A1147 Heaps (30) 堆

Cesium创建无人机视锥,同步视锥与鹰眼地图视角,实现视锥键盘控制。解决视锥初始化位置不准确问题等 本文描述了一个基于Cesium的3D视锥体可视化系统开发过程。系统实现了主视图与鹰眼视图的双屏联动,支持通过键盘控制视锥体的空间移动(WASD前后左右,Q/E旋转,C/Z升降),并实时同步更新两个视图。关键技术包括:1)基于主视图Scene创建Camera实现视锥体精确绘制;2)通过PerspectiveFrustum设置视锥参数;3)建立双视图的相机位置及视角同步机制。目前暂未实现视锥体far/near边界在鹰眼视图中的可视化,该系统可应用于无人机视角监控等三维场景交互需求。 阅读详情

本题思路:
层序遍历建树;
建树的同时通过两个flag的值,flag1为大顶堆成立,flag2为小顶堆成立,判断的时候如果既是大顶堆又是小顶堆,说明不是堆;
后序遍历节点;

#include<iostream>
#include<cstdio>
using namespace std;
const int maxn=1010;
int a[maxn],num,n,m;
bool flag,flag1,flag2;
struct node
{
	int data;
	node *lc,*rc;
};
node* level(int index)
{
	if(index>n)return NULL;
	node* root=new node;
	root->data=a[index];
	if(index*2<=n)
	{
		if(a[index]>=a[index*2])
		flag1=true;
		else
		flag2=true;
	}
	if(index*2+1<=n)
	{
		if(a[index]>=a[index*2+1])
		flag1=true;
		else
		flag2=true;
	}
	root->lc=level(2*index);
	root->rc=level(2*index+1);
	return root;
}
void post(node *root)
{
	if(root==NULL)return ;
	post(root->lc);
	post(root->rc);
	printf("%d",root->data);
	num++;
	if(num!=n)
	printf(" ");
	else 
	printf("\n");
}
int main()
{
	cin>>m>>n;
	for(int i=0;i<m;i++)
	{
		for(int j=1;j<=n;j++)
		{
			scanf("%d",&a[j]);
		}
		flag1=false;
		flag2=false;
		node* root=level(1);
		if(flag1==true&&flag2==false)
		{
			printf("Max Heap\n");
		}
		else if(flag2==true&&flag1==false)
		{
			printf("Min Heap\n");
		}
		else
		{
			printf("Not Heap\n");
		}
		num=0;
		post(root);
	}
}
Vivado ChipScope Debugging: Overcoming Common DRC Errors in FPGA Design 本文深入解析Vivado ChipScope调试中常见的DRC错误,特别是AVAL-244和AVAL-245等时钟相关问题,提供实用的Tcl命令解决方案和最佳实践。通过将DRC错误降级为警告的技巧,帮助FPGA设计者高效完成调试流程,同时强调何时应保留DRC检查以确保设计可靠性。 阅读详情

相关推荐

学习海康VisionMaster之圆查找

(1):ROI选中位置修正,并且修正信息需要选择刚刚创建的位置修正组件输出信息,这里的话后续的检测窗口一般都需要做这一步,因为大部分的应用场景都是需要用到实时跟踪的。圆查找模块是用于查找指定区域的圆,并且会输出该圆的相关参数信息,比如圆中心点,圆半径,圆是否存在等信息。7):剔除点数:和高级参数里面的剔除距离搭配使用的,就是过滤较远的边缘点,让拟合圆心出来的轮廓更加接近。中的圆查找,相当于就是拟合圆心,这个在视觉软件中的应用也是非常广泛的,通常用于定位,测量项目。

yuhouxiyang的专栏 2104

A1147 Heaps (30 分)

原题:https://pintia.cn/problem-sets/994805342720868352/problems/994805342821531648 首先建堆,然后判断是最大还是最小,之后post order traversal。但是我的代码有一个数据总是过不了,实在不知道为啥。最后我给出了柳婼大神的代码 #include <iostream> #include <v...

qq_44761480的博客 141

在win10系统上安装opnet14.5注意事项和步聚

opnet 14.5安装避坑!

miracle_eicont的专栏 1038

PAT 1147 Heaps(30 分)

In computer science, aheapis a specialized tree-based data structure that satisfies the heap property: if P is a parent node of C, then the key (the value) of P is either greater than or equal t...

weixin_30516243的博客 263

SAP BTP开发入门:从PaaS到CAP/UI5的30天学习路线

在云计算与数字化转型的大背景下,企业系统集成与云端扩展需求激增,传统的单点开发模式正面临深刻变革。PaaS(平台即服务)作为云原生的核心概念,将基础设施、开发环境与应用服务统一封装,为企业提供了灵活的扩展能力。SAP BTP正是SAP生态中的PaaS平台,它承载着系统集成、低代码开发、AI服务与数据管理,成为连接S/4HANA等核心业务系统与创新应用的桥梁。对于ABAP开发者而言,掌握BTP不仅是技能升级,更是应对从传统部署向云端订阅模式转型的必由之路。本文从平台基础出发,梳理BTP的核心服务与开发模型(如

weixin_31766003的博客 285

4、Vitis_HLS、Vivado 和 Vitis IDE 工具的设置与使用

本文详细介绍了使用Vitis_HLS、Vivado和Vitis IDE进行FPGA开发的完整流程,涵盖开发板选择、软件安装、IP设计与模拟、综合、导出以及在FPGA上的实现与验证。以Xilinx Artix-7和Zynq系列开发板为例,指导用户从环境搭建到最终硬件验证,完成一个32位加法器IP的开发,适用于嵌入式系统和硬件加速项目的学习与实践。

fff88的博客 181

PAT A1147 Heaps (30分) (堆的判定)

