nov 2012 22:01:47 +0800</pubDate><description><![CDATA[最近对AI突然来了兴趣，玩起了PACMAN(ai.net9.org)
刚玩上瘾，就听说了AI比赛马上就要开始，而那个OJ随时都有可能关闭，于是我就想让我的AI能在本地运行
我首先登入了windows7系统，可是令我失望的是，game.exe不管怎样都会崩溃，于是我果断登录了Ubuntu，首先，我在控制台里瞬间编译、模拟了游戏，可是，当我把生成好的result.txt用playerlocal播放]]></description><category></category></item><item><title><![CDATA[NOIP2012纪念]]></title><link>https://blog.youkuaiyun.com/yc5_yc/article/details/8188689</link><guid>https://blog.youkuaiyun.com/yc5_yc/article/details/8188689</guid><author>yc5_yc</author><pubDate>Thu, 15 Nov 2012 21:46:06 +0800</pubDate><description><![CDATA[这回参加了提高组。
说说初步思考后的算法。
Day1
第一题：
第二题：两个点相乘，排序（思路来自yjy）
第三题：模拟。。。
Day2
第一题：线性同余（好像算导上的算法算出的X0不是最小的？）
第二题：线段树(lazy思想)。。。
第三题：。。。。。。]]></description><category></category></item><item><title><![CDATA[迷宫游戏代码OpenGL]]></title><link>https://blog.youkuaiyun.com/yc5_yc/article/details/8108036</link><guid>https://blog.youkuaiyun.com/yc5_yc/article/details/8108036</guid><author>yc5_yc</author><pubDate>Wed, 24 Oct 2012 17:21:33 +0800</pubDate><description><![CDATA[#include  // Must have for Windows platform builds
//#include "gl\glee.h" // OpenGL Extension ¡°autoloader¡±
#include  // Microsoft OpenGL headers (version 1.1 by themselves)
#include  // OpenGL Utili]]></description><category></category></item><item><title><![CDATA[PHP登入登出系统]]></title><link>https://blog.youkuaiyun.com/yc5_yc/article/details/7984197</link><guid>https://blog.youkuaiyun.com/yc5_yc/article/details/7984197</guid><author>yc5_yc</author><pubDate>Sun, 16 Sep 2012 12:15:01 +0800</pubDate><description><![CDATA[------------------------------------~index.php-------------------------------------------


