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[10000];
void Init()
{
    int ok=1;
    for (int i=1; i<=64; i++)
    {
        if (ok) P[i].a=P[i].b=P[i].c=':';
        else P[i].a=P[i].b=P[i].c='.';
        if (i%8) ok=!ok;
    }
}
int main()
{
    char x,y;
    int i,j;
    string White,Black;
    string tmp="+---+---+---+---+---+---+---+---+";
    getline(cin,White);
    getline(cin,Black);
    Init();
    for (i=7; i<=White.length()-1; i++)
    {
        if (White[i]>='A'&&White[i]<='Z')
        {
            x=White[i+2],y=White[i+1];
            P[(x-48-1)*8+(y-'a'+1)].b=White[i];
        }
        else if (White[i-1]==',')
        {
            x=White[i+1],y=White[i];
            P[(x-48-1)*8+(y-'a'+1)].b='P';
        }
    }
    for (i=7; i<=Black.length()-1; i++)
    {
        if (Black[i]>='A'&&Black[i]<='Z')
        {
            x=Black[i+2],y=Black[i+1];
            P[(x-48-1)*8+(y-'a'+1)].b=Black[i]+32;
        }
        else if (Black[i-1]==',')
        {
            x=Black[i+1],y=Black[i];
            P[(x-48-1)*8+(y-'a'+1)].b='p';
        }
    }
    cout<<tmp<<endl;
    for (i=8; i>=1; i--)
    {
        cout<<'|';
        for (j=1; j<=8; j++)
            cout<<P[(i-1)*8+j].a<<P[(i-1)*8+j].b<<P[(i-1)*8+j].c<<'|';
        cout<<endl;
        cout<<tmp<<endl;
    }
    return 0;
}