PAT甲级:A1147 Heaps (30分) In computer science, a heap is a specialized tree-based data structure that satisfies the heap property: if P is a parent node of C, then the key (the value) of P is either gre...

只想用最简短的代码刷题 239

GEE数据融合——Landsat (collection 2,level 2 )4、5、7、8、9长时间序列影像数据融合和视频导出分析

长时间序列影像数据融合是指将Landsat影像数据集合2级2(Level 2)中的4、5、7、8和9这五个卫星的数据进行融合。具体来说,这包括将同一地点的多个卫星影像数据进行处理和整合,以产生一个单一的、具有更高质量的影像产品。在长时间序列影像数据融合中,首先需要对不同卫星的影像数据进行校正,以消除不同卫星之间的光谱差异、辐射度问题和几何纠正误差等。然后,通过使用融合算法,将多个卫星的影像数据进行组合,以生成一张全新的、高质量的合成影像。

此星光明博客 1735

PAT A1147 Heaps (30 分) 堆

题目大意:给出完全二叉树的层序遍历序列,判断这棵完全二叉树是否是最大或最小堆,并输出后序遍历的结果。 以下标为0为根节点,下标为k的节点的左孩子下标是 2*k+1, 右孩子下标是 2*k+2,最后一个非叶子节点的下标是(N-2)/2。因此遍历下标为 0 ~ (N-2) / 2 的节点,考察它们与左右孩子的大小关系,容易判断出是否是最大或最小堆。 AC代码: #include...

HackingGate的博客 191

Vite 任意文件读取漏洞复现(CVE-2026-39363)

Vite开发服务器 WebSocket中的fetchModule方法未实施server.fs访问控制,导致攻击者可通过WebSocket连接并读取服务器上的任意文件(CVE-2026-39363)

iSee857的博客 221

ZYNQ 7020 之 FPGA知识点重塑笔记一——串口通信

通信方式一般分为串行通信和并行通信。并行通信是指多比特数据同时通过并行线进行传送。这种传输方式通信线多、成本高,故不宜进行远距离通信,通常传输距离小于 30 米。串行通信是指数据在一条数据线上,一比特接一比特地按顺序传送的方式。这种运输方式通常节省传输线,大大降低使用成本,但数据传送速度慢。综上可知,串行通信主要应用于长距离、低速率的通信场合。本次实验我们主要讲解下串行通信。串行通信一般有2种通信方式:同步串行通信和异步串行通信。同步串行通信。

m0_51430584的博客 3095

pat 甲级 堆 A1147 Heaps (30分)

1147 Heaps (30分) In computer science, a heap is a specialized tree-based data structure that satisfies the heap property: if P is a parent node of C, then the key (the value) of P is either greater than or equal to (in a max heap) or less than or equal to

weixin_42358356的博客 184

ConvNeXt 系列改进:引入大核注意力(LKA)替换 7×7 深度卷积,涨点 2.1%

LSKA替换7×7深度卷积,在ImageNet分类任务上实现2.1%的显著提升,参数量仅增加1M。LSKA的四层1D分解将计算复杂度从O(k²)降至O(k),21×21大核的计算量仅相当于原版7×7的1.7倍,实现了感受野与计算成本的完美平衡。部署友好度极高:LSKA模块在TensorRT FP16量化后延迟仅1.2ms,精度损失<0.1%,远优于Transformer类模型。下游任务全面领先。

ZXSXJ的博客 700

A1147 Heaps (30 分| 堆| 树的遍历,附详细注释,逻辑分析)

写在前面 思路分析 判断是否为堆 从第2个节点开始遍历 比父节点小,不是小顶堆 比父节点大,不是大顶堆 后序遍历,根据index分别遍历index2和index2+1,即左右孩子 遍历完左右子树,最后输出根结点 有点儿经验,进阶ing 测试用例 input: 3 8 98 72 86 60 65 12 23 50 8 38 25 58 52 82 70 60 10 28 ...

kngines 540

jdk-8u144-linux-x64.tar.gz.

该资源为搭建Linux系统Java环境用,资源的后缀为tar.gz格式的压缩包

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")

1151 LCA in a Binary Tree (30 分)&1147 Heaps (30 分) &1143 Lowest Common Ancestor (30 分)&(二叉树、堆的判断)

The lowest common ancestor (LCA) of two nodes U and V in a tree is the deepest node that has both U and V as descendants. Given any two nodes in a binary tree, you are supposed to find their LCA. I...

