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[Other综合实验哈夫曼编码

Description: 1.n个叶结点,权分别为w1,w2,• • • ,wn的二叉树中,带权路径长度WPL最小的二叉树叫最优二叉树, 也叫霍夫曼树。 2.霍夫曼算法:1)根据给定的权值{w1,w2,• • • ,wn}构造n个二叉树F={T1,T2,• • • ,Tn}每个Ti只有一个根结点,权为wi。2)在F中选取两棵根结点的权值最小的树 构成一棵新的二叉树,其根的权值为左右子树根的权值的和。3)F中删去这两棵树,加上新得的树。4)重复2)3)直到只剩一棵树。 -1. N leaf nodes, the right to w1, w2, # 8226 # 8226, wn the binary tree, with the right path length WPL smallest binary tree called Optimal binary tree, the tree is also called Hoffman. 2. Hoffman algorithm : 1) According to the weights (w1, w2, # 8226 # 8226 # 8226, wn) n binary tree structure F = (T1, T2, # 8226 # 8226 # 8226, Tn) each only a Ti Root, the right to wi. 2) In 1984 two F Root in the value of the right to the smallest tree constitute a new binary tree, the roots of the right value for the son around the roots and the value of the right. 3) F deleted these two trees, coupled with the new tree. 4) repeat 2) 3) until only a tree.
Platform: | Size: 1654 | Author: yy | Hits:

[Other resourcehuffmancode111

Description: huffman完整源代码C语言实现,有本人超级详细解释(看不懂你去跳楼吧) 算法设计: 1、对给定的n个权值{W1,W2,W3,...,Wi,...,Wn}构成n棵二叉树的初始集合F={T1,T2,T3,...,Ti,...,Tn},其中每棵二叉树Ti中只有一个权值为Wi的根结点,它的左右子树均为空。(为方便在计算机上实现算法,一般还要求以Ti的权值Wi的升序排列。) 2、在F中选取两棵根结点权值最小的树作为新构造的二叉树的左右子树,新二叉树的根结点的权值为其左右子树的根结点的权值之和。 3、从F中删除这两棵树,并把这棵新的二叉树同样以升序排列加入到集合F中。 4、重复二和三两步,直到集合F中只有一棵二叉树为止。 -Huffman complete C source code language, I have super-detailed explanation (you do not understand it jumped) algorithm design : one, the right to the right values n (W1, W2, W3 ,..., Wi ,..., Wn) n trees constitute the binary tree initial pool F = (T1, T2, T3, ... Ti ,..., Tn), which is indeed a binary tree Ti only a right to the value of Wi Root. it's about subtrees are empty. (For the convenience of the computer algorithm, the general also demanded the right to Ti Wi value of ascending.) 2. in 1984 two F Root weights smallest tree as a new structure of the binary tree around subtrees, new Binary Tree Root in the right value for their son around the tree Root and the right value. 3, F deleted from this two trees, and how the new binary tree in the same ascending into the pool F. 4, re
Platform: | Size: 11903 | Author: 乐乐 | Hits:

[Other综合实验哈夫曼编码

Description: 1.n个叶结点,权分别为w1,w2,• • • ,wn的二叉树中,带权路径长度WPL最小的二叉树叫最优二叉树, 也叫霍夫曼树。 2.霍夫曼算法:1)根据给定的权值{w1,w2,• • • ,wn}构造n个二叉树F={T1,T2,• • • ,Tn}每个Ti只有一个根结点,权为wi。2)在F中选取两棵根结点的权值最小的树 构成一棵新的二叉树,其根的权值为左右子树根的权值的和。3)F中删去这两棵树,加上新得的树。4)重复2)3)直到只剩一棵树。 -1. N leaf nodes, the right to w1, w2,# 8226# 8226, wn the binary tree, with the right path length WPL smallest binary tree called Optimal binary tree, the tree is also called Hoffman. 2. Hoffman algorithm : 1) According to the weights (w1, w2,# 8226# 8226# 8226, wn) n binary tree structure F = (T1, T2,# 8226# 8226# 8226, Tn) each only a Ti Root, the right to wi. 2) In 1984 two F Root in the value of the right to the smallest tree constitute a new binary tree, the roots of the right value for the son around the roots and the value of the right. 3) F deleted these two trees, coupled with the new tree. 4) repeat 2) 3) until only a tree.
Platform: | Size: 1024 | Author: yy | Hits:

