<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[0x3f3f3f3f]]></title><description><![CDATA[我要一步一步往上爬，在最高点乘着叶片往前飞]]></description><link>https://blog.youkuaiyun.com/L_avender</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; L_avender]]></copyright><item><title><![CDATA[HDU 1010 —— Tempter of the Bone   DFS+剪枝]]></title><link>https://blog.youkuaiyun.com/L_avender/article/details/54948544</link><guid>https://blog.youkuaiyun.com/L_avender/article/details/54948544</guid><author>L_avender</author><pubDate>Thu, 09 Feb 2017 16:30:27 +0800</pubDate><description><![CDATA[原题：http://acm.hdu.edu.cn/showproblem.php?pid=1010
题意：从S走到D，其中X是墙，不能走；给定时间 t ，问能否从S到D正好花费 t 时间（一步即为一秒）；
思路：用DFS+剪枝；


此处有两种剪枝：
1、从S到D的最短时间是 tmp = abs(sx-dx)+abs(sy-dy)，如果 tmp > t，显然是“NO”；
2、如果 t]]></description><category></category></item><item><title><![CDATA[HDU 2177 —— 取(2堆)石子游戏]]></title><link>https://blog.youkuaiyun.com/L_avender/article/details/53440265</link><guid>https://blog.youkuaiyun.com/L_avender/article/details/53440265</guid><author>L_avender</author><pubDate>Fri, 02 Dec 2016 23:11:07 +0800</pubDate><description><![CDATA[原题：http://acm.hdu.edu.cn/showproblem.php?pid=2177


#include
#include
using namespace std;
int a, b;
int main(){
    while(cin>>a>>b){
        if(a == 0 && b == 0)    break;
        double tmp = (s]]></description><category></category></item><item><title><![CDATA[HDU 1527 —— 取石子游戏（入门水题）]]></title><link>https://blog.youkuaiyun.com/L_avender/article/details/53439940</link><guid>https://blog.youkuaiyun.com/L_avender/article/details/53439940</guid><author>L_avender</author><pubDate>Fri, 02 Dec 2016 22:30:15 +0800</pubDate><description><![CDATA[原题：http://acm.hdu.edu.cn/showproblem.php?pid=1527


#include
#include
using namespace std;
int a, b;
int main(){
    while(cin>>a>>b){
        int k = abs(a-b);
        double tmp = (sqrt(5)+1)/2.0]]></description><category></category></item><item><title><![CDATA[HDU 1907 —— John]]></title><link>https://blog.youkuaiyun.com/L_avender/article/details/53435194</link><guid>https://blog.youkuaiyun.com/L_avender/article/details/53435194</guid><author>L_avender</author><pubDate>Fri, 02 Dec 2016 16:01:53 +0800</pubDate><description><![CDATA[原题：http://acm.hdu.edu.cn/showproblem.php?pid=1907
题意：取法和Nim游戏一样，但是最后拿完的那个人是loser；
思路：需要考虑特殊情况 —— 最开始每堆的个数都为1，这个时候对堆数进行奇偶判断；


#include
using namespace std;
const int maxn = 50;
int cas, n;
int a[]]></description><category></category></item><item><title><![CDATA[HDU 1849 —— Rabbit and Grass]]></title><link>https://blog.youkuaiyun.com/L_avender/article/details/53433391</link><guid>https://blog.youkuaiyun.com/L_avender/article/details/53433391</guid><author>L_avender</author><pubDate>Fri, 02 Dec 2016 12:18:27 +0800</pubDate><description><![CDATA[原题：http://acm.hdu.edu.cn/showproblem.php?pid=1849
思路：以两个棋子为例，如果当前它们在不同的位置，那么先手只需要将两个棋子变成在同一个位置，就能获胜，与Nim博弈类似；


#include
using namespace std;
int m;
int main(){
    while(cin>>m){
        if(m == ]]></description><category></category></item><item><title><![CDATA[POJ 2234 —— Matches Game 入门水题]]></title><link>https://blog.youkuaiyun.com/L_avender/article/details/53433354</link><guid>https://blog.youkuaiyun.com/L_avender/article/details/53433354</guid><author>L_avender</author><pubDate>Fri, 02 Dec 2016 12:08:10 +0800</pubDate><description><![CDATA[原题：http://poj.org/problem?id=2234


#include
using namespace std;
int m;
int main(){
    while(cin>>m){
        int x = 0;
        while(m--){
            int y;
            cin>>y;
            x =]]></description><category></category></item><item><title><![CDATA[Nim博弈讲解篇]]></title><link>https://blog.youkuaiyun.com/L_avender/article/details/53433350</link><guid>https://blog.youkuaiyun.com/L_avender/article/details/53433350</guid><author>L_avender</author><pubDate>Fri, 02 Dec 2016 12:06:38 +0800</pubDate><description><![CDATA[Nim游戏的概述：

