POJ_2488——骑士遍历棋盘,字典序走法

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

 1 #include <cstdio>

 2 #include <iostream>

 3 #include <string>

 4 #include <cstring>

 5 //lexicographically first path

 6 //题目output中出现了以上几个单词

 7 //就是说骑士的移动步子是按照字典顺序来排列的

 8 //首先对列进行排序较小的在前面，然后按行进行排序也是较小的在前面 

 9 //我构造的xy坐标是按照SDL中的图形坐标，x往下递增，y往右递增

10 //转换成数学上的xy坐标就要先对列x做排序然后再对行y做排序 

11 int dirx[8]={-2,-2,-1,-1,1,1,2,2};

12 int diry[8]={-1,1,-2,2,-2,2,-1,1};

13 

14 int T,p,q,mark[27][27];

15 std::string way;

16 bool DFS(int x,int y,int step)

17 {

18    if(step==p*q)

19       {

20          return true;

21       }

22    for(int i=0;i<8;i++)

23       {

24          int dx=x+dirx[i];

25          int dy=y+diry[i];

26          if((dx>=1 && dx<=q) && (dy>=1 && dy<=p) && mark[dx][dy]!=1)

27             {

28                mark[dx][dy]=1;

29                //当找到路径才会插入操作，所以在main函数循环中不需要清空string 

30                if(DFS(dx,dy,step+1))

31                   {

32                      //在第0个位置插入1个字符

33                      //先插入数字然后再插入字母，否则顺序颠倒 

34                      way.insert(0,1,dy+'0');

35                      way.insert(0,1,dx+'A'-1);

36                      return true;   

37                   }

38                mark[dx][dy]=0;

39             }

40       }

41    return false;

42 }

43 int main()

44 {

45    scanf("%d",&T);

46    for(int i=1;i<=T;i++)

47       {

48          scanf("%d%d",&p,&q);

49          way.clear();//每次都要清空string

50          bool find=false;

51          for(int x=1;x<=q && !find;x++)

52             {

53                for(int y=1;y<=p;y++)

54                   {

55                      memset(mark,0,sizeof(mark));

56                      mark[x][y]=1;

57                      if(DFS(x,y,1))

58                         {

59                            //找到的话插入起始位置 

60                            way.insert(0,1,y+'0');

61                            way.insert(0,1,x+'A'-1);

62                            find=true;

63                            break;

64                         }

65                   }

66             }

67          std::cout<<"Scenario #"<<i<<":"<<std::endl;

68          if(find)

69             {

70                //整体输出就行

71                std::cout<<way<<std::endl<<std::endl;

72             }

73          else

74             printf("impossible\n\n");

75       }

76    return 0;   

77 }