Poj 2993 Emag eht htiw Em Pleh

1.Link：

http://poj.org/problem?id=2993

2.Content：

Emag eht htiw Em Pleh

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2801		Accepted: 1861

Description

This problem is a reverse case of the  problem 2996. You are given the output of the problem H and your task is to find the corresponding input.

Input

according to output of  problem 2996.

Output

according to input of  problem 2996.

Sample Input

White: Ke1,Qd1,Ra1,Rh1,Bc1,Bf1,Nb1,a2,c2,d2,f2,g2,h2,a3,e4

Black: Ke8,Qd8,Ra8,Rh8,Bc8,Ng8,Nc6,a7,b7,c7,d7,e7,f7,h7,h6

Sample Output

+---+---+---+---+---+---+---+---+

|.r.|:::|.b.|:q:|.k.|:::|.n.|:r:|

+---+---+---+---+---+---+---+---+

|:p:|.p.|:p:|.p.|:p:|.p.|:::|.p.|

+---+---+---+---+---+---+---+---+

|...|:::|.n.|:::|...|:::|...|:p:|

+---+---+---+---+---+---+---+---+

|:::|...|:::|...|:::|...|:::|...|

+---+---+---+---+---+---+---+---+

|...|:::|...|:::|.P.|:::|...|:::|

+---+---+---+---+---+---+---+---+

|:P:|...|:::|...|:::|...|:::|...|

+---+---+---+---+---+---+---+---+

|.P.|:::|.P.|:P:|...|:P:|.P.|:P:|

+---+---+---+---+---+---+---+---+

|:R:|.N.|:B:|.Q.|:K:|.B.|:::|.R.|

+---+---+---+---+---+---+---+---+

Source

CTU Open 2005

3.Method:

模拟题，我的方法是想把棋盘初始化好，然后再往里面添棋子

由于很简单，就没有把white和black统一为一种情况，代码不怎么精简

不过对于水题，已经足够了

4.Code:

 1 #include <iostream>

 2 #include <string>

 3 

 4 using namespace std;

 5 

 6 int main()

 7 {

 8     //freopen("D://input.txt","r",stdin);

 9 

10     int i,j;

11 

12     //init output

13     string str_out[8 + 9];

14     str_out[0] = "+---+---+---+---+---+---+---+---+";

15     for(i = 0; i < 8; ++i)

16     {

17         str_out[i * 2 + 1] = "|";

18         for(j = 0; j < 8; ++j)

19         {

20             if((i + j) % 2 == 0) str_out[i * 2 + 1] += "...";

21             else str_out[i * 2 + 1] += ":::";

22             str_out[i * 2 + 1] += "|";

23         }

24         str_out[(i + 1) * 2] = "+---+---+---+---+---+---+---+---+";

25     }

26 

27 

28     string str;

29 

30     //white:

31     cin >> str;

32     cin >> str;

33 

34     str = "," + str;

35 

36     //cout << str << endl;

37     int arr_ch[] = {'K','Q','R','B','N'};

38 

39     string::size_type str_i;

40     for(str_i = 0; str_i != str.size(); ++str_i)

41     {

42         if(str[str_i] == ',')

43         {

44             for(i = 0; i < 5; ++i) if(str[str_i + 1] == arr_ch[i]) break;

45             if(i < 5)

46             {

47                 str_out[(8 - (str[str_i + 3] - '0')) * 2 + 1][2 + 4 * (str[str_i + 2] - 'a')] = arr_ch[i];

48             }

49             else

50             {

51                 str_out[(8 - (str[str_i + 2] - '0')) * 2 + 1][2 + 4 * (str[str_i + 1] - 'a')] = 'P';

52             }        

53         }

54     }

55 

56     //black:

57     cin >> str;

58     cin >> str;

59     str = "," + str;

60 

61     //cout << str << endl;

62 

63     for(str_i = 0; str_i != str.size(); ++str_i)

64     {

65         if(str[str_i] == ',')

66         {

67             for(i = 0; i < 5; ++i) if(str[str_i + 1] == arr_ch[i]) break;

68             if(i < 5)

69             {

70                 str_out[(8 - (str[str_i + 3] - '0')) * 2 + 1][2 + 4 * (str[str_i + 2] - 'a')] = arr_ch[i] + ('a' - 'A');

71             }

72             else

73             {

74                 str_out[(8 - (str[str_i + 2] - '0')) * 2 + 1][2 + 4 * (str[str_i + 1] - 'a')] = 'p';

75             }        

76         }

77     }

78 

79     

80     for(i = 0; i < 17; ++i) cout << str_out[i] << endl;

81 

82     //fclose(stdin);

83 

84     return 0;

85 }

 

5.Reference: