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

3

1 1

2 3

4 3

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:\n",++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("\n");

52         }

53         else

54             printf("impossible\n");

55         printf("\n");

56 

57     }

58     return 0;

59 }

View Code