poj 3525 二分+半平面交

poj3525 计算凸多边形最大内接圆(二分+半平面交 ) 题意 给你一个凸包问你凸包的某一点到每条边的最短距离最大? 思路 题目的实质就是多边形的最大内接圆半径! 二分圆的半径mid,用半平面交验证,每条直线向内平移mid的距离,之后跑一遍半平面, 判断有没有凸核,若有凸核,l=mid(进一步扩大范围),else r=mid 将逆时针凸包中每条边向内推进R的实现:直线平移! 把直线表示成Ax + By + C = 0后(已考虑顺逆时针所带来的正负号问题),把直线向内推进d时,只需要写: c += d * sqrt(a*a + b*b); .. 阅读详情

2011-05-02

POJ 3525 Most Distant Point from the Sea (半平面交+二分求最大内切圆) Most Distant Point from the Sea Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 3476 Accepted: 1596 Special Judge Description The main l 阅读详情

相关推荐

OpenCV快速入门:绘制图形、图像金字塔和感兴趣区域

OpenCV是一个强大的计算机视觉库,广泛应用于图像处理、计算机视觉和机器学习等领域。本文将介绍OpenCV中的一些基础功能,包括如何绘制图形、创建图像金字塔以及如何处理感兴趣区域(ROI)。

吾家有猫名探花的博客 1962

poj3525 二分+半平面交

题目链接:http://poj.org/problem?id=3525 题意:给定一个凸多边形,求多边形中距离边界最远的点到边界的距离。 思路 : 每次将凸多边形每条边往里平移d,判断是否存在核;二分d即可。 #include #include #include #include using namespace std; const double eps = 1e-10; co

4019

训练AI:从数据收集到模型部署的完整指南

如果你刚开始接触AI,建议先学习一些基础的编程和机器学习知识,还可以使用现成的AI平台和工具来帮助你训练AI模型。在这篇文章中,我们将详细探讨从数据收集到模型部署的完整步骤,并提供一些有用的技巧和工具,以帮助你成功地训练自己的AI模型。在收集数据时,你需要从可靠的数据源(如数据仓库、数据集市、API接口)获取数据,确保数据的准确性和完整性。常用的划分比例是70%的数据集用于训练模型、15%的数据集用于验证模型,5%的数据集用于超参数调整,最后10%的数据集用于测试模型的性能。测试数据集用于测试模型的性能。

a_wsnd的博客 4166

POJ-3525 半平面交,二分答案

Description The main land of Japan called Honshu is an island surrounded by the sea. In such an island, it is natural to ask a question: “Where is the most distant point from the sea?” The answer to t...

刷题记录 287

Chromatix_AEC—深入解析曝光表配置与传感器参数计算

本文深入解析高通Chromatix AEC模块中的曝光表配置与传感器参数计算,详细介绍了自动曝光控制(AEC)的核心概念、首行曝光时间计算、行时间与曝光模式对比、帧率计算逻辑以及曝光表高级配置技巧。通过实战案例展示如何优化传感器参数联调,帮助开发者掌握Chromatix工具在摄像头系统中的关键应用。

266

傻瓜教程:使用GeoServer发布离线地图服务

一、离线地图下载 离线地图分为两种:一种叫“金字塔瓦片”数据,一种叫“矢量地图数据”。我们看的在线地图比如,百度,谷歌,高德等等网页上的地图,都是金字塔瓦片;另外一种是手机上用来导航的,比如高德地图APP,也可以下载离线地图数据,这个就是矢量的。 卫星影像地图都是金字塔瓦片,无矢量数据。 因此,我们所说的离线地图下载通常都是指离线地图金字塔瓦片数据下载。 离线地图金字塔瓦片搭建的离线地图服务分很多种:一般有WEB方式,TMS、WMTS/WMS等等,不同的应用有不同...

