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[8 + 9];
    str_out[0] = "+---+---+---+---+---+---+---+---+";
    for(i = 0; i < 8; ++i)
    {
        str_out[i * 2 + 1] = "|";
        for(j = 0; j < 8; ++j)
        {
            if((i + j) % 2 == 0) str_out[i * 2 + 1] += "...";
            else str_out[i * 2 + 1] += ":::";
            str_out[i * 2 + 1] += "|";
        }
        str_out[(i + 1) * 2] = "+---+---+---+---+---+---+---+---+";
    }


    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 = 0; str_i != str.size(); ++str_i)
    {
        if(str[str_i] == ',')
        {
            for(i = 0; i < 5; ++i) if(str[str_i + 1] == arr_ch[i]) break;
            if(i < 5)
            {
                str_out[(8 - (str[str_i + 3] - '0')) * 2 + 1][2 + 4 * (str[str_i + 2] - 'a')] = arr_ch[i];
            }
            else
            {
                str_out[(8 - (str[str_i + 2] - '0')) * 2 + 1][2 + 4 * (str[str_i + 1] - 'a')] = 'P';
            }        
        }
    }

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

    //cout << str << endl;

    for(str_i = 0; str_i != str.size(); ++str_i)
    {
        if(str[str_i] == ',')
        {
            for(i = 0; i < 5; ++i) if(str[str_i + 1] == arr_ch[i]) break;
            if(i < 5)
            {
                str_out[(8 - (str[str_i + 3] - '0')) * 2 + 1][2 + 4 * (str[str_i + 2] - 'a')] = arr_ch[i] + ('a' - 'A');
            }
            else
            {
                str_out[(8 - (str[str_i + 2] - '0')) * 2 + 1][2 + 4 * (str[str_i + 1] - 'a')] = 'p';
            }        
        }
    }

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

    //fclose(stdin);

    return 0;
}

5.Reference: