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棋盘布局解析

最新推荐文章于 2021-10-25 16:20:40 发布

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本文介绍了一个从特定格式的棋子布局描述中解析并还原国际象棋棋盘布局的方法。通过输入白方和黑方棋子的位置信息，程序能够正确地绘制出对应的棋盘布局。此方法适用于需要从文字描述中还原棋盘状态的应用场景。

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

/*
棋盘的还原;
*/
#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
//const int Maxn=1100000;
char str[]="+---+---+---+---+---+---+---+---+";//奇数行输出;
char s1[]="|:::|...|:::|...|:::|...|:::|...|";//4的倍数行的选择;
char s2[]="|...|:::|...|:::|...|:::|...|:::|";//2的倍数行的选择;
char Map[55][55];//标记每行每列的字符;
void Trans(char s[])//位置坐标转化为字母;
{
bool flag=false;
if(s[0]=='B')//黑棋;
flag=true;
for(int i=7;s[i];)
{
if(s[i]>='A'&&s[i]<='Z')
{
int u=s[i+1]-'a'+1;//纵坐标的标记;
int v=s[i+2]-'0';//横坐标的标记;
if(flag)
{
Map[v][u]=s[i]+32;//黑棋代表的是小写字母;
}
else
{
Map[v][u]=s[i];//白棋即大写字母;
}
i+=4;
}
else if(s[i]>='a'&&s[i]<='z')//p位置的确定;
{
int u=s[i]-'a'+1;//纵坐标
int v=s[i+1]-'0';//横坐标;
if(flag)
Map[v][u]='p';//黑棋;
else
Map[v][u]='P';//白棋;
i+=3;
}
}
}
int main()
{
char white[110];
char black[110];
memset(white,'\0',sizeof(white));
memset(black,'\0',sizeof(black));
//getchar();
gets(white);
gets(black);
memset(Map,0,sizeof(Map));
Trans(white);
Trans(black);
int u=8;//共八行;;
for(int i=17;i>=1;i--)
{
if(i%2)
{
printf("%s\n",str);
}
else
{
int ans=2;
int v=1;
if((i/2)%2)//
{
for(int j=0;j<33;j++)
{
if(j!=ans)//不是棋子;
{
printf("%c",s1[j]);
}
else
{
if(Map[u][v])//有棋子;
{
printf("%c",Map[u][v]);
}
else
printf("%c",s1[j]);
v++;//纵坐标加一;
ans+=4;
}
}
}
else
{
for(int j=0;j<33;j++)
{
if(j!=ans)
{
printf("%c",s2[j]);
}
else
{
if(Map[u][v])
{
printf("%c",Map[u][v]);
}
else
{
printf("%c",s2[j]);
}
v++;
ans+=4;
}
}
}
printf("\n");
u--;
}
}
return 0;
}