米兰的小码农的博客 1万+

poj3525——二分+半平面交

题目链接:http://poj.org/problem?id=3525 The main land of Japan called Honshu is an island surrounded by the sea. In such an island, it is natural to ask a question: “Where is the most distant point from ...

mumei的博客 257

Flightmare实战:从零搭建强化学习无人机控制环境

本文详细介绍了从零开始搭建Flightmare强化学习无人机控制环境的完整流程。内容包括环境准备、核心库安装、依赖冲突解决,以及如何运行PPO算法示例进行训练与可视化测试。文章还提供了环境配置、自定义任务修改及常见问题的避坑指南,帮助开发者快速上手这一模块化的无人机RL仿真平台。

weixin_29081101的博客 356

POJ 3525 二分答案,推进多边形和半平面交

题目在问这样一个问题:给定一个凸多边形,找到一个

暗黑ZX Cyberwind 860

Paddle账户与支付宝功能快速开通指南:全球化支付解决方案

本文为开发者提供了Paddle账户与支付宝功能集成的详细开通与配置指南。通过Paddle这一全球化支付与销售平台,开发者可一站式管理全球支付与税务合规,而集成支付宝则能无缝覆盖中国用户,实现“一个平台,覆盖全球主流支付方式”,有效提升支付转化率与收款效率。

qq_27496129的博客 1121

实战交易策略 篇二十二:情绪流龙头交易策略

在股票市场中,机构主导趋势,游资驱动热点,散户易被情绪裹挟,形成“金字塔”式生态链,并存在情绪传递、强者恒强、物极必反的规律特点。股市是资金驱动的情绪博弈场,价格波动由群体心理、资金博弈和预期差共同驱动,核心是资本逐利性与人性贪婪恐惧的交织,兼具资源配置、财富再分配与风险定价功能。涨停加速与逃逸缺口(主升强化)凹字板/平台突破(趋势延续)天量弱转强(分歧买点)不参与首波(二波策略)情绪周期(道术合一)主线思维(聚焦核心)开盘抢筹(竞价博弈)一字助攻(情绪强化)分批进出(滚动操作)

生活在别处 1500

POJ 3525 二分+半平面交

Most Distant Point from the Sea Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 3812   Accepted: 1779   Special Judge Description The main land of Japan ca

_rabbit的专栏 1120

C#【通用方法】字节数组添加、移除、获取的方法类(妙用List<byte>,开发中常用)

C#【通用方法】添加移除字节的方法类(妙用List,开发中常用)

星Yeah 的博客 3359

POJ3525(二分+半平面交)

题目:题目链接   题目:就是给你一个多边形,让你判断这个多边形的最大内切圆   分析:二分半径r,把每条边向内收缩r,半平面交判可行   半平面交还是太弱:   代码:   #include #include #include #include #include #include #include #include #include #include #i

