Description Ted has a new house with a huge window. In this big summer, Ted decides to decorate the window with some posters to prevent the glare outside. All things that Ted can find are rectangle posters. However, Ted is such a picky guy that in ev…

Description Students often have problems taking up seats. When two students want the same seat, a quarrel will probably begin. It will have very bad effect when such subjects occur on the BBS.  So, we urgently need a seat-taking-up rule. After severa…

传送门:http://acm.hdu.edu.cn/showproblem.php?pid=3262 题意:教室有n*m个座位,每个座位有一个舒适值,有K个学生在不同时间段进来,要占t个座位,必须是连续的并且自己坐在最左边,如果有多个的话,找最舒适的座位,如果没有连续t个,那么只给自己找个最舒适的位子,如果都满的话,输出-1. 题解:一个简单的搜索模拟,注意的是,要排序每个同学进来的时间,而且输出要按照给的顺序输出,被坑了几次,样例数据太弱了. AC代码: #include <iostream>…

题目链接: http://poj.org/problem?id=3829 题意描述: 输入矩阵的大小n和m,以及来占位置的人数k 输入n*m的教室座位矩阵,每个值表示该座位的满意度 输入每个人来占位置的时间和需要几个位置h,m,q 计算并输出每个来占位置的人根据占位规则得到的坐标zx,zy,若一个位置都没有了,输出-1. 规则: 首先看有没有在同一行上的连续的k个座位,有则这个占位的同学坐在这连续k个座位的最左边,如果有多行上都存在连续的k个位置,则该占位的同学会选择最左边值最大的位置(坑点就在…

Description Facer is addicted to a game called "Tidy is learning to swim". But he finds it too easy. So he develops a new game called "Facer is learning to swim" which is more difficult and more interesting. In the new game, a robot na…

Description The city of M is a famous shopping city and its open-air shopping malls are extremely attractive. During the tourist seasons, thousands of people crowded into these shopping malls and enjoy the vary-different shopping. Unfortunately, the…

Description On the evening of 3 August 1492, Christopher Columbus departed from Palos de la Frontera with a few ships, starting a serious of voyages of finding a new route to India. As you know, just in those voyages, Columbus discovered the America…

Problem Description Peer-to-peer(P2P) computing technology has been widely used on the Internet to exchange data. A lot of P2P file sharing systems exist and gain much popularity in nowadays. Let's consider a simplified model of P2P file sharing syst…

题目链接: 黑书 P116 HDU 2157 棋盘分割 POJ 1191 棋盘分割 分析:  枚举所有可能的切割方法. 但如果用递归的方法要加上记忆搜索, 不能会超时... 代码: #include<iostream> #include<cstdio> #include<cstring> #include<cmath> using namespace std; const int inf=6400*6400; const int N=8; int sum[1…

标题效果 有着n巫妖.m精灵.k木.他们都有自己的位置坐标表示.冷却时间,树有覆盖范围. 假设某个巫妖攻击精灵的路线(他俩之间的连线)经过树的覆盖范围,表示精灵被树挡住巫妖攻击不到.求巫妖杀死所有精灵的时间.若无法所有杀死输出-1: 解题思路: 推断巫妖能否打到精灵用线段与点的最短距离来推断,若最短距离小于树的覆盖范围,就攻击不到. 最小时间能够跑费用流来解决,也能够二分图的最优匹配. 以下是代码: #include <set> #include <map> #include &l…

GCC Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 3867    Accepted Submission(s): 1272 Problem Description The GNU Compiler Collection (usually shortened to GCC) is a compiler system produc…

Description Could you imaging a monkey writing computer programs? Surely monkeys are smart among animals. But their limited intelligence is no match for our human beings. However, there is a theorem about monkeys, and it states that monkeys can write…

Description Math Olympiad is called “Aoshu” in China. Aoshu is very popular in elementary schools. Nowadays, Aoshu is getting more and more difficult. Here is a classic Aoshu problem: ABBDE __ ABCCC = BDBDEIn the equation above, a letter stands for a…

WHUgirls Time Limit: 3000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 2068    Accepted Submission(s): 785 Problem Description There are many pretty girls in Wuhan University, and as we know, every girl lo…

以为有啥牛逼定理,没推出来,随便写写就A了----题非常水,可是wa了一次 n>=m  则n!==0 注意的一点,最后 看我的凝视 #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <iostream> using namespace std; const int maxn = 115; #define ll long lo…

