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

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Emag eht htiw Em Pleh

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 3492 Accepted: 2294

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,e4Black: 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）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,e4Black:Ke8,Qd8,Ra8,Rh8,Bc8,Ng8,Nc6,a7,b7,c7,d7,e7,f7,h7,h6*/

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