Location:
Search - b tree
Search list
Description: An Object Explorer Written in a Map/Filter style with C# Partial Application Functions
Recently I wanted to visually inspect the structure of an object created by some service, so I wrote a function that reflects over an object and builds a tree of the various class properties and drills into List<T> collections. Simple enough, but I thought, gee, this might be useful, I'd like to expand it to handle Dictionary<A, B>, enums, nullable property types, etc. The main recursive function ended up being a 130 line long sort of organized mess. Thinking about refactoring it into more manageable pieces, it occurred to me that all this code was basically an imperative implementation of map and filter operations.
Platform: |
Size: 925696 |
Author: xps1530
|
Hits:
Description: ip2region是准确率99.9%的ip地址定位库，0.0x毫秒级查询，数据库文件大小只有2.7M，提供了java, php, c, python,nodejs,golang查询绑定和Binary,B树,内存三种查询算法，妈妈再也不同担心我的ip地址定位！(Ip2region is a IP library address location accuracy of 99.9% millisecond, 0.0x query, database file size is only 2.7M, PHP, C provides Java, python, nodejs, golang and Binary B binding query, tree, three kinds of query algorithm of memory, my mother no longer worry about my IP address location!)
Platform: |
Size: 3136512 |
Author: Chunhui1012
|
Hits:
Description: 给定一颗有权无根树，修改任意一条边，求最少的任意两点间的距离和:修改一条边是指删除树中的一条边，然后选择两个端点连线，形成一课新的树，并且不改变这条边的权值，在这棵树的基础上，求最少的任意两点间的距离和是多少
任意两点的距离和指：sum(a到b的距离，a<b,a和b都是树上的点)
输入：第一行是整数n，接下来n-1行，每行包括3个整数ai,bi,wi,表示节点ai与bi之间有一条权重为wi的边。
输出：最小的任意两点间的距离和
(Given a right to unrooted tree, modify any one side (you can not modify), the number of required minimum distance between any two points and is? Modify an edge means an edge to delete the tree, then the two endpoints of the connection, the formation of a new class of trees, and does not change the weight of this edge, in this tree, based on the minimum requirements of any two the distance between the points and how much distance between any two points and means: sum (a b distance, a <b,a和b都是树上的点)
输入：第一行是整数n，接下来n-1行，每行包括3个整数ai,bi,wi,表示节点ai与bi之间有一条权重为wi的边。
输出：最小的任意两点间的距离和)
Platform: |
Size: 1024 |
Author: qzm |
Hits:
Description: 建立树的孩子兄弟链表存储结构，求树的深度。
要求：
1）自上而下输入树中每条边（例如#A,AB,AC,BD,##），不要转换成二叉树输入。
2）按树中由根至叶子结点层次遍历顺序（每层中自左至右输入），输入结点序列及每个结点的度，构造孩子兄弟链表。(Establish the storage structure of the tree's child and brother list, and find the depth of the tree.
Requirement:
1) Input each edge in the tree from top to bottom (for example, a, AB, AC, BD, b). Do not convert it to binary tree input.
2) According to the traversal order from root to leaf node in the tree (input from left to right in each layer), input the node sequence and the degree of each node, and construct the child brother list.)
Platform: |
Size: 1536000 |
Author: lmt19990807 |
Hits:
«
1
2
...
26
27
28
29
30
31»