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 相关搜索： knight
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