还记得这个游戏吗？
给出n列珍珠，两人轮流取珍珠，每次在某一列中取至少1颗珍珠，但不能在两列中取。最后拿光珍珠的人输。
后来，在一份资料上看到，这种游戏称为“拈（Nim）”。据说，它源自中国，经由被贩卖到美洲的奴工们外传。辛苦的工人们，在工作闲暇之余，用石头玩游戏以排遣寂寞。后来流传到高级人士，则用便士（Pennies），在酒吧柜台上玩。
最有名的玩法，是把十二枚便]]></description><category></category></item><item><title><![CDATA[Codeforces Beta Round #17 D. Notepad]]></title><link>https://blog.youkuaiyun.com/L_avender/article/details/53380016</link><guid>https://blog.youkuaiyun.com/L_avender/article/details/53380016</guid><author>L_avender</author><pubDate>Mon, 28 Nov 2016 16:52:35 +0800</pubDate><description><![CDATA[原题：http://codeforces.com/contest/17/problem/D
题意：求[(b-1)*b^(n-1)]%c, 如果ans = 0, 则ans = c;
思路：指数循环节 ——


（其中Φ为欧拉函数）；




#include  
#define ll __int64  
using namespace std;  
const int maxn]]></description><category></category></item><item><title><![CDATA[HDU 1069 —— Monkey and Banana]]></title><link>https://blog.youkuaiyun.com/L_avender/article/details/51811622</link><guid>https://blog.youkuaiyun.com/L_avender/article/details/51811622</guid><author>L_avender</author><pubDate>Sat, 02 Jul 2016 16:52:18 +0800</pubDate><description><![CDATA[原题：http://acm.hdu.edu.cn/showproblem.php?pid=1069


#include
#include
using namespace std;
const int maxn = 1100;
int n;
int dp[maxn];
int T = 0;

struct node
{
	int x, y, z;
	
	friend bool operato]]></description><category></category></item><item><title><![CDATA[Codeforces Round #260 (Div. 2)]]></title><link>https://blog.youkuaiyun.com/L_avender/article/details/51532641</link><guid>https://blog.youkuaiyun.com/L_avender/article/details/51532641</guid><author>L_avender</author><pubDate>Sun, 29 May 2016 17:02:34 +0800</pubDate><description><![CDATA[A. Laptops


题意：有n台电脑，给出每台电脑的价格 p 和质量q ；问是否存在这样的电脑——价格比另一台低并且质量比另一台好；
如果存在就输出 “Happy Alex”，否则输出“Poor Alex”；


#include
#include
using namespace std;
const int maxn = 1e5+5;
int n;
int p[maxn],]]></description><category></category></item><item><title><![CDATA[POJ 1087 —— A Plug for UNIX]]></title><link>https://blog.youkuaiyun.com/L_avender/article/details/51357168</link><guid>https://blog.youkuaiyun.com/L_avender/article/details/51357168</guid><author>L_avender</author><pubDate>Mon, 09 May 2016 23:14:24 +0800</pubDate><description><![CDATA[原题：http://poj.org/problem?id=1087


题意：n个插座，m个电器及其对应的插座，k种转化器，转换器（u，v）表示可以把原本需要u插座的电器转接到v插座上，问最少有多少设备没有插座用，每种转换器数量不限;；


#include
#include
#include
#include
#include
#include
#include
#define in]]></description><category></category></item><item><title><![CDATA[HDU 3605 —— Escape 状态压缩+最大流]]></title><link>https://blog.youkuaiyun.com/L_avender/article/details/51352833</link><guid>https://blog.youkuaiyun.com/L_avender/article/details/51352833</guid><author>L_avender</author><pubDate>Mon, 09 May 2016 14:29:00 +0800</pubDate><description><![CDATA[原题：http://acm.hdu.edu.cn/showproblem.php?pid=3605




#include
#include
#include
#include
#include
#include
#include
#define inf 1e9
using namespace std;
const int maxn = 1500;
const int maxm = 1]]></description><category></category></item><item><title><![CDATA[HDU 2883 —— kebab]]></title><link>https://blog.youkuaiyun.com/L_avender/article/details/51348450</link><guid>https://blog.youkuaiyun.com/L_avender/article/details/51348450</guid><author>L_avender</author><pubDate>Sun, 08 May 2016 23:06:53 +0800</pubDate><description><![CDATA[原题：http://acm.hdu.edu.cn/showproblem.php?pid=2883