[Data structshuffmancode111

Description: huffman完整源代码C语言实现,有本人超级详细解释(看不懂你去跳楼吧) 算法设计: 1、对给定的n个权值{W1,W2,W3,...,Wi,...,Wn}构成n棵二叉树的初始集合F={T1,T2,T3,...,Ti,...,Tn},其中每棵二叉树Ti中只有一个权值为Wi的根结点,它的左右子树均为空。(为方便在计算机上实现算法,一般还要求以Ti的权值Wi的升序排列。) 2、在F中选取两棵根结点权值最小的树作为新构造的二叉树的左右子树,新二叉树的根结点的权值为其左右子树的根结点的权值之和。 3、从F中删除这两棵树,并把这棵新的二叉树同样以升序排列加入到集合F中。 4、重复二和三两步,直到集合F中只有一棵二叉树为止。 -Huffman complete C source code language, I have super-detailed explanation (you do not understand it jumped) algorithm design : one, the right to the right values n (W1, W2, W3 ,..., Wi ,..., Wn) n trees constitute the binary tree initial pool F = (T1, T2, T3, ... Ti ,..., Tn), which is indeed a binary tree Ti only a right to the value of Wi Root. it's about subtrees are empty. (For the convenience of the computer algorithm, the general also demanded the right to Ti Wi value of ascending.) 2. in 1984 two F Root weights smallest tree as a new structure of the binary tree around subtrees, new Binary Tree Root in the right value for their son around the tree Root and the right value. 3, F deleted from this two trees, and how the new binary tree in the same ascending into the pool F. 4, re
Platform: | Size: 11264 | Author: 乐乐 | Hits:

[Software Engineering2008638621518

Description: 树是由n(n≥0)个结点组成的有限集合T。n=0的树称为空树;对n>0的树,有:(1)仅有一个特殊的结点称为根结点,根结点没有前驱结点;(2)当n>1时,除根结点外其余的结点分为m(m>0)个互不相交的有限集合T1,T2,…,Tm,其中每个集合Ti本身又是一棵结构和树类似的子树。 -Tree is n (n ≥ 0) consisting of nodes finite set T. n = 0 of the tree known as the empty tree for n> 0 of the tree, there is: (1) has only one special node called the root node, root node there is no precursor node (2) when n> 1 when, with the exception of root node except the node is divided into m (m> 0) a finite set of mutually exclusive cross-T1, T2, ..., Tm, one of each set Ti is itself a structure and are similar subtree tree .
Platform: | Size: 146432 | Author: 李南 | Hits:

[GUI Developtree7

Description: tree-7实现了树型的注释(tooltips),经典源码-tree-7 achieved a tree comments (tooltips), the classic source
Platform: | Size: 16384 | Author: women2009 | Hits:

[Data structsHuffman

Description: 用vc++实现用最短路径查找哈弗曼树。n个权值构成n棵二叉树的集合F={T1, T2, …, Tn},其中每棵二叉树Ti中只有一个带树为Ti的根结点。-Achieved with the vc++ to find the shortest path tree Havermann. n-weight constitutes a collection of n binary trees F = (T1, T2, ..., Tn), which is only a Ti binary Each species of tree with the root node for the Ti.
Platform: | Size: 1024 | Author: liangsi | Hits:

[Othertournament

Description: This assignment about data structure. This is write for linked-list and binary tree. This is useful for new-bie, who is begin ti learn about data structure.-This is assignment about data structure. This is write for linked-list and binary tree. This is useful for new-bie, who is begin ti learn about data structure.
Platform: | Size: 2048 | Author: hs_diego | Hits:

