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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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astar
A*算法 1、将开始节点放入开放列表(开始节点的F和G值都视为0)
2、重复以下步骤:
在开放列表中查找具有最小F值的节点,并把查找到的节点作为当前节点
把当前节点从开放列表删除, 加入到封闭列表.
(A* algorithm 1, will begin to node placed in the and opening up list of (the began to node of the F and G values are regarded as 0) 2, repeat the the following steps: to Find a the has a the the smallest F value of the node in the the and opening up list of, and put Find a to the node as the current node current node is removed from the open list, added to the closed list.)
- 2013-04-04 11:06:42下载
- 积分：1
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Monkeys-and-peach
Monkeys and peach,Monkeys and peach(Monkeys and peach)
- 2013-09-01 22:06:42下载
- 积分：1
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1013_digital_root
数字根算法，用最简单的代码计算数字根。如果把一个大数的各位数字相加得到一个和，再把这个和的各位数字相加又得一个和，再继续作数字和，直到最后的数字和是个位数为止，这最后的数称为最初那个数的“数字根”。(Digital root algorithm, with the most simple code to calculate the digital root. If you figure in a large numbers is the sum of a, then this and you figure the sum is again a, and then continue for digital and until the final figure and the single digits, the last number called initially that the number of "digital root".)
- 2012-04-27 10:07:18下载
- 积分：1
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point_to_line
计算点到直线的距离．输入两点坐标确定一个直线，再输入一个点的坐标，计算该点到直线的距离．(Calculated point to the straight line distance. Enter the coordinates of two points determine a straight line, and then input the coordinates of a point, to calculate the straight-line distance between points.)
- 2020-11-30 15:09:27下载
- 积分：1
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maze
迷宫求解一般采用“穷举法”，逐一沿顺时针方向查找相邻块（一共四块－东(右)、南(下)，西(左)、北(上)）是否可通，即该相邻块既是通道块，且不在当前路径上。用一个栈来记录已走过的路径栈是限定仅在表尾(top)进行插入或删除操作的线性表。(Maze solving generally use the " exhaustive" one by one to find adjacent blocks in a clockwise direction (a total of four- East (right), South (down), West (left), North (on)) whether to pass, that the both channel blocks adjacent blocks, and not in the current path. Use a stack to record the path traversed stack is limited only in the tail (top) to insert or delete a linear form.)
- 2013-12-13 17:55:45下载
- 积分：1
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ds3
单向链表的创建与操作
设单向链表中节点的数据域的数据类型为整型，编写函数实现以下操作：
（1）实现单向链表的创建（包括初始化）与输出操作，节点的个数及节点的数据由用户输入。
（源代码：ds3-1.c）
（2）查找给定的单链表中的第i个节点，并将其地址返回。若不存在第i个节点，则返回空地址。
（源代码：ds3-2.c）
（3）查找给定的单链表中值为n的节点，并将其地址返回。若不存在值为n的节点，则返回空地址。同时，还应通过参数传回该节点的序号。
（源代码：ds3-3.c）
（4）删除给定的单链表中的第i个节点，成功返回1，失败返回0。
（源代码：ds3-4.c）
（5）删除给定的单链表中值为n的节点，成功返回1，失败返回0。
（源代码：ds3-5.c）
（6）在给定的单链表的第i位上插入值为n的节点。
（源代码：ds3-6.c）
（7）在给定单链表的值为m的节点的前面插入一个值为n的节点。
（源代码：ds3-7.c）
(Creation and operation of a one-way linked list
Set up a one-way linked list data type node integer data fields , write a function to achieve the following:
( 1 ) achieve the creation of a one-way linked list ( including initialization ) and output operation , the number of nodes and node data entered by the user .
( Source : ds3-1.c)
( 2 ) Find a single list given in the i-th node and returns its address . Without the presence of the i-th node , returns an empty address.
( Source : ds3-2.c)
( 3 ) Find a given node in a given value of n single list , and return address . Without the presence of the value of n nodes , returns an empty address. Meanwhile, the number should be returned by the parameters of the node .
( Source : ds3-3.c)
( 4 ) Delete the given singly linked list in the i-th node , the successful return 1, else return 0 .
( Source : ds3-4.c)
( 5 ) to delete a single node in the list is given n , the successful return 1, else return 0 .
( Source : ds3-5.c)
( 6 ) )
- 2014-05-11 19:19:41下载
- 积分：1
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baiji
百鸡算法是一种简单的逻辑算法，通过不同的鸡不同的价钱，得出大鸡小鸡的个数(Typical code for white chicken algorithm)
- 2012-06-02 16:31:30下载
- 积分：1
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Birch
这是birch C语言的源代码。里面有详细的代码注释方便学习birch算法(birch c)
- 2012-05-08 09:00:44下载
- 积分：1
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box-nesting
最长d维箱嵌套问题的贪心算法，
采用贪心算法不能得到整体最优解,
但其最终结果也可以是最优解的很好的近似(The Greedy Algorithm of the Longest Nesting of d一Dimenson Boxes)
- 2012-06-11 16:47:23下载
- 积分：1
