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xianduan
输入俩条线段的首尾坐标,用于判断俩条线段是否相交的(Used to determine whether the intersection of two line segments)
- 2011-05-22 22:47:16下载
- 积分:1
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11组表达式求值
设计一个表达式求值的程序。该程序必须可以接受包含(,),+,-,*,/的中缀表达式,并求出结果。如果表达式正确,则输出表达式的结果;如果表达式非法,则不输出或输出其他字符。(Design an expression evaluation procedures. The program must accept (,), including +, -, *, / the infix expression, and the results obtained. If the expression is correct, the output of the expression is output; if the expression is illegal, no other characters are output or exported.)
- 2017-06-14 15:06:29下载
- 积分:1
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function
提供字符串方法,大家面试可以用,包括找出对称字串(Some string function)
- 2011-05-14 23:14:49下载
- 积分:1
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class
给定m个n维向量a , a , ,am 1 2 ,向量分类问题要求将相同的向量划分为同一类。试用
抽象数据类型表设计解向量分类问题的有效算法。(Given m a n-dimensional vector a, a,, am 1 2 )
- 2008-05-05 22:03:41下载
- 积分:1
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order
通过冒泡排序法和选择排序法实现对若干数字的排序。(To achieve the order of several numbers)
- 2014-06-08 18:35:10下载
- 积分:1
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SeqQueue
数据结构循环队列的实现 在visual c++6.0运行(Circular queue data structure to achieve the visual c++6.0 run)
- 2013-06-15 16:27:32下载
- 积分:1
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K_Merge
文件的多路归并排序算法, C++描述(要求待排序文件已经事先有序)(C++ implementation of K-Merge Algorithm)
- 2010-08-03 10:24:48下载
- 积分:1
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delete
数据结构 实验练习 删除相同的节点 希望会对大家有所帮助(Data structures lab exercises to remove the same node want to be helpful to everyone)
- 2012-11-14 12:59:15下载
- 积分:1
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ListOrder
链表,输入一定的数据,通过链表进行储存,并排序后输出。(List。Input numbers,then output in order~~)
- 2011-11-24 11:02:04下载
- 积分:1
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Figure-topological-sort
说明: 拓扑排序
对一个 有向无环图 G进行拓扑排序,是将G中所有顶点排成一个线性序列,使得图中任
意一对顶点u和v,若u,v ∈E,则u在线性序列中出现在v之前。
通常,这样的线性序列称为满足拓扑次序的序列,简称 拓扑序列 。
注意:
①若将图中顶点按拓扑次序排成一行,则图中所有的有向边均是从左指向右的。
②若图中存在有向环,则不可能使顶点满足拓扑次序。
③一个DAG的拓扑序列通常表示某种方案切实可行。 (Topological sort
In a directed acyclic graph G topological sort, is all vertices will G formed a linear sequence, make diagram post
Meaning a pair of vertex u and v, if u, v ∈ E, then u in linear sequence in which they appear in v before.
Usually, such linear sequence sequence of topological order called meet, abbreviation topology sequence.
Note:
(1) if the diagram according to the topology order line vertices, then figure all from left to right edge are pointing.
(2) if the figure to exist in the ring, is impossible to have the vertex meet topological sequence.
(3) a DAG topological sequence usually expressed some scheme is feasible.
)
- 2011-03-22 17:19:14下载
- 积分:1