ACM_devil 999

efm32 入门.pdf

本PDF共计22页,介绍的是操作驱动API,下载与否自己斟酌。

深度学习工具包 DeepLearningToolbox

深度学习工具包Deprecation notice.-----This toolbox is outdated and no longer maintained.There are much better tools available for deep learning than this toolbox, e.g. [Theano](http://deeplearning.net/software/theano/), [torch](http://torch.ch/) or [tensorflow](http://www.tensorflow.org/)I would suggest you use one of the tools mentioned above rather than use this toolbox.Best, Rasmus.DeepLearnToolbox================A Matlab toolbox for Deep Learning.Deep Learning is a new subfield of machine learning that focuses on learning deep hierarchical models of data.It is inspired by the human brain's apparent deep (layered, hierarchical) architecture.A good overview of the theory of Deep Learning theory is[Learning Deep Architectures for AI](http://www.iro.umontreal.ca/~bengioy/papers/ftml_book.pdf)For a more informal introduction, see the following videos by Geoffrey Hinton and Andrew Ng.* [The Next Generation of Neural Networks](http://www.youtube.com/watch?v=AyzOUbkUf3M) (Hinton, 2007)* [Recent Developments in Deep Learning](http://www.youtube.com/watch?v=VdIURAu1-aU) (Hinton, 2010)* [Unsupervised Feature Learning and Deep Learning](http://www.youtube.com/watch?v=ZmNOAtZIgIk) (Ng, 2011)If you use this toolbox in your research please cite [Prediction as a candidate for learning deep hierarchical models of data](http://www2.imm.dtu.dk/pubdb/views/publication_details.php?id=6284)```@MASTERSTHESIS\{IMM2012-06284, author = "R. B. Palm", title = "Prediction as a candidate for learning deep hierarchical models of data", year = "2012",}```Contact: rasmusbergpalm at gmail dot comDirectories included in the toolbox-----------------------------------`NN/` - A library for Feedforward Backpropagation Neural Networks`CNN/` - A library for Convolutional Neural Networks`DBN/` - A library for Deep Belief Networks`SAE/` - A library for Stacked Auto-Encoders`CAE/` - A library for Convolutional Auto-Encoders`util/` - Utility functions used by the libraries`data/` - Data used by the examples`tests/` - unit tests to verify toolbox is workingFor references on each library check REFS.mdSetup-----1. Download.2. addpath(genpath('DeepLearnToolbox'));Example: Deep Belief Network---------------------```matlabfunction test_example_DBNload mnist_uint8;train_x = double(train_x) / 255;test_x = double(test_x) / 255;train_y = double(train_y);test_y = double(test_y);%% ex1 train a 100 hidden unit RBM and visualize its weightsrand('state',0)dbn.sizes = [100];opts.numepochs = 1;opts.batchsize = 100;opts.momentum = 0;opts.alpha = 1;dbn = dbnsetup(dbn, train_x, opts);dbn = dbntrain(dbn, train_x, opts);figure; visualize(dbn.rbm{1}.W'); % Visualize the RBM weights%% ex2 train a 100-100 hidden unit DBN and use its weights to initialize a NNrand('state',0)%train dbndbn.sizes = [100 100];opts.numepochs = 1;opts.batchsize = 100;opts.momentum = 0;opts.alpha = 1;dbn = dbnsetup(dbn, train_x, opts);dbn = dbntrain(dbn, train_x, opts);%unfold dbn to nnnn = dbnunfoldtonn(dbn, 10);nn.activation_function = 'sigm';%train nnopts.numepochs = 1;opts.batchsize = 100;nn = nntrain(nn, train_x, train_y, opts);[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.10, 'Too big error');```Example: Stacked Auto-Encoders---------------------```matlabfunction test_example_SAEload mnist_uint8;train_x = double(train_x)/255;test_x = double(test_x)/255;train_y = double(train_y);test_y = double(test_y);%% ex1 train a 100 hidden unit SDAE and use it to initialize a FFNN% Setup and train a stacked denoising autoencoder (SDAE)rand('state',0)sae = saesetup([784 100]);sae.ae{1}.activation_function = 'sigm';sae.ae{1}.learningRate = 1;sae.ae{1}.inputZeroMaskedFraction = 0.5;opts.numepochs = 1;opts.batchsize = 100;sae = saetrain(sae, train_x, opts);visualize(sae.ae{1}.W{1}(:,2:end)')% Use the SDAE to initialize a FFNNnn = nnsetup([784 100 10]);nn.activation_function = 'sigm';nn.learningRate = 1;nn.W{1} = sae.ae{1}.W{1};% Train the FFNNopts.numepochs = 1;opts.batchsize = 100;nn = nntrain(nn, train_x, train_y, opts);[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.16, 'Too big error');```Example: Convolutional Neural Nets---------------------```matlabfunction test_example_CNNload mnist_uint8;train_x = double(reshape(train_x',28,28,60000))/255;test_x = double(reshape(test_x',28,28,10000))/255;train_y = double(train_y');test_y = double(test_y');%% ex1 Train a 6c-2s-12c-2s Convolutional neural network %will run 1 epoch in about 200 second and get around 11% error. %With 100 epochs you'll get around 1.2% errorrand('state',0)cnn.layers = { struct('type', 'i') %input layer struct('type', 'c', 'outputmaps', 6, 'kernelsize', 5) %convolution layer struct('type', 's', 'scale', 2) %sub sampling layer struct('type', 'c', 'outputmaps', 12, 'kernelsize', 5) %convolution layer struct('type', 's', 'scale', 2) %subsampling layer};cnn = cnnsetup(cnn, train_x, train_y);opts.alpha = 1;opts.batchsize = 50;opts.numepochs = 1;cnn = cnntrain(cnn, train_x, train_y, opts);[er, bad] = cnntest(cnn, test_x, test_y);%plot mean squared errorfigure; plot(cnn.rL);assert(er<0.12, 'Too big error');```Example: Neural Networks---------------------```matlabfunction test_example_NNload mnist_uint8;train_x = double(train_x) / 255;test_x = double(test_x) / 255;train_y = double(train_y);test_y = double(test_y);% normalize[train_x, mu, sigma] = zscore(train_x);test_x = normalize(test_x, mu, sigma);%% ex1 vanilla neural netrand('state',0)nn = nnsetup([784 100 10]);opts.numepochs = 1; % Number of full sweeps through dataopts.batchsize = 100; % Take a mean gradient step over this many samples[nn, L] = nntrain(nn, train_x, train_y, opts);[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.08, 'Too big error');%% ex2 neural net with L2 weight decayrand('state',0)nn = nnsetup([784 100 10]);nn.weightPenaltyL2 = 1e-4; % L2 weight decayopts.numepochs = 1; % Number of full sweeps through dataopts.batchsize = 100; % Take a mean gradient step over this many samplesnn = nntrain(nn, train_x, train_y, opts);[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.1, 'Too big error');%% ex3 neural net with dropoutrand('state',0)nn = nnsetup([784 100 10]);nn.dropoutFraction = 0.5; % Dropout fraction opts.numepochs = 1; % Number of full sweeps through dataopts.batchsize = 100; % Take a mean gradient step over this many samplesnn = nntrain(nn, train_x, train_y, opts);[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.1, 'Too big error');%% ex4 neural net with sigmoid activation functionrand('state',0)nn = nnsetup([784 100 10]);nn.activation_function = 'sigm'; % Sigmoid activation functionnn.learningRate = 1; % Sigm require a lower learning rateopts.numepochs = 1; % Number of full sweeps through dataopts.batchsize = 100; % Take a mean gradient step over this many samplesnn = nntrain(nn, train_x, train_y, opts);[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.1, 'Too big error');%% ex5 plotting functionalityrand('state',0)nn = nnsetup([784 20 10]);opts.numepochs = 5; % Number of full sweeps through datann.output = 'softmax'; % use softmax outputopts.batchsize = 1000; % Take a mean gradient step over this many samplesopts.plot = 1; % enable plottingnn = nntrain(nn, train_x, train_y, opts);[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.1, 'Too big error');%% ex6 neural net with sigmoid activation and plotting of validation and training error% split training data into training and validation datavx = train_x(1:10000,:);tx = train_x(10001:end,:);vy = train_y(1:10000,:);ty = train_y(10001:end,:);rand('state',0)nn = nnsetup([784 20 10]); nn.output = 'softmax'; % use softmax outputopts.numepochs = 5; % Number of full sweeps through dataopts.batchsize = 1000; % Take a mean gradient step over this many samplesopts.plot = 1; % enable plottingnn = nntrain(nn, tx, ty, opts, vx, vy); % nntrain takes validation set as last two arguments (optionally)[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.1, 'Too big error');```[![Bitdeli Badge](https://d2weczhvl823v0.cloudfront.net/rasmusbergpalm/deeplearntoolbox/trend.png)](https://bitdeli.com/free "Bitdeli Badge")

