poj1556The Doors】的更多相关文章

链接 枚举两点 若不和任何线段相交 建边为dis(i,j) floyd求最短路 #include <iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<stdlib.h> #include<vector> #include<cmath> #include<queue> #include<set> usin…
两道Bellman Ford解最短路的范例,Bellman Ford只是一种最短路的方法,两道都可以用dijkstra, SPFA做. Bellman Ford解法是将每条边遍历一次,遍历一次所有边可以求得一点到任意一点经过一条边的最短路,遍历两次可以求得一点到任意一点经过两条边的最短路...如 此反复,当遍历m次所有边后,则可以求得一点到任意一点经过m条边后的最短路(有点类似离散数学中邻接矩阵的连通性判定) POJ1556-The Doors 初学就先看POJ2240吧 题意:求从(0,5)到…
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 =…
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 =…
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])…
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…
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).…
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…
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…
A. The Doors time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Three years have passes and nothing changed. It is still raining in London, and Mr. Black has to close all the doors in his home…
或许提到本书甚至本书的作者Johanna Rothman我们会感到些许陌生,那么提起她的另一本获得素有软件界奥斯卡之称的Jolt生产效率大奖的名著<项目管理修炼之道>,会不会惊讶的发现,原来是她! 很久以前就买了英文版的Behind Closed Doors: Secrets of Great Management一书(中文版为<门后的秘密:卓越管理的故事>),可惜当时并没有引起足够的重视,草草翻阅后便束之高阁.开始的印象就是一本以故事叙述为主的工作场景再现,配以总结的若干管理技巧…
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“…
IBM Rational DOORS 可实现对整个产品的全生命周期需求管理,覆盖从需求.到设计以及测试阶段.是一款具有广泛使用的企业级专业需求管理工具.DOORS 可以将项目开发过程中产生的各级需求和与需求相关的文件进行链接管理,同时能够对需求进行影响分析.DOORS 自带数据库,可以在多个项目间共享文件,便于文件的保存.备份及项目复用.此外DOORS 还支持可疑链接的自动检测.基于需求条目的权限管理等.目前国内外主要汽车OEM 和Tier1,以及工业.交通.电子等行业主要的主机厂所.成品厂均采…
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…
The  Boy Next   Doors 题意:给定一个固定大小的房间($x,y$的范围都是$[0,10]$),有$n$个墙壁作为障碍(都与横坐标轴垂直),每个墙壁都有两扇门分别用四个点来描述,起点终点固定在$(0,5)$和$(10,5)$,求起点到终点的最短路长度,$n<=18$ 题解: 我们把每堵墙的每一"段"作为一条线段,对任意两点$u,v$,如果两点间的连线不和其他线段相交,那我们从$u$走到$v$的最短距离就是他们的欧几里得距离,对所有点对都这么做一遍,处理出所有能够…
思路:暴力判断每个点连成的线段是否被墙挡住,构建图.求最短路. 思路很简单,但是实现比较复杂,模版一定要可靠. #include<stdio.h> #include<string.h> #include<math.h> #include<iostream> using namespace std; ,M=N*N; const double INF=0x3f3f3f3f; ; int sgn(double x){ ; ) ; ; } struct point{…
黑书上的一道例题:如果走最短路则会碰到点,除非中间没有障碍. 这样把能一步走到的点两两连边,然后跑SPFA即可. #include<cmath> #include<cstdio> #include<cstring> #include<algorithm> #define N 100003 using namespace std; struct Point { double x, y; Point(double _x = 0, double _y = 0) :…
题意: 给一个正方形,从左边界的中点走到右边界的中点,中间有一些墙,问最短的距离是多少. 解法: 将起点,终点和所有墙的接触到空地的点存下来,然后两两之间如果没有线段(墙)阻隔,就建边,最后跑一个最短路SPFA,即可得出答案. 代码: #include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <cmath> #include <a…
题目传送门 题意:从(0, 5)走到(10, 5),中间有一些门,走的路是直线,问最短的距离 分析:关键是建图,可以保存所有的点,两点连通的条件是线段和中间的线段都不相交,建立有向图,然后用Dijkstra跑最短路.好题! /************************************************ * Author :Running_Time * Created Time :2015/10/24 星期六 09:48:49 * File Name :POJ_1556.cpp…
http://poj.org/problem?id=1556 首先路径的每条线段一定是端点之间的连线.证明?这是个坑...反正我是随便画了一下图然后就写了.. 然后re是什么节奏?我记得我开够了啊...然后再开大点才a...好囧啊. #include <cstdio> #include <cstring> #include <cmath> #include <string> #include <iostream> #include <alg…
#include <iostream> #include <stdio.h> #include <string.h> #include <algorithm> #include <queue> #include <map> #include <vector> #include <set> #include <string> #include <math.h> using namespac…
题意: 一个正方形中有n道竖直的墙,每道墙上开两个门.求从左边中点走到右边中点的最短距离. 分析: 以起点终点和每个门的两个端点建图,如果两个点可以直接相连(即不会被墙挡住),则权值为两点间的欧几里得距离. 然后求起点到终点的最短路即可. #include <cstdio> #include <cmath> #include <cstring> #include <vector> #include <algorithm> using namesp…