POJ 2488 A Knight's Journey (DFS)

A Knight's Journey

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 30656		Accepted: 10498

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

Source

TUD Programming Contest 2005, Darmstadt, Germany

题目思路很清晰就是说如果能遍历整个图则打印路径，如果不行，则输出impossible

这道题按照字典序去搜索，想清楚行列关系再写，不然会WA

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<stdlib.h>
 4 #include<algorithm>
 5 using namespace std;
 6 const int MAXN=200+10;
 7 const int INF=0x3f3f3f3f;
 8 int n,m,flag;
 9 int vis[MAXN][MAXN],vx[MAXN],vy[MAXN];
10 int dir[8][2]={{-2,-1},{-2,1},{-1,-2},{-1,2},{1,-2},{1,2},{2,-1},{2,1}};//字典序
11 void DFS(int x,int y,int sum)
12 {
13     vx[sum]=x;//列，字母
14     vy[sum]=y;//行，数字
15     if(sum==n*m)
16     {
17         flag=1;
18         return ;
19     }
20     for(int i=0;i<8;i++)
21     {
22         int xx=x+dir[i][0];
23         int yy=y+dir[i][1];
24         if((1<=xx&&xx<=m)&&(1<=yy&&yy<=n)&&!vis[xx][yy]&&!flag)
25         {
26             vis[xx][yy]=1;
27             DFS(xx,yy,sum+1);
28             vis[xx][yy]=0;
29         }
30     }
31 }
32 int main()
33 {
34     //freopen("in.txt","r",stdin);
35     int kase,cnt=0;
36     scanf("%d",&kase);
37     while(kase--)
38     {
39         flag=0;
40         scanf("%d %d",&n,&m);
41         memset(vis,0,sizeof(vis));
42         memset(vx,0,sizeof(vx));
43         memset(vy,0,sizeof(vy));
44         vis[1][1]=1;
45         DFS(1,1,1);
46         printf("Scenario #%d:
",++cnt);
47         if(flag)
48         {
49             for(int i=1;i<=n*m;i++)
50                 printf("%c%d",'A'+vx[i]-1,vy[i]);
51             printf("
");
52         }
53         else
54             printf("impossible
");
55         printf("
");
56 
57     }
58     return 0;
59 }

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