POJ 2993：Emag eht htiw Em Pleh

Time Limit: 1000MS

Memory Limit: 65536KB

64bit IO Format: %I64d & %I64u

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.|

+---+---+---+---+---+---+---+---+

题意就是把2996的那道题的输出变输入，输入变输出。我觉得一下子变简单了很多，起码没那么多的麻烦事了。

思路就是现在是一个空棋盘，然后题目给了各个棋子的位置，拿这些棋子去占相应的位置，之后再输出占好各个位置的棋盘。

代码：

#include <iostream>

#include <algorithm>

#include <cmath>

#include <vector>

#include <string>

#include <cstring>

#pragma warning(disable:4996)

using namespace std;

const string temp="+---+---+---+---+---+---+---+---+";

string test[20];

string cal[5];

int i,len;

char row,col;

int main()

{

	test[1]="|...|:::|...|:::|...|:::|...|:::|";

	test[2]="|:::|...|:::|...|:::|...|:::|...|";

	test[3]="|...|:::|...|:::|...|:::|...|:::|";

	test[4]="|:::|...|:::|...|:::|...|:::|...|";

	test[5]="|...|:::|...|:::|...|:::|...|:::|";

	test[6]="|:::|...|:::|...|:::|...|:::|...|";

	test[7]="|...|:::|...|:::|...|:::|...|:::|";

	test[8]="|:::|...|:::|...|:::|...|:::|...|";

	cin>>cal[0];

	cin>>cal[1];

	cin>>cal[3];

	cin>>cal[2];

	len=cal[1].length();

	for(i=0;i<len;i++)

	{

		if(cal[1][i]==',')continue;

		else if(cal[1][i]=='K')

		{

			row=cal[1][i+1];

			col=cal[1][i+2];

			test[9-(col-'0')][(row-'a')*4+2]='K';

			i=i+2;

		}

		else if(cal[1][i]=='Q')

		{

			row=cal[1][i+1];

			col=cal[1][i+2];

			test[9-(col-'0')][(row-'a')*4+2]='Q';

			i=i+2;

		}

		else if(cal[1][i]=='R')

		{

			row=cal[1][i+1];

			col=cal[1][i+2];

			test[9-(col-'0')][(row-'a')*4+2]='R';

			i=i+2;

		}

		else if(cal[1][i]=='B')

		{

			row=cal[1][i+1];

			col=cal[1][i+2];

			test[9-(col-'0')][(row-'a')*4+2]='B';

			i=i+2;

		}

		else if(cal[1][i]=='N')

		{

			row=cal[1][i+1];

			col=cal[1][i+2];

			test[9-(col-'0')][(row-'a')*4+2]='N';

			i=i+2;

		}

		else if(cal[1][i]>='a'&&cal[1][i]<='h')

		{

			row=cal[1][i];

			col=cal[1][i+1];

			test[9-(col-'0')][(row-'a')*4+2]='P';

			i=i+1;

		}

	}

	len=cal[2].length();

	for(i=0;i<len;i++)

	{

		if(cal[2][i]==',')continue;

		else if(cal[2][i]=='K')

		{

			row=cal[2][i+1];

			col=cal[2][i+2];

			test[9-(col-'0')][(row-'a')*4+2]='k';

			i=i+2;

		}

		else if(cal[2][i]=='Q')

		{

			row=cal[2][i+1];

			col=cal[2][i+2];

			test[9-(col-'0')][(row-'a')*4+2]='q';

			i=i+2;

		}

		else if(cal[2][i]=='R')

		{

			row=cal[2][i+1];

			col=cal[2][i+2];

			test[9-(col-'0')][(row-'a')*4+2]='r';

			i=i+2;

		}

		else if(cal[2][i]=='B')

		{

			row=cal[2][i+1];

			col=cal[2][i+2];

			test[9-(col-'0')][(row-'a')*4+2]='b';

			i=i+2;

		}

		else if(cal[2][i]=='N')

		{

			row=cal[2][i+1];

			col=cal[2][i+2];

			test[9-(col-'0')][(row-'a')*4+2]='n';

			i=i+2;

		}

		else if(cal[2][i]>='a'&&cal[2][i]<='h')

		{

			row=cal[2][i];

			col=cal[2][i+1];

			test[9-(col-'0')][(row-'a')*4+2]='p';

			i=i+1;

		}

	}

	cout<<temp<<endl;

	for(i=1;i<=8;i++)

	{

		cout<<test[i]<<endl;

		cout<<temp<<endl;

	}

	return 0;

}