长路径 meaning in Chinese
long-path
Examples
- It is the longest path through the project
它是通过项目的最长路径。 - In my experience , this approach will be at least 2 - 3 times as fast in most situations , and geometrically faster ( 10 + times as fast ) on longer paths
以我的经验，在大多数解决方案中，这个方法会快至少2 - 3倍，在长路径上更快（ 10倍以上） 。 - This is what the professionals do , using large areas for long paths , and then switching to finer searches using smaller squares / areas when you get close to the target
这就是专业人士的做法，对长路径使用大的区域，然后当接近目标时切换到使用小一些的方块/区域的精确搜索。 - And because the xml storing is an indispensable component in cms , the article proposed an efficiently xml storing strategy from the perspective of constraint schema of xml instance based on non - equal path analysis
另外，在内容管理中xml的存储已经成为一个必要的组成部分。文章提出了一种新颖的存储策略，这种方法基于非等长路径的xmlschema分析，最后通过一系列的测试证明了其高效性。 - The paper will study the upper questions based on the two interconnection networks . first , we give a fault - tolerant routing algorithm under the connectivity of the crossed cube in o ( n ) time and the length of the longest routing path ; second , with the rapid progress in vlsi , the failing probability of processors and links is very low , the traditional connectivity underestimates the resilience of large networks / here by applying the concept " conditional connectivity " introduce by harary , we show that the n - crossed cube can tolerate up to 2n - 3 ( n > 2 ) processors failure and remain connected provide that all the neighbors of each processor do not ' fail at the same time , the result is the same as the hypercube . we also give a related algorithm in o ( n ) time , and the length of the longest path ; third , we apply cluster faun tolerance introduced by q . - p
根据menger定理， n -维交叉立方体可以容纳n - 1个故障顶点，我们给出了它的时间复杂度为o ( n )的容错路由选择算法及其最长路径长度分析；在此基础上本文证明， n -维交叉立方体的条件连通度为2n - 2 ( n 2 ) ，并给出了相应时间复杂度为o ( n )的算法及其最长路径长度；除此之外，本文还证明当n -维交叉立方体中的故障簇个数不大于n - 1 ，其直径不大于1 ，故障顶点总数不超过2n - 3 ( n 2 )时，交叉立方体中任两个无故障顶点都至少有一条可靠路径。