题目链接:Doors Breaking and Repairing 题目大意:有n个门,先手攻击力为x(摧毁),后手恢复力为y(恢复),输入每个门的初始“生命值”,当把门的生命值攻为0时,就无法恢复了.问:最多可以把几个门的生命值攻为0. 思路:(1)当 x>y 的时候肯定所有的门的生命值都能降为0: (2)当 x<=y 的时候,先手的最优策略就是每次去攻击那些当前“生命值”比自己攻击力小的门,使它们的生命值降为0: 后手的最优策略就是去提高那些“生命值”比先手小的门的“生命值”,来减少先手“…

You are policeman and you are playing a game with Slavik. The game is turn-based and each turn consists of two phases. During the first phase you make your move and during the second phase Slavik makes his move. There are nn doors, the ii-th door ini…

Descirbe You are policeman and you are playing a game with Slavik. The game is turn-based and each turn consists of two phases. During the first phase you make your move and during the second phase Slavik makes his move. There are n doors, the i-th d…

题意:有\(n\)扇门,你每次可以攻击某个门,使其hp减少\(x\)(\(\le 0\)后就不可修复了),之后警察会修复某个门,使其hp增加\(y\),问你最多可以破坏多少扇门? 题解:首先如果\(x>y\),那么我肯定全部都能破坏,否则,统计\(hp\le x\)的门的个数,谁先碰门谁先赢,而我是先手,所以能破坏的门的个数就是\(\lceil \frac{cnt}{2} \rceil\). 代码: int n,x,y; int a[N]; int main() { //ios::sync_wi…

Codeforces Round #531 (Div. 3) 题目总链接:https://codeforces.com/contest/1102 A. Integer Sequence Dividing 题意: 给一个数n,然后要求你把1,2.....n分为两个集合,使得两个集合里面元素的和的差的绝对值最小. 题解: 分析可以发现,当n%4==0 或者 n%3==0,答案为0:其余答案为1.之后输出一下就好了. 代码如下: #include <bits/stdc++.h> using name…

Reinvent the Wheel Often Jason P. Sage Just use something that exists-it's silly to reinvent the wheel-. HAVE YOU EVER HEARD THiS OR SOME VARiATiON THEREOF? Sure you have! Every developer and student probably hears comments like this fre- quently. Wh…

How can I protect derived classes from breaking when I change the internal parts of the base class? A class has two distinct interfaces for two distinct sets of clients: It has a public interface that serves unrelated classes It has a protected inter…

The Doors Time Limit: 1000 MS Memory Limit: 10000 KB 64-bit integer IO format: %I64d , %I64u   Java class name: Main [Submit] [Status] [Discuss] Description You are to find the length of the shortest path through a chamber containing obstructing wall…

The Doors Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 5210   Accepted: 2124 Description You are to find the length of the shortest path through a chamber containing obstructing walls. The chamber will always have sides at x = 0, x =…

http://poj.org/problem?id=3216 Repairing Company Time Limit: 1000MS   Memory Limit: 131072K Total Submissions: 6776   Accepted: 1822 Description Lily runs a repairing company that services the Q blocks in the city. One day the company receives M repa…

Question There are 100 doors in a row that are all initially closed. You make 100 passes by the doors. The first time through, visit every door and toggle the door (if the door is closed,open it;if it is open,close it). The second time, only visit ev…

  The Doors Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 7430   Accepted: 2915 Description You are to find the length of the shortest path through a chamber containing obstructing walls. The chamber will always have sides at x = 0, x…

The Doors Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 6734   Accepted: 2670 Description You are to find the length of the shortest path through a chamber containing obstructing walls. The chamber will always have sides at x = 0, x =…

POJ 1556 - The Doors题意:    在 10x10 的空间里有很多垂直的墙,不能穿墙,问你从(0,5) 到 (10,5)的最短距离是多少.    分析:        要么直达,要么一定是墙的边缘点之间以及起始点.终点的连线.        所以先枚举墙上每一点到其他点的直线可达距离,就是要判定该线段是否与墙相交(不含端点).        然后最短路. #include <iostream> #include <cstdio> #include <cmat…

POJThe Doors AND NYIST 有趣的问题 题目链接:pid=227" target="_blank">Click Here~ 题目分析: 给你横纵坐标分别表示门所在的位置.叫你求出从起点到终点的最短距离. 算法分析: 该题好像有多种解法,我仅仅说我做的. 我用的是几何+图论. 建模分析: 1.先推断两个点之间能否够连接. 2.推断两个点能否够链接的方法是用是否推断墙与这两点连成的线段是否相交. 3.假设没有相交则直接连接. 4.最后最短路求出结果就好了…

引言 IBM Rational DOORS,简称DOORS,是被业界广泛认可的需求管理工具,在国内外需求管理领域具有较高的市场占有率.需求管理作为传统的工程领域,理论发展相对成熟和健全.随着越来越多的企业开始注重在需求管理工程层面的投入,企业的需求管理成熟度也在逐步提高.在需求管理实施过程中,不可避免的会依托相关的需求管理工具支撑.而实施过程中所存在的关键困难之一就是:工具与业务如何紧密结合.很多企业虽然购买了相应的产品,但是工具层面的操作培训不足以使得企业的项目在需求管理工具中落地.鉴于此种情…