[Othertree3

Description: thies is a tree tree info file withw is ti veri slow form but enyqway fun
Platform: | Size: 20480 | Author: toma | Hits:

[Software EngineeringBBB

Description: This is the device tree for the beaglebone from the TI
Platform: | Size: 7168 | Author: harsha443 | Hits:

[Linux-Unixcvmx-l2t-defs

Description: Device tree bindings for TI s ADS7843, ADS7845, ADS7846, ADS7873, TSC2046 SPI driven touch screen controllers.
Platform: | Size: 2048 | Author: pfxueheng | Hits:

[Linux-Unixti-keystone-pllctrl

Description: Device tree bindings for Texas Instruments keystone pll controller.
Platform: | Size: 13312 | Author: rhcapd | Hits:

[Linux-Unixti-aemif

Description: Device tree bindings for Texas instruments AEMIF controller.
Platform: | Size: 7168 | Author: fhcggk | Hits:

[Linux-Unixti-keystone-devctrl

Description: Device tree bindings for Texas Instruments keystone device state control.
Platform: | Size: 2048 | Author: nqlonzen | Hits:

[Linux-Unixti-keystone-timer

Description: Device tree bindings for Texas instruments Keystone timer.
Platform: | Size: 2048 | Author: zfbacy | Hits:

[Linux-Unixclk-master

Description: Device tree bindings for Ethernet chip connected to TI GPMC.
Platform: | Size: 3072 | Author: minnikie | Hits:

[Othercnmb

Description: 在本实验中,程序的输入是一个表示树结构的广义表。假设树的根为 root ,其子树森林 F = ( T1 , T2 , ... , Tn ),设与该树对应的广义表为 L ,则 L =(原子,子表 1 ,子表 2 , ... ,子表 n ),其中原子对应 root ,子表 i ( 1<i< n )对应 Ti 。例如:广义表 (a,(b,(c),(d)),(f,(g),(h ),(i))) 表示的树如图所示: 程序的输出为树的层次结构、树的度以及各种度的结点个数。 在输出树的层次结构时,先输出根结点,然后依次输出各个子树,每个子树向里缩进 4 个空格,如:针对上图表示的树,输出的内容应为:-In this experiment, the input to the program is a table showing the generalized tree structure. Suppose root of the tree is root, its sub-tree forest F (T1, T2, ..., Tn), is set to L and the corresponding generalized tree table, L (atom, a child in Table 1, Table 2 child, ... child table n), in which the atoms correspond root, the child table i (1 <i< n )对应 Ti 。例如:广义表 (a,(b,(c),(d)),(f,(g),(h ),(i))) 表示的树如图所示: 程序的输出为树的层次结构、树的度以及各种度的结点个数。 在输出树的层次结构时,先输出根结点,然后依次输出各个子树,每个子树向里缩进 4 个空格,如:针对上图表示的树,输出的内容应为:
Platform: | Size: 1024 | Author: 王浩东 | Hits:

[Windows DevelopPATA1053

Description: 给定一个带有根R的非空树,并赋给每个树节点TI的权重WI。从R到L的路径的重量被定义为从R到任何叶节点L的路径上所有节点的权重之和。(Given a non-empty tree with root R, and with weight Wi assigned to each tree node Ti. The weight of a path from R to L is defined to be the sum of the weights of all the nodes along the path from R to any leaf node L. Now given any weighted tree, you are supposed to find all the paths with their weights equal to a given number. For example, let's consider the tree showed in Figure 1: for each node, the upper number is the node ID which is a two-digit number, and the lower number is the weight of that node. Suppose that the given number is 24, then there exists 4 different paths which have the same given weight: {10 5 2 7}, {10 4 10}, {10 3 3 6 2} and {10 3 3 6 2}, which correspond to the red edges in Figure 1.)
Platform: | Size: 3181568 | Author: lld | Hits:

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