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Description: 骑士遍历问题，在一个n*n个方格的国际象棋棋盘上，马（骑士）从任意指定方格出发，按照横1 步竖2 步，或横2 步竖1步的跳马规则，走遍棋盘的每一个格子，且每个格子只走1次。-Knight, in an n * n box at the international chess board, Ma (Knight) from the arbitrary designation box and, in accordance with a further horizontal two-step vertical, horizontal or a vertical two-step further vaulting horse rules traveled a checkerboard lattice of each and every one lattice just take time.
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Size: 1357 |
Author: 张三 |
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Description: 求解Knight Tour Problem，马的走法是“日”字形路线，例如当马在位置15的时候，它可以到达2、4、7、11、19、23、26和28。但是规定马是不能跳出棋盘外的，例如从位置1只能到达9和14
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Size: 966 |
Author: happydabu@126.com |
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Description: 一个经典的acm题-a classic that acm
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Size: 1024 |
Author: 李超 |
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Description: 著名的马周游问题的代码，可选择6×6，8×8，10×10大小的棋盘，找到结果后动态运行-on famous knight code. The chessboard with size 6*6,8*8,10*10 can be chosen.It will run automaticly after finding the result.
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Size: 241664 |
Author: zqm |
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Description: 本程序可模拟中国象棋中马步走法
对你输入的一个点快速遍步整个棋盘-this program can simulate Chinese Chess Knight law to which you input a point fast times throughout the chessboard step
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Size: 2048 |
Author: 钟毓秀 |
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Description: “八皇后”问题归朔法求解。八皇后问题是一个古老而著名的问题，该问题是十九世纪著名的数学家高斯1850年提出：在8X8格的国际象棋上摆放八个皇后，使其不能互相攻击，即任意两个皇后都不能处于同一行、同一列或同一斜线上，问有多少种摆法。 高斯认为有76种方案。1854年在柏林的象棋杂志上不同的作者发表了40种不同的解，后来有人用图论的方法解出92种结果。-"8 Queen" The problem Schomburg method. 8 Queen's problem is an ancient and famous, the problem is the famous 19th century mathematician Gauss 1850 : in the 8x8 grid placed on the international chess 8 Queen's, making it unable to attack each other. that is arbitrary two Queens are not at the same trip, the same series or in the same slash and asked how many pendulum. Gaussian that 76 species program. In 1854 in Berlin on different chess magazine published by the author of 40 different solutions, Later, someone using graph theory methods to come up with 92 kinds of results.
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Size: 162816 |
Author: yefeng |
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Description: 问题算法源代码：骑士遍历、万年历、N皇后问题回溯算法、动态计算网络最长最短路线、货郎担分枝限界图形演示、货郎担限界算法、矩阵乘法动态规划、网络最短路径Dijkstra算法-problems algorithm source code : Knight traversal, calendar, N Queens backtracking algorithms, Dynamic computing network longest shortest routes, traveling salesman Branch and Bound graphic demonstration, traveling salesman Bound algorithm, Matrix multiplication dynamic programming, network Dijkstra shortest path algorithm
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Size: 23552 |
Author: 安德里周 |
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Description: 这是一个骑士问题的程式，虽然作法不是很好～有点暴力～但是跑出来还OK！-This is a knight of the program, although the practice was not very good- a little violence- but also come out OK!
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Size: 1024 |
Author: 林建达 |
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Description: 如意令外挂原代码，能实现骑士挂机，攻击100％成功。-wishful so extrapolated original code, able to hook Knight, 100% successful attack.
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Size: 839680 |
Author: hwk |
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Description: Spring in Action 的代码-Spring in Action code
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Size: 900096 |
Author: 豆豆 |
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Description: 骑士漫游，最常见的递规问题，大家看看吧。呵呵，初次，请多见谅-Knight roaming, the most common delivery regulatory issues, we look at it. Oh, first, please excuse me
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Size: 1024 |
Author: 马硕 |
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Description: 骑士遍历问题，在一个n*n个方格的国际象棋棋盘上，马（骑士）从任意指定方格出发，按照横1 步竖2 步，或横2 步竖1步的跳马规则，走遍棋盘的每一个格子，且每个格子只走1次。-Knight, in an n* n box at the international chess board, Ma (Knight) from the arbitrary designation box and, in accordance with a further horizontal two-step vertical, horizontal or a vertical two-step further vaulting horse rules traveled a checkerboard lattice of each and every one lattice just take time.
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Size: 1024 |
Author: 张三 |
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Description: 解决骑士巡游(在国际象棋里，骑士能遍历所有格子的方法)的问题-Solve the knight parade (in chess, the knight can traverse all the lattice method) the problem of
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Size: 5120 |
Author: alone |
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Description: 骑士巡游问题的源代码，做算法课作业必备资料。-Knight tour problem' s source code, doing class work necessary information algorithm.
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Size: 263168 |
Author: xnhcx |
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Description: Java源代码案例--骑士巡游问题
展示了一个KT(Knight’s Tour)小程序， 用来演示一个限制版的骑士巡游问题。 骑士并不是从任何一个方格开始， 而是从角落上的四个方格之一开始。
附有详细的源码及讲解。-Java Source Code Case- knight tour problem shows a KT (Knight' s Tour) a small program used to demonstrate a restricted version of the knight tour problem. Knight does not start from any one box, but from one corner of the start of the four squares. The source code with detailed and explained.
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Size: 23552 |
Author: 孟涛 |
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Description: 骑士漫游的改进算法，回溯算法加贪心选择策略，n最大可以到128-Knight improved roaming algorithms, backtracking algorithms greedy selection strategy increases, n the greatest can go to 128
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Size: 929792 |
Author: 李珍 |
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Description: 在一个n*n个方格的国际象棋棋盘上，马（骑士）从任意指定方格出发，按照横1 步竖2 步，或横2 步竖1步的跳马规则，走遍棋盘的每一个格子，且每个格子只走1次。这样的跳马步骤称为1 个成功的骑士征途。例如，当n=5 时的1 个成功的骑士征途如下图所示。
1 2 3 4 5
1 25 14 1 8 19
2 4 9 18 13 2
3 15 24 3 20 7
4 10 5 22 17 12
5 23 16 11 6 21
算法设计：
对于给定的n和n*n方格的起始位置x和y。用分支限界法找出从指定的方格(x,y)出发的一条成功的骑士征途。
数据输入：
第一行有1 个正整数n (1≤n≤10)；第二行有2 个正整
数x 和y，表示骑士的起始位置为(x,y)。
结果输出:
如果不存在从(x,y)出发的成功的骑士征途则输出’No Solution!’。
输入：
5
1 3
输出
25 14 1 8 19
4 9 18 13 2
15 24 3 20 7
10 5 22 17 12
23 16 11 6 21-In an n* n squares on the chess board, horse (knight) starting from any given square, erected in accordance with step 1 step 2 horizontal, vertical or horizontal-step 1 step 2 vault rules, traveled to every board lattice, and each grid only take 1. This step is called a successful vault knights journey. For example, when n = 5 when a successful knight journey as shown below.
12345
125,141,819
24,918,132
315,243,207
4105221712
5231611621
Algorithm Design:
For a given n and n* n grid starting position of the x and y. Using branch and bound method to find out from the specified grid (x, y) starting a successful journey Knight.
Data entry:
The first line has a positive integer n (1 ≤ n ≤ 10) the second line has two positive integers
Number of x and y, said Cleveland s starting position (x, y).
The resulting output:
If not from the (x, y) the success of the Knights start the journey is the output No Solution! .
Input:
5
13
Export
25,141,819
4918132
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Size: 881664 |
Author: wakaka |
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Description: 骑士巡游问题:从国际象棋棋盘上任意给定的方格开始移动骑士，相继地到达所有的64个方格，进入每个方格一次且仅进入一次。回溯法实现-Knight Parade problem: the chess board from any given knight box began to move, have to reach all the 64 squares, each square once and only entered into once. Backtracking to achieve
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Size: 396288 |
Author: chenb |
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Description: source code knight online ,information: the cpp file server and client
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Size: 38781952 |
Author: toi la ai |
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Description: 在一个N*N的棋盘上指点任意一点的坐标，以马走日的方式走完整个棋盘且没有重复，试分别用DFS、BFS方法求解并输出全部可能的路径。(knight tour problem solved by the method of BFS and DFS.)
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Size: 43008 |
Author: 急速蜗牛 |
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