POJ2993——Emag eht htiw Em Pleh(字符串处理+排序)

Emag eht htiw Em Pleh

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.|
+---+---+---+---+---+---+---+---+
题目大意：POJ2996的反向。输入2996的输出，输出2996的输入。
解题思路：
　　　　1.定义结构体存每个格子的状态
　　　　2.初始化结构体数组为“：：：”||“...”　　　　
　　　　3.读取字符串,根据数据改变结构体的第二个变量
　　　　4.格式输出
Code：

 #include<iostream>
 #include<string>
 using namespace std;
 struct Point
 {
     char a,b,c;
 } P[];
 void Init()
 {
     int ok=;
     for (int i=; i<=; i++)
     {
         if (ok) P[i].a=P[i].b=P[i].c=':';
         else P[i].a=P[i].b=P[i].c='.';
         if (i%) ok=!ok;
     }
 }
 int main()
 {
     char x,y;
     int i,j;
     string White,Black;
     string tmp="+---+---+---+---+---+---+---+---+";
     getline(cin,White);
     getline(cin,Black);
     Init();
     for (i=; i<=White.length()-; i++)
     {
         if (White[i]>='A'&&White[i]<='Z')
         {
             x=White[i+],y=White[i+];
             P[(x--)*+(y-'a'+)].b=White[i];
         }
         else if (White[i-]==',')
         {
             x=White[i+],y=White[i];
             P[(x--)*+(y-'a'+)].b='P';
         }
     }
     for (i=; i<=Black.length()-; i++)
     {
         if (Black[i]>='A'&&Black[i]<='Z')
         {
             x=Black[i+],y=Black[i+];
             P[(x--)*+(y-'a'+)].b=Black[i]+;
         }
         else if (Black[i-]==',')
         {
             x=Black[i+],y=Black[i];
             P[(x--)*+(y-'a'+)].b='p';
         }
     }
     cout<<tmp<<endl;
     for (i=; i>=; i--)
     {
         cout<<'|';
         for (j=; j<=; j++)
             cout<<P[(i-)*+j].a<<P[(i-)*+j].b<<P[(i-)*+j].c<<'|';
         cout<<endl;
         cout<<tmp<<endl;
     }
     return ;
 }