Uva-227 - Puzzle-AC

解体思路：二维数组储存原型，记录空格位置，按顺序移动，移动越界即为This puzzle has no final configuration.
注意：1、这个题输出格式卡的我老是WA，注意具体那个空行的位置

TRGSJ
XDOKI
M VLN
WPABE
UQHCF
ARRBBL0
Puzzle #1:
T R G S J
X O K L I
M D V B N
W P   A E
U Q H C F
ABCDE
FGHIJ
KLMNO
PQRS
TUVWX
AAA
LLLL0

 

Puzzle #2:
  A B C D
F G H I E
K L M N J
P Q R S O
T U V W X
ABCDE
FGHIJ
KLMNO
PQRS
TUVWX
AAAAABBRRRLL0

 

Puzzle #3:
This puzzle has no final configuration.
Z

Process returned 0 (0x0)   execution time : 16.359 s
Press any key to continue.
2、这个题读取原型时，也应注意不要用scanf，因为有空格！

 

AC代码：

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int i,j,x,y,t=1,flag;
    char a[10][10];
    while(1)
    {
        flag=0;
        for (i=0;i<5;i++)
        {
            for (j=0;j<5;j++)
            {
                a[i][j]=getchar();
                if (a[0][0]=='Z') return 0;
                if (a[i][j]==' ') {x=i; y=j;}
            }
            getchar();
        }
        char commd;
        while((commd=getchar())&&commd!='0')
        {
            if (flag) continue;
            if (commd=='A')
            {
                if (x==0) {flag=1;continue;}
                a[x][y]=a[x-1][y];
                a[x-1][y]=' ';
                x--;
            }
            else if (commd=='B')
            {
                if (x==4) {flag=1;continue;}
                a[x][y]=a[x+1][y];
                a[x+1][y]=' ';
                x++;
            }
            else if (commd=='L')
            {
                if (y==0) {flag=1;continue;}
                a[x][y]=a[x][y-1];
                a[x][y-1]=' ';
                y--;
            }
            else if (commd=='R')
            {
                if (y==4) {flag=1;continue;}
                a[x][y]=a[x][y+1];
                a[x][y+1]=' ';
                y++;
            }
        }
        if (t-1) putchar('\n');
        printf("Puzzle #%d:\n",t++);
        if (!flag)
        {
            for (i=0;i<5;i++)
            {
                for (j=0;j<5;j++)
                {
                    if (!j) printf("%c",a[i][j]);
                    else printf(" %c",a[i][j]);
                }
                putchar('\n');
            }
        }
        else
                printf("This puzzle has no final configuration.\n");
        getchar();
    }
    return 0;
}