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目录 　　 使用非对称卷积分解大filters 　　 重新设计pooling层 　　 辅助构造器 　　 使用标签平滑 　　 参考资料 在《 深度学习面试题20：GoogLeNet(Inception V1) 》和《 深度学习面试题26：GoogLeNet(Inception V2) 》中对前两个Inception版本做了介绍，下面主要阐述V3版本的创新点 使用非对称卷积分解大filters InceptionV3中在网络较深的位置使用了非对称卷积，他的好处是在不降低模型效果的前提下，缩减模型的参数规模，在《 深度学习面试题27：非对称卷积(Asymmetric Convolutions) 》中介绍过。 end_point = 'Mixed_6d' with tf.variable_scope(end_point): with tf.variable_scope('Branch_0'): branch_0 = slim.conv2d(net, depth(192), [1, 1], scope='Conv2d_0a_1x1') with tf.variable_scope('Branch_1'): branch_1 = slim.conv2d(net, depth(160), [1, 1], scope='Conv2d_0a_1x1') branch_1 = slim.conv2d

I've built a crawler that had to run on about 5M pages (by increasing the url ID) and then parses the pages which contain the info' I need. after using an algorithm which run on the urls (200K) and saved the good and bad results I found that the I'm wasting a lot of time. I could see that there are a a few returning subtrahends which I can use to check the next valid url. you can see the subtrahends quite fast (a little ex' of the few first "good IDs") - 510000011 # +8 510000029 # +18 510000037 # +8 510000045 # +8 510000052 # +7 510000060 # +8 510000078 # +18 510000086 # +8 510000094 # +8

I am using a dataset in which it has images where each pixel is a 16 bit unsigned int storing the depth value of that pixel in mm. I am trying to visualize this as a greyscale depth image by doing the following: cv::Mat depthImage; depthImage = cv::imread("coffee_mug_1_1_1_depthcrop.png", CV_LOAD_IMAGE_ANYDEPTH | CV_LOAD_IMAGE_ANYCOLOR ); // Read the file depthImage.convertTo(depthImage, CV_32F); // convert the image data to float type namedWindow("window"); float max = 0; for(int i = 0; i < depthImage.rows; i++){ for(int j = 0; j < depthImage.cols; j++){ if(depthImage.at<float>(i,j) > max){

纪中17日T1 2321. 方程 (File IO): input:cti.in output:cti.out 时间限制: 1000 ms 空间限制: 262144 KB 具体限制 Goto ProblemSet 题目描述 输入 输出 样例输入 样例输出 数据范围限制 提示 吐槽 这些图片太模糊了吧…… 还有那吓人的 mod 998244353 都使得我们对这道题的恐惧感叠加了998244353层…… 没想到……只有三种答案！（ 三进制 呵呵哒） Solution （约定：用line[i]表示第i个输入的数据，man[i]表示在第i位是否为男生（i from 1 to n）） step1 先对line[1]进行判断： 　　若line[1]==0 　　　　直接dfs 　　若line[1]==1 　　　　进行两遍dfs，其中一遍man[1]=1，另一边man[2]=1； 　　若line[1]==2 　　　　一遍dfs，man[1]=man[2]=1； if(line[1]==1){ man[1]=1; dfs(1); memset(man,0,sizeof(man)); man[2]=1; dfs(1); } if(line[1]==0) dfs(1); if(line[1]==2){ man[1]=man[2]=1; dfs(1); } 处理完在边缘位置的line[1]后

I am trying to convert Kinect depth map to distance in meters. The problem is that for depthmap value '1080' and around it, distance is too large because the term in denominator becomes very close to '0'. and for values above '1090', distance is negative. if (depthValue < 2047) { depthM = 1.0 / (depthValue*-0.0030711016 + 3.3309495161); } The correct answer is actually a comment on your question. The number given is actually a distance in millimetres. To get this number, you either need to use a skeleton joint and call DepthImageFrame's MapFromSkeletonPoint or shift the raw short value right

原来旧的办法是，在主相机上加一个Panel，把3D模型显示在Panel上面，感觉这个方法不怎么好，现在进行改进: 现在用了两个相机，一个相机显示3D模型，另外一个是主相机。还需要GUITexture来作为背景 1,选择背景图片，创建一个GUITexture， 2.添加一个Layout 命名为Product 3.创建一个Camera 4.设置Camera的Depth为1，ClearFlags为Depth only, Culling Mask为Product 5.主相机的设置: Culling Mask中的Product去掉。 6.创建一个Cube，设置Layout 为Product 这样 Cube就可以在GUI之上了，效果图如下: 转载于:https://www.cnblogs.com/WilliamJiang/p/3223490.html 来源： https://blog.csdn.net/weixin_30252155/article/details/99617300

public class BST<E extends Comparable<E>>{ private class Node{ public E e; public Node left, right; public Node(E e){ this.e = e; this.left = null; this.right = null; } } private Node root; private int size; // 节点个数 public BST(){ root = null; size = 0; } public int size(){ return size; } public boolean isEmpty(){ return 0 == size; } // 向二分搜索树中添加新的元素 e public void add(E e){ root = add(root, e); } // 向以 node为根的二分搜索树中插入元素 e，递归算法 // 返回插入新节点后二分搜索树的根 private Node add(Node node, E e){ if(null == node){ size++; return new Node(e); } if(e.compareTo(node.e) < 0){ node.left = add(node.left, e); }else if

import java.util.*; /* public class TreeNode { int val = 0; TreeNode left = null; TreeNode right = null; public TreeNode(int val) { this.val = val; } } */ /* 层序遍历，但是使用start、end来记录每层的节点数 注意end值一直在更新，当每一层的结点值被取出时，end更新，更新之后为下一层的节点存储 做准备。第一层1个，end = 1 */ public class Solution { ArrayList<ArrayList<Integer> > Print(TreeNode pRoot) { ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>(); if(pRoot == null){ return result; } Queue<TreeNode> layer = new LinkedList<TreeNode>(); ArrayList<Integer> layerList = new ArrayList<Integer>(); layer.add(pRoot); int start = 0, end = 1; while(

When you have two planes in Three.js / WebGL and one or both of them are transparent, sometimes the plane behind will be hidden by the transparent plane above. Why is this? Toji This is not a bug, it's just how OpenGL (and, hence, WebGL) works. Transparent surfaces don't play well with the z-buffer, and as such must be manually sorted and rendered back-to-front. Three JS is attempting to do this for you (which is why the problem goes away when you set the X value > 0) but cannot robustly handle the case of intersecting geometry like you're showing. I've explained the issue more in-depth in a