POJ 3525 Most Distant Point from the Sea [二分+半平面交]

POJ 3525 Most Distant Point from the Sea [二分+半平面交]

ACTerminate的博客 711

IEM模型,地表粗糙度模拟

利用微波反演土壤水分时候,利用IEM模型可以实现对随机地表的模拟

POJ 3525 Most Distant Point from the Sea 半平面交+二分

半平面交+二分 二分距离,判断是否有半平面交,精度要求很高 Most Distant Point from the Sea Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 4163   Accepted: 1942   Special Judge

码代码的猿猿 1209

SIMATIC S7-1200, S7-1500 PID 控制[手册]

SIMATIC S7-1200, S7-1500 PID 控制[手册]

poj 3525 求凸多边形的最大内切圆

【题意】 凸多边形的最大内切圆 【题解】 二分+半平面交判定。 【代码】 #include #include #define eps 1e-8 #define oo 1e5 using namespace std; const int maxn=105; struct point { double x,y; point() {} poi

AChunter的专栏 5054

uva1396,poj3525半平面交+二分

链接:http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=533&page=show_problem&problem=4142 提议:海中一个凸多边形小岛,求出岛上离海最远点到海的距离。 思路:就是求这个凸多边形内切圆的最大半径。对半径进行二分,将多边形所有边...

weixin_30451709的博客 135

POJ 3525 Most Distant Point from the Sea 二分+半平面交

链接:http://poj.org/problem?id=3525 题意:给一个凸多边形的海岛,寻找海岛之中距离海边距离最长的一个点的距离。 思路:求凸多边形的最大内切圆。做法是二分半径加半平面交,将凸多边形的每条边向内部(垂直方向)收缩半径r,看每条边的半平面是否还会交出凸多边形。 P.S. 无意间找到这道题,顺便检验一下自己敲的模板。找一个好的模板真的很重要。 代码: #includ

看曙光 1120

【POJ3525】Most Distant Point from the Sea(半平面交+二分+直线平移)

题目地址:http://poj.org/problem?id=3525 题目: 给出一个形状为凸多边形的小岛,四周环海,求小岛内一点距海的最远距离,输出距离。 解题思路: 相当于在凸多边形内找一点,使得这个点到所有边的垂直距离的最小值最大。也即找到一个最大的圆,使得圆在凸多边形内和凸多边形相切。 半平面:有向直线左侧的区域。 可以二分答案,判断是否有距离凸多边形的每条边长度为d的...

Cassie_zkq的博客 397

poj 3525 半平面交求多边形内切圆最大半径【半平面交】+【二分】

<题目链接> 题目大意:给出一个四面环海的凸多边形岛屿,求出这个岛屿中的点到海的最远距离。 解题分析: 仔细思考就会发现,其实题目其实就是让我们求该凸多边形内内切圆的最大半径是多少。但是,这个最大半径,没有什么比较好的求法,于是,我们可以想到二分答案求半径。对于二分的半径,我们可以将该凸多边形的边界向内平移 r 的距离,然后再用半平面交法,用这些平移后的直线去切割原凸多边形,如果...

weixin_30551947的博客 180

POJ - 3525 Most Distant Point from the Sea (半平面交+二分)

题目: The main land of Japan called Honshu is an island surrounded by the sea. In such an island, it is natural to ask a question: “Where is the most distant point from the sea?” The answer to this que...

slark_的博客 308

半平面交 poj 3335(判断是否有内核) poj 3525 (半平面交+二分)

poj 3335 Description This year, ACM/ICPC World finals will be held in a hall in form of a simple polygon. The coaches and spectators are seated along the edges of the polygon. We want to p

winycg的博客 729

POJ 3525 Most Distant Point from the Sea 二分+半平面交

题目就是求多变形内部一点。 使得到任意边距离中的最小值最大。 那么我们想一下,可以发现其实求是看一个圆是否能放进这个多边形中。 那么我们就二分这个半径r,然后将多边形的每条边都往内退r距离。 求半平面交看是否存在解即可 #include #include #include #include #include #include #include #include #inc

水 1265

POJ 3525 Most Distant Point from the Sea (二分答案+半平面交)

DescriptionThe main land of Japan called Honshu is an island surrounded by the sea. In such an island, it is natural to ask a question: “Where is the most distant point from the sea?” The answer to thi

Ab.Ever 847
上一篇: codeforces 2011 悲剧
下一篇: 计算几何 模板
mindmb
博客等级 码龄16年 16粉丝 86原创
评论
成就一亿技术人!
拼手气红包6.0元
还能输入1000个字符
 
 条评论被折叠 查看
添加红包

请填写红包祝福语或标题

个

红包个数最小为10个

元

红包金额最低5元

当前余额3.43元 前往充值 >
需支付:10.00元
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付元
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值