欢迎访问yc综合系统



    if(isset($_GET["debug"]))
        setcookie("login","",time()-3600);
    if(isset]]></description><category></category></item><item><title><![CDATA[树状数组]]></title><link>https://blog.youkuaiyun.com/yc5_yc/article/details/7909563</link><guid>https://blog.youkuaiyun.com/yc5_yc/article/details/7909563</guid><author>yc5_yc</author><pubDate>Sun, 26 Aug 2012 17:55:48 +0800</pubDate><description><![CDATA[树状数组支持部分线段树的操作。
主要工作：求前缀和
lowbit函数：f(x)=x&(x^(x-1)) 负责转移
习题：SPOJ NKTEAM 详见PDF
           POJ 3321 求后序遍历二叉树顺序，子树即为前面的N项（N为包括自己的子节点个数）]]></description><category></category></item><item><title><![CDATA[一个重要的公式]]></title><link>https://blog.youkuaiyun.com/yc5_yc/article/details/7802636</link><guid>https://blog.youkuaiyun.com/yc5_yc/article/details/7802636</guid><author>yc5_yc</author><pubDate>Mon, 30 Jul 2012 16:37:56 +0800</pubDate><description><![CDATA[max(abs(a),abs(b))=(abs(a+b)+abs(a-b))/2来源：POJ 3212]]></description><category></category></item><item><title><![CDATA[刷进前5000!]]></title><link>https://blog.youkuaiyun.com/yc5_yc/article/details/7784354</link><guid>https://blog.youkuaiyun.com/yc5_yc/article/details/7784354</guid><author>yc5_yc</author><pubDate>Wed, 25 Jul 2012 13:37:58 +0800</pubDate><description><![CDATA[]]></description><category></category></item><item><title><![CDATA[矩阵乘法模板二（递归 例题3613）]]></title><link>https://blog.youkuaiyun.com/yc5_yc/article/details/7624587</link><guid>https://blog.youkuaiyun.com/yc5_yc/article/details/7624587</guid><author>yc5_yc</author><pubDate>Fri, 01 Jun 2012 20:54:08 +0800</pubDate><description><![CDATA[____ DFS(int a)
{
    ____ ret1;
    if(a==1) return G2;
    ret1=DFS(a/2);
    ret1=Cheng(ret1,ret1);
    if(a%2)ret1=Cheng(ret1,G2);
    return ret1;
}
注：____是矩阵结构体的类型，Cheng代表两个矩阵相乘返回的值，DFS]]></description><category></category></item><item><title><![CDATA[矩阵模版（例题3070、3744）]]></title><link>https://blog.youkuaiyun.com/yc5_yc/article/details/7617937</link><guid>https://blog.youkuaiyun.com/yc5_yc/article/details/7617937</guid><author>yc5_yc</author><pubDate>Wed, 30 May 2012 21:36:22 +0800</pubDate><description><![CDATA[struct matrix 
{
    double A[NMax][NMax];
    matrix operator*(const matrix & m)
    {
        matrix ret={0,0,0,0};
        for(int i=0;i<NMax;i++)
            for(int k=0;k<NMax;k++)]]></description><category></category></item><item><title><![CDATA[最大费用最大流模版(POJ 3422)]]></title><link>https://blog.youkuaiyun.com/yc5_yc/article/details/7607488</link><guid>https://blog.youkuaiyun.com/yc5_yc/article/details/7607488</guid><author>yc5_yc</author><pubDate>Sun, 27 May 2012 21:34:19 +0800</pubDate><description><![CDATA[struct edge
{
    int num,len,C;
    edge *next,*rev;
}*S[nnMax],pool[MMax];
void build(int x,int y,int z,int C)
{
    edge *p=&pool[L++],*q=&pool[L++];
    p->num=y;p->len=z;p->C=C;p->next=S[]]></description><category></category></item><item><title><![CDATA[最小费用最大流模版]]></title><link>https://blog.youkuaiyun.com/yc5_yc/article/details/7607266</link><guid>https://blog.youkuaiyun.com/yc5_yc/article/details/7607266</guid><author>yc5_yc</author><pubDate>Sun, 27 May 2012 20:03:41 +0800</pubDate><description><![CDATA[struct edge
{
    int num,len,C;
    edge *next,*rev;
}*S[nnMax],pool[MMax];
int d[nnMax];
edge *P[nnMax];
bool inq[nnMax];
queue Q;
void build_edge(int x,int y,int z,int C)
{
    edge *p=&]]></description><category></category></item><item><title><![CDATA[DP算法入门（3）——最大子段和]]></title><link>https://blog.youkuaiyun.com/yc5_yc/article/details/7561098</link><guid>https://blog.youkuaiyun.com/yc5_yc/article/details/7561098</guid><author>yc5_yc</author><pubDate>Sat, 12 May 2012 21:40:06 +0800</pubDate><description><![CDATA[令b[j]表示以位置 j 为终点的所有子区间中和最大的一个子问题:如j为终点的最大子区间包含了位置j-1,则以j-1为终点的最大子区间必然包括在其中如果b[j-1] >0, 那么显然b[j] = b[j-1] + a[j]，用之前最大的一个加上a[j]即可，因为a[j]必须包含如果b[j-1]

int max = 0;
int b[n+1];
int start = 0;
int end =]]></description><category></category></item><item><title><![CDATA[DP算法入门（2）——完全背包问题（POJ2063题解）]]></title><link>https://blog.youkuaiyun.com/yc5_yc/article/details/7555132</link><guid>https://blog.youkuaiyun.com/yc5_yc/article/details/7555132</guid><author>yc5_yc</author><pubDate>Thu, 10 May 2012 20:49:56 +0800</pubDate><description><![CDATA[昨天阅读了完全背包问题，顺便做了点题。
完全背包问题超简单，尤其是O(VN)的算法，很容易就能懂，把01背包的第二重循环倒过来即可。
2063这题就是按年走，每年根据当前的余额使用完全背包算出最大获利加到余额上，没什么可说的，注意都要除以1000，memset慎用，附属代码：

#include 
#include 
#include 
using namespace std;
const]]></description><category></category></item></channel></rss>