#include
#include
#include
#include
#include
#include
#define inf 1e9
using namespace std;
const int maxn = 41000;
const int maxm = 8040000;
int ]]></description><category></category></item><item><title><![CDATA[HDU 2732 —— Leapin' Lizards]]></title><link>https://blog.youkuaiyun.com/L_avender/article/details/51345654</link><guid>https://blog.youkuaiyun.com/L_avender/article/details/51345654</guid><author>L_avender</author><pubDate>Sun, 08 May 2016 14:51:31 +0800</pubDate><description><![CDATA[原题：http://acm.hdu.edu.cn/showproblem.php?pid=2732


#include
#include
#include
#include
#include
#include
#define inf 1e9
using namespace std;
const int maxn = 850;
const int maxm = 200000;
int cas]]></description><category></category></item><item><title><![CDATA[poj 2289 —— Jamie's Contact Groups 二分+最大流]]></title><link>https://blog.youkuaiyun.com/L_avender/article/details/51345367</link><guid>https://blog.youkuaiyun.com/L_avender/article/details/51345367</guid><author>L_avender</author><pubDate>Sun, 08 May 2016 13:43:50 +0800</pubDate><description><![CDATA[原题：http://poj.org/problem?id=2289


#include
#include
#include
#include
#include
#include
#include
#define inf 1e9
using namespace std;
const int maxn = 2000;
const int maxm = 510000;
int n, m, num]]></description><category></category></item><item><title><![CDATA[POJ 2391 —— Ombrophobic Bovines 二分+Floyd+最大流]]></title><link>https://blog.youkuaiyun.com/L_avender/article/details/51335220</link><guid>https://blog.youkuaiyun.com/L_avender/article/details/51335220</guid><author>L_avender</author><pubDate>Fri, 06 May 2016 22:39:51 +0800</pubDate><description><![CDATA[原题：http://poj.org/problem?id=2391


#include
#include
#include
#include
#include
#define inf 1LL<<60
#define ll long long
using namespace std;
const int maxn = 500;
const int maxm = 50000;
int f, p]]></description><category></category></item><item><title><![CDATA[POJ 2112 —— Optimal Milking 二分+Floyd+最大流]]></title><link>https://blog.youkuaiyun.com/L_avender/article/details/51332309</link><guid>https://blog.youkuaiyun.com/L_avender/article/details/51332309</guid><author>L_avender</author><pubDate>Fri, 06 May 2016 15:54:34 +0800</pubDate><description><![CDATA[原题：http://poj.org/problem?id=2112


#include
#include
#include
#include
#include
#include
#define inf 0x3f3f3f3f
using namespace std;
const int maxn = 250;
int dis[maxn][maxn];
int k, c, m;
int num]]></description><category></category></item><item><title><![CDATA[POJ 3498 —— March of the Penguins （枚举+最大流）]]></title><link>https://blog.youkuaiyun.com/L_avender/article/details/51308293</link><guid>https://blog.youkuaiyun.com/L_avender/article/details/51308293</guid><author>L_avender</author><pubDate>Tue, 03 May 2016 22:19:19 +0800</pubDate><description><![CDATA[#include
#include
#include
#include
#include
#include
#define inf 0x3f3f3f3f
using namespace std;
const int maxn = 205;
int num_nodes;
int n;
double m;
struct Edge
{
	int from, to, flow, cap;
}edge[ma]]></description><category></category></item><item><title><![CDATA[POJ 1698 —— Alice's Chance]]></title><link>https://blog.youkuaiyun.com/L_avender/article/details/51308083</link><guid>https://blog.youkuaiyun.com/L_avender/article/details/51308083</guid><author>L_avender</author><pubDate>Tue, 03 May 2016 21:54:05 +0800</pubDate><description><![CDATA[#include
#include
#include
#include
#include
#include
#define inf 0x3f3f3f3f
using namespace std;
const int maxn = 50000;
int num_nodes;

struct Edge
{
	int from, to, flow, cap;
}edge[maxn*2];

vector]]></description><category></category></item><item><title><![CDATA[POJ 3281 —— Dining]]></title><link>https://blog.youkuaiyun.com/L_avender/article/details/51255126</link><guid>https://blog.youkuaiyun.com/L_avender/article/details/51255126</guid><author>L_avender</author><pubDate>Tue, 26 Apr 2016 22:55:25 +0800</pubDate><description><![CDATA[原题：http://poj.org/problem?id=3281


题意：有F种食物，D种饮料，N头牛；下面N行给出每头牛喜欢的食物种数和饮料种数，并且分别给出喜欢的食物的种类和饮料的种类；
问最多可以满足几头牛吃到自己喜欢的食物和饮料；
每种食物或者饮料只能提供给一头牛，每头牛只能吃一种食物和饮料；
思路：建立一个超级源点 s 和一个超级汇点 t ，因为食物和饮料之间没有直接关系]]></description><category></category></item></channel></rss>