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:

 #include <iostream>
 #include <string>

 using namespace std;

 int main()
 {
     //freopen("D://input.txt","r",stdin);

     int i,j;

     //init output
     string str_out[ + ];
     str_out[] = "+---+---+---+---+---+---+---+---+";
     for(i = ; i < ; ++i)
     {
         str_out[i *  + ] = "|";
         for(j = ; j < ; ++j)
         {
             if((i + j) %  == ) str_out[i *  + ] += "...";
             else str_out[i *  + ] += ":::";
             str_out[i *  + ] += "|";
         }
         str_out[(i + ) * ] = "+---+---+---+---+---+---+---+---+";
     }

     string str;

     //white:
     cin >> str;
     cin >> str;

     str = "," + str;

     //cout << str << endl;
     int arr_ch[] = {'K','Q','R','B','N'};

     string::size_type str_i;
     for(str_i = ; str_i != str.size(); ++str_i)
     {
         if(str[str_i] == ',')
         {
             for(i = ; i < ; ++i) if(str[str_i + ] == arr_ch[i]) break;
             if(i < )
             {
                 str_out[( - (str[str_i + ] - '')) *  + ][ +  * (str[str_i + ] - 'a')] = arr_ch[i];
             }
             else
             {
                 str_out[( - (str[str_i + ] - '')) *  + ][ +  * (str[str_i + ] - 'a')] = 'P';
             }
         }
     }

     //black:
     cin >> str;
     cin >> str;
     str = "," + str;

     //cout << str << endl;

     for(str_i = ; str_i != str.size(); ++str_i)
     {
         if(str[str_i] == ',')
         {
             for(i = ; i < ; ++i) if(str[str_i + ] == arr_ch[i]) break;
             if(i < )
             {
                 str_out[( - (str[str_i + ] - '')) *  + ][ +  * (str[str_i + ] - 'a')] = arr_ch[i] + ('a' - 'A');
             }
             else
             {
                 str_out[( - (str[str_i + ] - '')) *  + ][ +  * (str[str_i + ] - 'a')] = 'p';
             }
         }
     }

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

     //fclose(stdin);

     return ;
 }

5.Reference: