HDU1372:Knight Moves(BFS) 解题心得

原题：

Description

 

A friend of you is doing research on the Traveling Knight Problem (TKP) where you are to find the shortest closed tour of knight moves that visits each square of a given set of n squares on a chessboard exactly once. He thinks that the most difficult part of the problem is determining the smallest number of knight moves between two given squares and that, once you have accomplished this, finding the tour would be easy.

Of course you know that it is vice versa. So you offer him to write a program that solves the "difficult" part.

 

Your job is to write a program that takes two squares a and b as input and then determines the number of knight moves on a shortest route from a to b.

 

Input Specification

The input file will contain one or more test cases. Each test case consists of one line containing two squares separated by one space. A square is a string consisting of a letter (a-h) representing the column and a digit (1-8) representing the row on the chessboard.

 

Output Specification

For each test case, print one line saying "To get from xx to yy takes n knight moves.".

 

Sample Input

 

e2 e4

a1 b2

b2 c3

a1 h8

a1 h7

h8 a1

b1 c3

f6 f6

 

Sample Output

 

To get from e2 to e4 takes 2 knight moves.

To get from a1 to b2 takes 4 knight moves.

To get from b2 to c3 takes 2 knight moves.

To get from a1 to h8 takes 6 knight moves.

To get from a1 to h7 takes 5 knight moves.

To get from h8 to a1 takes 6 knight moves.

To get from b1 to c3 takes 1 knight moves.

To get from f6 to f6 takes 0 knight moves.

 

 

 我的代码 ：

 

 1 #include<iostream>

 2 #include<cstdio>

 3 #include<cstring>

 4 #include<string.h>

 5 #include<queue>

 6 using namespace std;

 7 

 8 

 9 struct point

10 {

11     int x;

12     int y;

13     int step;

14     point ()

15     {

16         x = 0;

17         y = 0;

18         step = 0;

19     }

20 

21     bool operator ==(point &b)

22     {

23         return (x == b.x&&y == b.y);

24     }

25 }first,target;

26 

27 int times[8][8];

28 

29 int directionX[8] = { 1, 2, 2, 1, -1, -2, -2, -1 };

30 int directionY[8] = { 2, 1, -1, -2, -2, -1, 1, 2 };

31 

32 

33 bool check(point &p)

34 {

35     if (p.x<0 || p.y<0 || p.x>7 || p.y>7 )

36     {

37         return 0;

38     }

39     if (times[p.x][p.y] != 0)

40     {

41         return 0;

42     }

43     return 1;

44 }

45 

46 int bfs(point & first)

47 {

48     queue<point> q;

49     times[first.x][first.y]++;

50     q.push(first);                //因为没有到终点，所以加入搜索队列

51     while(!q.empty())

52     {

53         point cur = q.front();                //cur  是当前的对象

54         q.pop(); 

55         point temp;                        //构建当前对象副本

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

57         {

58             //进行变换

59             temp.x = cur.x + directionX[i];

60             temp.y = cur.y + directionY[i];

61             //变换结束，检查

62             if (!check(temp))    continue;

63             if (temp == target)        return cur.step+1;

64             times[temp.x][temp.y]++;

65             temp.step = cur.step + 1;

66             q.push(temp);

67         }

68     }

69     return -1;

70     

71     

72 }

73 

74 int main()

75 {

76     char x1, x2, y1, y2; 

77     int min;

78     while (cin >> x1>>y1)

79     {

80         cin >> x2 >> y2;

81         first.x = x1 - 'a';

82         first.y = y1-'1';

83         target.x = x2 - 'a';

84         target.y = y2-'1';

85         if (first == target)

86         {

87             min = 0;

88         }

89         else

90         {

91             memset(times, 0, sizeof(times));

92             min = bfs(first);

93         }

94         printf("To get from %c%c to %c%c takes %d knight moves.\n",x1, y1,x2, y2, min);

95     }

96 

97     return 0;

98 }