poj 2488 A Knight's Journey

题意转载于：優YoU http://user.qzone.qq.com/289065406/blog/1303350143

大致题意:

给出一个国际棋盘的大小，判断马能否不重复的走过所有格，并记录下其中按字典序排列的第一种路径。

解题思路：

难度不大，但要注意的地方有3点：

1、题目要求以"lexicographically"方式输出，也就是字典序...要以字典序输出路径，那么搜索的方向（我的程序是path()函数）就要以特殊的顺序排列了...这样只要每次从dfs(A,1)开始搜索，第一个成功遍历的路径一定是以字典序排列...

下图是搜索的次序，马的位置为当前位置，序号格为测试下一步的位置的测试先后顺序

按这个顺序测试，那么第一次成功周游的顺序就是字典序

2、国际象棋的棋盘，行为数字p；列为字母q

大神说这是水题一道，可对于本菜鸟来说还是挺难的，纠结了半天，还是参考的各大神的解题报告，主要是对Dfs理解不够，革命道路依旧漫长啊！！

这是题目：

A Knight's Journey

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 26044		Accepted: 8888

Description

Background The knight is getting bored of seeing the same black and white squares again and again and has decided to make a journey around the world. Whenever a knight moves, it is two squares in one direction and one square perpendicular to this. The world of a knight is the chessboard he is living on. Our knight lives on a chessboard that has a smaller area than a regular 8 * 8 board, but it is still rectangular. Can you help this adventurous knight to make travel plans?
Problem Find a path such that the knight visits every square once. The knight can start and end on any square of the board.

Input

The input begins with a positive integer n in the first line. The following lines contain n test cases. Each test case consists of a single line with two positive integers p and q, such that 1 <= p * q <= 26. This represents a p * q chessboard, where p describes how many different square numbers 1, . . . , p exist, q describes how many different square letters exist. These are the first q letters of the Latin alphabet: A, . . .

Output

The output for every scenario begins with a line containing "Scenario #i:", where i is the number of the scenario starting at 1. Then print a single line containing the lexicographically first path that visits all squares of the chessboard with knight moves followed by an empty line. The path should be given on a single line by concatenating the names of the visited squares. Each square name consists of a capital letter followed by a number.
If no such path exist, you should output impossible on a single line.

Sample Input

Sample Output

Scenario #1:
A1

Scenario #2:
impossible

Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4

经过大神“指点”之后的代码：

 1 #include<stdio.h>
 2 #include<string.h>
 3 #include<stdlib.h>
 4 int t,n,m;
 5 int s[26][26];//标记数组
 6 char p[60];
 7 static int dis[8][2]={-2, -1, -2, 1, -1, -2, -1, 2, 1, -2, 1, 2, 2, -1, 2, 1};//方向
 8 int dfs(int x,int y,int mark)
 9 {
10     if(mark==n*m)  return 1;
11     int x1,y1;
12     for(int i=0;i<8;++i)
13     {
14         x1=x+dis[i][0];
15         y1=y+dis[i][1];
16         if(x1>=0&&x1<m&&y1>=0&&y1<n&&s[y1][x1]==0)
17         {
18             s[y1][x1]=1;
19             p[(mark<<1)]=x1+'A';//记录列
20             p[(mark<<1)+1]=y1+'1';//记录行
21             if(dfs(x1,y1,mark+1))
22                  return 1;
23             s[y1][x1]=0;
24         }
25     }
26     return 0;
27 }
28 
29 int main()
30 {
31 
32     scanf("%d",&t);
33     for(int i=1;i<=t;i++)
34     {
35         scanf("%d %d",&n,&m);
36         memset(s,0,sizeof(s));
37         memset(p,0,sizeof(p));
38         s[0][0]=1;
39         p[0]='A';
40         p[1]='1';
41         if(dfs(0,0,1))
42         {
43             printf("Scenario #%d:
",i);
44             for(int j=0;j<strlen(p);j++)
45                 printf("%c",p[j]);
46             printf("

");
47         }
48         else
49         {
50              printf("Scenario #%d:
impossible

",i);
51         }
52     }
53     return 0;
54 }

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