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迷宫问题 队列求解 界面还不完善 可以自己添加(Maze problem solving interface queue is not perfect you can add your own)

/*快速排序：以待排序元素中的第一个元素作为基准将带排序的元素分为三组 *大于基准元素为一组置于基准值的右边，小于基准值的元素为一组置于基准值的左边， *基准值为一组，这样就找到了基准元素的位置就确定了 *如此递归调用。直到每一组的low==high，即实现了排序的要求。 */ (/* Rapid Sort: the first element to be sorted in the element as a basis with with sorting the elements are divided into three groups* is greater than the reference element of a group is placed on the right side of the reference value, smaller than the reference value elements for a group of placed on the left side of the reference value,* reference value, a group, so that the location of the reference element is found it is determined* so recursive call. Until each group of low == high to achieve the requirements of the sort.* /)

双链表的定义和插入算法，双链表的定义，删除，插入和查找算法(Doubly linked list of definitions and insertion algorithm, the definition of double-linked list, delete, insert, and search algorithm)

使用全排列实现hash，看看吧相信对大家会有些帮助的(Full array hash)

本程序实现一元多项式合并。本程序输入方式为按项输入，每输入一个项，会先提示输入项的幂，再提示输入项的系数。(This procedure combined to achieve a dollar polynomial. This procedure according to entry input methods for the input, one for each input item, it will first prompt entry of the power, and then prompts entry of the coefficient.)

约瑟夫环（约瑟夫问题）是一个数学的应用问题：已知n个人（以编号1，2，3...n分别表示）围坐在一张圆桌周围。从编号为k的人开始报数，数到m的那个人出列；他的下一个人又从1开始报数，数到m的那个人又出列；依此规律重复下去，直到圆桌周围的人全部出列。通常解决这类问题时我们把编号从0~n-1，最后结果+1即为原问题的解。 (Josephus (Josephus problem) is the application of a mathematical problem: Given n individuals (with numbers 1,2,3 ... n respectively) sitting around a round table. From the number of people gettin k, number of the m man out of the line he s the next person and a number of gettin number to m the man was out of the line and so the law is repeated until the round table were all out of the column. We numbered 0 ~ n-1, the final result is the original problem solution+1 usually solve these problems.)

利用栈完成算术表达式求值：从键盘或文件中输入算术表达式，计算其结果并显示。要求处理过程为：（1）转换为后缀表达式并输出；（2）对后缀表达式求值并输出。 输入的表达式中可以有整数、实数、括号，运算符包括+、-、*、/、#（代表单目负）。可以多次输入不同的表达式进行计算，直到用户选择“退出”。栈的基本操作可以自己实现，也可以使用系统提供的STL实现。 (Done using an arithmetic expression evaluation stack: file input from the keyboard or an arithmetic expression, calculate and display the results. Requirements process: (1) to postfix expression and output (2) evaluation of the postfix expression and output. Enter the expression can be integers, real numbers, parentheses, operators including+,-,*,/,#( representatives unary minus). You can enter several different expressions are evaluated until the user selects " Exit." Stack can implement the basic operations, you can also use the system implementation of the STL.)

最小生成树算法 克鲁斯法尔算法 球最佳路径(kulru method class maiun void cx )

数据结构课程作业，查找算法。用哈希算法下实行查找(Data structure course work, search algorithm. Under the hash algorithm used to find)

参加运动会的 个学校编号为 。比赛分成 个男子项目和 个女子项目，项目编号分别为 和 。由于各项目参加人数差别较大，有些项目取前五名，得分顺序为7，5，3，2，1；还有些项目只取前三名，得分顺序为5，3，2。写一个统计程序产生各种成绩单和得分报表。 2、要求产生各学校的成绩单，内容包括各校所取得的每项成绩的项目号、名次（成绩）、姓名和得分；产生团体总分报表，内容包括校号、男子团体总分、女子团体总分和团体总分。 3、测试数据：对于 ， ， ，编号为奇数的项目取前五名，编号为偶数的项目取前三名，设计一组实例数据。 (The school serial number participating in Games is. Competition divides into men s event and women s event , the project serial number parts for the sum. The difference is bigger since every project participates in a number , some projects choose the first five , score order is 7 , 5 , 3 , 2 , 1 Still have some of projects taking the first three places , score only being 5 , 3 , 2 in proper order. Write a form for report counting procedure producing the various school report card and score. 2, demands the school report card producing every school , content includes every achievement project number , position in a name list (achievement) , full name and score per got by school Produce the group total points form for report , content including the school number , male person group total points , woman group total points and group total points. 3, testing data: Be that the odd number project chooses the first five to the serial number, the serial number is that the even number project c)