图论中得kruskal 算法 求解最小生成树 算法为图论中得经典算法...

1． 问题描述设计一个国际象棋的马踏遍棋盘的演示程序。2． 基本要求将马随机放在国际象棋8*8棋盘的某个方格中，马按走棋规则进行移动。编制非递归程序，求出马的行走路线，将数字1，2，3，4，…，64依次填入一个8*8方阵。-1. Problem description design of a chess board Ma travels in the demonstration program. 2. Ma basic requirements on random chess 8* 8 chessboard in a box, according to Ma go mobile chess rules. Preparation of non- recursive procedure, and seeking stepped routes, the number 1,2,3,4 ..., 64 were filled an 8* 8 matrix.

一个用C写的迷宫小游戏，旨在对C中的指针及数据结构中的单链表和栈的使用的训练。-Using C to write a maze game, designed for C of the pointer and data structure of a single linked list and stack the use of training.

这是一个简单的动态现实的约瑟夫环 用于排除制定圈内的某个人-this is a simple system for yuesefu loop

用C实现的基于O-tree的ic设计布图布线工具源代码,有相关文档-C realized based on O-tree-id wiring layout design tools source code, related documentation

以一个Maze(m, n)的长方阵表示迷宫, 迷宫的入口是Maze(1,1), 出口是Maze(m, n), Maze(i,j)等于0表示通路, Maze(i,j)等于1表示障碍。设计一个程序，对任意设定的迷宫，求出一条从入口到出口的通路，或得出没有通路的结论。各个命令以及相关数据的输入格式如下： 输入迷宫：C，接下来一行是m n，其中m是迷宫矩阵行数，n是迷宫矩阵列数，接下来的m行数据是迷宫矩阵，每行n个值，表示一个m*n的迷宫矩阵 求迷宫的一条通路：Q 当输入的命令为E时，程序结束-With a Maze (m, n), said the long square maze, the maze entrance is a Maze (1,1), exports is the Maze (m, n), Maze (i, j) is equal to 0, said channel, Maze (i, j) is equal to 1, said barrier. Designing a program, for any set of the maze, find a path from the entrance to the export, or come to the conclusion there is no path. The various commands and data input format is as follows: Enter the Labyrinth: C, followed by his party mn, where m is the maze of matrix rows, n is a maze of matrix columns, the next line of data is a maze m matrix, each row n- value, indicating an m* n matrix of the labyrinth a maze of pathways: Q when the input command for the E, the end of the process

prim算法的原理利用 prim算法构造最小生成树。有机的应用prim和数组存储生成树。-prim algorithm using the principle of Minimum Spanning Tree Algorithm prim. Application of organic prim and storage arrays spanning tree.

VB6 的小程序，接收来自 YHT3 规模的重量，保存日期来访问数据库。

多种排序算法（冒泡排序，插入排序，堆排序等等）效率分析-Efficiency analysis of a variety of sorting algorithms

可以实现两个大整数（不限数位）相加减！ 在C＋＋中，加减法的两个数必须在系统规定内，本程序突破了这一界限！-can achieve two large integers (Open-digit) is modified! In C++, and subtract the number two must be within the system, the process broke the limits!

informix错误信息-informix wrong message