Description Aliens on planet Pandora also write computer programs like us. Their programs only consist of capital letters (‘A’ to ‘Z’) which they learned from the Earth. On planet Pandora, hackers make computer virus, so they also have anti-virus sof…

hdu 2844 poj 1742 Coins 题目相同,但是时限不同,原本上面的多重背包我初始化为0,f[0] = 1;用位或进行优化,f[i]=1表示可以兑成i,0表示不能. 在poj上运行时间正好为时限3000ms....太慢了,hdu直接TLE(时限1s); 之 后发现其实并不是算法的问题,而是库函数的效率没有关注到.我是使用fill()按量初始化的,但是由于memset()可能是系统底层使用了四个字节拷 贝的函数(远比循环初始化快),效率要高得多..这就是为什么一直TLE的原因,fil…

Description Too worrying about the house price bubble, poor Mike sold his house and rent an apartment in a 50-floor building several months ago. This building has only one elevator because it is a so called “rotten tail building”. There are always a…

学习链接:http://blog.csdn.net/lwt36/article/details/48908031 学习扫描线主要学习的是一种扫描的思想,后期可以求解很多问题. 扫描线求矩形周长并 hdu 1928 Picture Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4795    Accepted Submission(s):…

Tunnel Warfare                                                             Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)                                                                                            …

Coins HDU - 2844 POJ - 1742 多重背包可行性 当做一般多重背包,二进制优化 #include<cstdio> #include<cstring> int n,m,anss; ],c[],f[]; int main() { int i,j,t; scanf("%d%d",&n,&m); ||m!=) { anss=; memset(f,,sizeof(f)); ;i<=n;i++) scanf("%d&qu…

Description The 15-puzzle has been around for over 100 years; even if you don't know it by that name, you've seen it. It is constructed with 15 sliding tiles, each with a number from 1 to 15 on it, and all packed into a 4 by 4 frame with one tile mis…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4772 题面: Zhuge Liang's Password Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1404    Accepted Submission(s): 926 Problem Description In the anc…

Description Students often have problems taking up seats. When two students want the same seat, a quarrel will probably begin. It will have very bad effect when such subjects occur on the BBS. So, we urgently need a seat-taking-up rule. After several…

树链剖分是一个很固定的套路 一般用来解决树上两点之间的路径更改与查询 思想是将一棵树分成不想交的几条链 并且由于dfs的顺序性 给每条链上的点或边标的号必定是连着的 那么每两个点之间的路径都可以拆成几条链 那么就是对一群区间进行更改 这时候基本是用线段树进行logn的操作 做了三道基础题 都属于比较好想的 也就是线段树比较麻烦 需要写相当长一段时间... HDU 3966 给出一棵树的连接状况和边的大小 每次可以对a-b的路径的边的权值取反 或者改变指定边的值 或者求a-b路径的最大值 每次取反…

一条边<u,v>表示u选那么v一定被选. #include <iostream> #include <cstring> #include <cstdio> #include <algorithm> using namespace std; ; ; struct EDGE{int to,next;}edge[Maxm]; int T,m,Stack[Maxn],head[Maxn],Belong[Maxn],Id[Maxn],Dfn[Maxn],L…

1.POJ 1733 Parity game Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5744   Accepted: 2233 Description Now and then you play the following game with your friend. Your friend writes down a sequence consisting of zeroes and ones. You cho…

Eight Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 30176   Accepted: 13119   Special Judge Description The 15-puzzle has been around for over 100 years; even if you don't know it by that name, you've seen it. It is constructed with 15…

做这道题之前,建议先做POJ 1151  Atlantis,经典的扫描线求矩阵的面积并 参考连接: http://www.cnblogs.com/scau20110726/archive/2013/04/13/3018702.html 线段树辅助——扫描线法计算矩形周长并(轮廓线):http://www.cnblogs.com/scau20110726/archive/2013/04/13/3018687.htmlhttp://blog.csdn.net/ophunter/article/det…

题意:一条线上有n个点,D x是破坏这个点,Q x是表示查询x所在的最长的连续的点的个数,R是恢复上一次破坏的点. 思路:这题的关键是查询. 将被毁的村庄看成空位,当查询某个点的时候,如果我们知道它左边最近的空位a和右边最近的空位b, 那么我们只要查询区间[a,b]中的个数,即为答案,因为[a,b]之间不可能有空位存在了. 那么如何获取这样的a和b呢,这个就和HDU 4302 Holedox Eating 差不多了. 对每个节点,存储该区间中 空位的最大位置 和 空位的最小位置,还有 该区间村庄…