Star's blogs 113

文档:亲测可用解决vnc灰屏,黑屏,鼠标大黑叉情况

主要解决vnc-viewer灰屏,黑屏,鼠标大黑叉等现象,详细见文档。

PAT甲级A1147 Heaps(30 分)

1147 Heaps(30 分) In computer science, a heap is a specialized tree-based data structure that satisfies the heap property: if P is a parent node of C, then the key (the value) of P is either greater t...

vir_lee的博客 330

Outlook Express 6 (OE6) for Win7/Win8/Win10

英文版 Outlook Express6.0, 兼容Win7/Win10/WinXP,解压就可用。

1147 Heaps (30 分)(层序转后序,堆,树的遍历)

In computer science, a heap is a specialized tree-based data structure that satisfies the heap property: if P is a parent node of C, then the key (the value) of P is either greater than or equal to (i...

qq_42068614的博客 218

1147 Heaps (30 分)根据堆的层次遍历判断堆的类型并后序遍历

In computer science, a heap is a specialized tree-based data structure that satisfies the heap property: if P is a parent node of C, then the key (the value) of P is either greater than or equal to (i...

凡心所向,素履所往,生如逆旅,一苇以航 463

1147 Heaps (30分)[heap]

By Jalan 1147 Heaps (30分) 文章目录**By Jalan**知识工具需求数据结构和算法题干题解第一次思路编写用时代码CPP运行用时结尾 知识工具需求 数据结构和算法 heap 题干 输入一个堆检测是最大堆最小堆还是不是堆,之后要输出堆的dfs 30分里算水的了吧,可能是那次前面那个出的比较模糊做的补偿. 题解 第一次 思路 输入. 检测是不是最大堆,是则输出 检测是不是最小堆是则输出 输出不是堆. 编写用时 20分钟 代码 CPP #include <bits/stdc

JalanChao的博客 529

1147 Heaps (30分)

1147 Heaps (30分) AC代码 #include <iostream> #include <cstdio> #include <cstring> using namespace std; int m,n; bool isHeap(int u,int nums[],int len,int index){ if(u>=len) return...

renly123的博客 226

PAT A1147 Heaps (30point(s))

PAT A1147 Heaps Sample Input: 3 8 98 72 86 60 65 12 23 50 8 38 25 58 52 82 70 60 10 28 15 12 34 9 8 56 Sample Output: Max Heap 50 60 65 72 12 23 86 98 Min Heap 60 58 52 38 82 70 25 8 Not Heap 56 12 ...

qq_42347617的博客 252

【PAT】A1147 Heaps【堆的判定】

1147 Heaps (30 分) In computer science, a heap is a specialized tree-based data structure that satisfies the heap property: if P is a parent node of C, then the key (the value) of P is either greater t...

elegant coder 995

PAT 1147 Heaps C++版

PAT 1147 Heaps C++版 1.题意 给出一个完全二叉树,让你判断是大根堆还是小根堆,并输出这棵二叉树的后序输出。 2.分析 step1:因为是完全二叉树,所以很好判断是否是大根堆小根堆。 step2:那如何输出后序序列呢?可以通过构建一棵静态二叉树的方法,然后遍历输出即可。 3.代码 #include<cstdio> #include<queue> #i...

说文科技,做有态度的研究。 710

PAT-A 1147 Heaps

PAT-A 1147 Heaps 题意: 给定一个完全二叉树序列,判断该完全二叉树是否是是堆(大顶堆或小顶堆),最后输出完全二叉树的后序遍历。 代码: #include<cstdio> #include<string> #include<iostream> using namespace std; const int MAX_N = 1010; int tree[2*MAX_N];//用于存放完全二叉树 int T,n; //0表示属于大顶堆,1表示属于小顶堆,2表示不

weixin_44702215的博客 217

1147 Heaps (30 分)

PAT1147 题意,给出m个完全二叉树的层序遍历,树的结点时固定n个,要求判断是Min_Heap or Max_Heap or Not_Heap; 这个题目简单,用树的后序遍历,判断根是否大于左右儿子 需要注意的是,叶子结点的处理,那么如果判断max_heap 则将不存在的结点当作无穷小即一个很大的负数(因为题目说是integer) 我开始用的0,挂3个点,原来题目说的是整数,那么最小值不用用0...

qq_43645188的博客 170

PAT甲级 1147 Heaps (30分) 后序遍历+堆的判断/C++

1147 Heaps (30分) 题目大意:给定一个完全二叉树的层序遍历,要求判断其是否为堆,并输出后序遍历结果。 判断是否为堆,先确定为最大堆还是最小堆(比较开始的两个元素),再比较各个元素与其左右孩子的大小关系。 柳婼大神只用了25行代码解决!(传送门,学到了,还是要多看看大神的题解哒~) #include<iostream> //输入输出流头文件 #include<vector> //变长数组容器 usin

qq_43955154的博客 268
上一篇: A1146 Topological Order (25) 拓扑排序,vector用法
下一篇: A1149 Dangerous Goods Packaging (25)
coder_Kim
博客等级 码龄8年 2粉丝 160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