The Doors Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 8334   Accepted: 3218 Description You are to find the length of the shortest path through a chamber containing obstructing walls. The chamber will always have sides at x = 0, x =…

确定Breaking force合适的值同样很单调,但是按照下面的方法也并不是完全不可能: 输入一个随意的值,比如说100 检查实际场景中关节是否能承受住物理物体,在完美的情况下物理物体将保持静止. 如果关节断裂了,将该值翻倍(100*2),然后再试. 重复翻倍直到关节不再断裂,你现在知道实际Breaking force值得上限和下限了. 继续缩小范围,用下限和上限的差值除以2(类似于一种2分查找法),举个栗子来说 ;) 如果关节在5000值上断裂,但是稳定在10000值上,你可以接下来尝试75…

在SpriteBuilder中三种物理关节都包含Breaking force区域在属性框中. 该属性被设置成关节可以承受的压力临界值.如果关节的压力超出了Breaking force中设置的值,则关节将会被删除(removed). 注意:Max force必须被设置为无限的(unlimited)或者一个大于等于Breaking force的值.…

The Doors 签到题 #include <iostream> using namespace std; int a[200005]; int main() { int n; scanf("%d",&n); for(int i=0;i<n;i++) { scanf("%d",&a[i]); } int x=a[n-1]; int pos=n-2; for(int i=n-2;i>=0;i--) { if(x!=a[i])…

The Doors You are to find the length of the shortest path through a chamber containing obstructing walls. The chamber will always have sides at x = 0, x = 10, y = 0, and y = 10. The initial and final points of the path are always (0, 5) and (10, 5).…

ElasticSearch 2.1.1 (3) - Breaking Changes Search Changes search_type = scan Deprecated GET /my_index/_search?scroll=2m { "sort": [ "_doc" ] } search_type = count Deprecated GET /my_index/_search { "aggs": {...}, "size&…

本季Breaking Bad Season 5(2012)看点:故事紧接着上一季,通过一场精心策划的大爆炸,沃尔特(布莱恩·科兰斯顿 Bryan Cranston 饰)终于除掉了长久以来的威胁古斯塔沃(吉安卡洛·埃斯珀西多 Giancarlo Esposito 饰),也将古斯塔沃一手建立的毒品帝国连根拔起.然而,“事业”上的成功并没有为沃尔特带来太多的快乐,因为他和妻子斯凯勒(安娜·古恩 Anna Gunn 饰)之间的关系再次跌落到了冰点. <绝命毒师>第5季于北美当地时间上周日回归,收获了2…

本季Breaking Bad Season 1(2008)看点:新墨西哥州的高中化学老师沃尔特·H·怀特(布莱恩·科兰斯顿 Bryan Cranston 饰)是拮据家庭的唯一经济来源.他大半生安分守己,兢兢业业,却在50岁生日之际突然得知自己罹患肺癌晚期的噩耗,原本便不甚顺意的人生顿时雪上加霜. 沃尔特是个平凡的高中化学老师,在这个平凡的岗位上工作了25年.他本可以一直这样平凡下去,直到有一天,一纸绝症诊断彻底改变了沃尔特的生活.为了使自己的妻子儿子在自己死后也可以不为生计发愁,沃尔特用他的化学…

POJ 3216 Repairing Company id=3216">题目链接 题意:有m项任务,每项任务的起始时间,持续时间,和它所在的block已知,且往返每对相邻block之间的时间也知道,问最少须要多少个工人才干完毕任务,即x最少是多少 思路:先floyd求出每两个block之间的最小距离,然后就是最小路径覆盖问题,一个任务之后能赶到还有一个任务就建边 代码: #include <cstdio> #include <cstring> #include &l…

The Doors http://poj.org/problem?id=1556 Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 10466   Accepted: 3891 Description You are to find the length of the shortest path through a chamber containing obstructing walls. The chamber will…

The Doors Time Limit: 1000MS Memory Limit: 10000K Description You are to find the length of the shortest path through a chamber containing obstructing walls. The chamber will always have sides at x = 0, x = 10, y = 0, and y = 10. The initial and fina…

http://poj.org/problem?id=3216 n个地点,m个任务 每个任务有工作地点,开始时间,持续时间 最少派多少人可以完成所有的任务 传递闭包之后最小路径覆盖 #include<cstdio> #include<cstring> #include<iostream> using namespace std; #define N 21 #define M 201 int n,m; int f[N][N]; bool mp[M][M]; int pos[…

DOORS的引用类型包括:Project: 项目引用Folder: 文件夹引用Item: 项目或文件夹之内的项(项目.文件夹.模块)Module: 打开模块的引用Object: 对象的引用Link: 打开的源模块的链接引用LinkRef: 源模块关闭的链接引用AttrDef: 属性定义记录引用AttrType: 属性类型定义记录引用…

题目链接: http://codeforces.com/problemset/problem/507/E E. Breaking Good time limit per test2 secondsmemory limit per test256 megabytes 问题描述 Breaking Good is a new video game which a lot of gamers want to have. There is a certain level in the game that…