Poj 2993 Emag eht htiw Em Pleh

1.Link：

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:

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