POJ 2993-Emag eht htiw Em Pleh（模拟-根据棋子位置还原棋盘）

分手后的思念是犯贱 2022-07-11 08:10 137阅读 0赞

Emag eht htiw Em Pleh

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;65536K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;3492</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;2294</td> 
  </tr> 
 </tbody> 
</table>

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][]

## 题目意思： ##

与[POJ 2996][]（[http://blog.csdn.net/mikasa3/article/details/54630409][POJ 2996]）刚好相反，本题给定棋子位置，输出还原后的棋盘。

## 解题思路： ##

模拟题，白棋黑棋分别搞一搞，注意数组的范围不能开小了。

这题我本来用的C++输入输出，一直异常中断，错误提示如下：

*** glibc detected *** ./prog: free(): invalid pointer: 0x08fc0028 ***
    ======= Backtrace: =========
    /lib/libc.so.6[0x557f41f5]
    /lib/libc.so.6(cfree+0x9c)[0x557f5acc]
    /usr/lib/libstdc++.so.6(_ZdlPv+0x21)[0x5571f2e1]
    /usr/lib/libstdc++.so.6(_ZNSs4_Rep10_M_destroyERKSaIcE+0x1d)[0x556fbb2d]
    /usr/lib/libstdc++.so.6(_ZNSs7reserveEj+0xad)[0x556fd73d]
    /usr/lib/libstdc++.so.6(_ZNSs6appendEPKcj+0x8d)[0x556fd93d]
    /usr/lib/libstdc++.so.6(_ZStrsIcSt11char_traitsIcESaIcEERSt13basic_istreamIT_T0_ES7_RSbIS4_S5_T1_E+0x169)[0x556d6a89]
    ./prog(__gxx_personality_v0+0x453)[0x8048dbb]
    ./prog(__gxx_personality_v0+0x79)[0x80489e1]
    ======= Memory map: ========

看了蛮久不知道哪儿错了，遂改成C的输入输出，运行正常，AC。

#include<iostream>
    #include<cstdio>
    #include<cstdlib>
    #include<cstring>
    #include<algorithm>
    #include<cmath>
    using namespace std;
    
    int col(char c)//输出列
    {
        if(c=='a')return 2;
        else if(c=='b') return 6;
        else if(c=='c') return 10;
        else if(c=='d') return 14;
        else if(c=='e') return 18;
        else if(c=='f') return 22;
        else if(c=='g') return 26;
        else if(c=='h') return 30;
        return 0;
    }
    int row(char c)//输出行
    {
        if(c=='1') return 15;
        else if(c=='2') return 13;
        else if(c=='3') return 11;
        else if(c=='4') return 9;
        else if(c=='5') return 7;
        else if(c=='6') return 5;
        else if(c=='7') return 3;
        else if(c=='8') return 1;
        return 0;
    }
    char ma[18][34];
    void Init()//将边框和.和:插入到相应位置
    {
        char frame[34]= {'+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-',
                         '+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+'
                        };
        for(int i=0; i<17; i+=2)
            for(int j=0; j<33; j++)
                ma[i][j]=frame[j];
        for(int i=1; i<17; i+=2)
            for(int j=0; j<33; j+=4)
                ma[i][j]='|';
        int t;
        for(int i=1; i<17; i+=4)
            for(int j=1; j<33; j+=8)
            {
                for(int k=0; k<3; ++k)
                {
                    t=j;
                    ma[i][j+k]='.';
                }
                j=t;
            }
        for(int i=3; i<17; i+=4)
            for(int j=1; j<33; j+=8)
            {
                for(int k=0; k<3; ++k)
                {
                    t=j;
                    ma[i][j+k]=':';
                }
                j=t;
            }
        for(int i=1; i<17; i+=4)
            for(int j=5; j<33; j+=8)
            {
                for(int k=0; k<3; ++k)
                {
                    t=j;
                    ma[i][j+k]=':';
                }
                j=t;
            }
        for(int i=3; i<17; i+=4)
            for(int j=5; j<33; j+=8)
            {
                for(int k=0; k<3; ++k)
                {
                    t=j;
                    ma[i][j+k]='.';
                }
                j=t;
            }
    }
    void output()//输出棋盘
    {
        for(int i=0; i<17; i++)
        {
            for(int j=0; j<33; j++)
                printf("%c",ma[i][j]);
            printf("\n");
        }
    }
    int main()
    {
        Init();
        int cas=2;
        char color[10];
    #ifdef ONLINE_JUDGE
    #else
        freopen("F:/cb/read.txt","r",stdin);
        //freopen("F:/cb/out.txt","w",stdout);
    #endif
        while(cas--)
        {
            scanf("%s",color);
            if(!strcmp(color,"White:"))//比较结果为0则匹配成功
            {
                char str[50];
                scanf("%s",str);
                char s[3];
                int cnt=0;
                for(int i=0; i<strlen(str); ++i)
                {
                    if(str[i]==',')//以逗号为分隔点分别处理各个位置
                    {
                        if(s[0]>='A'&&s[0]<='Z')
                            ma[row(s[2])][col(s[1])]=s[0];
                        else if(s[0]>='a'&&s[0]<='z')
                            ma[row(s[1])][col(s[0])]='P';
                        cnt=0;
                    }
                    else s[cnt++]=str[i];
                }
                if(s[0]>='a'&&s[0]<='z')//注意最后一个位置的后面是没有逗号的所以要单独处理
                    ma[row(s[1])][col(s[0])]='P';
            }
            else if(!strcmp(color,"Black:"))
            {
                char str[70];
                scanf("%s",str);
                char s[3];
                memset(s,'\0',sizeof(s));
                int cnt=0,pos;
                for(int i=0; i<strlen(str); ++i)
                {
                    if(str[i]==',')
                    {
                        if(s[0]>='A'&&s[0]<='Z')
                            ma[row(s[2])][col(s[1])]=tolower(s[0]);
                        else if(s[0]>='a'&&s[0]<='z')
                            ma[row(s[1])][col(s[0])]='p';
                        cnt=0;
                        memset(s,'\0',sizeof(s));
                    }
                    else s[cnt++]=str[i];
                }
                if(s[0]>='a'&&s[0]<='z')
                    ma[row(s[1])][col(s[0])]='p';
            }
        }
        output();
        return 0;
    }
    /*
    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
    */

[problem 2996]: http://acm.pku.edu.cn/JudgeOnline/problem?id=2996
[CTU Open 2005]: http://poj.org/searchproblem?field=source&key=CTU+Open+2005
[POJ 2996]: http://blog.csdn.net/mikasa3/article/details/54630409