A children’s puzzle that was popular 30 years ago consisted of a 5×5 frame which contained 24 small
squares of equal size. A unique letter of the alphabet was printed on each small square. Since there
were only 24 squares within the frame, the frame also contained an empty position which was the same
size as a small square. A square could be moved into that empty position if it were immediately to the
right, to the left, above, or below the empty position. The object of the puzzle was to slide squares
into the empty position so that the frame displayed the letters in alphabetical order.
The illustration below represents a puzzle in its original configuration and in its configuration after
the following sequence of 6 moves:
- The square above the empty position moves.
- The square to the right of the empty position moves.
- The square to the right of the empty position moves.
- The square below the empty position moves.
- The square below the empty position moves.
- The square to the left of the empty position moves.
Write a program to display resulting frames given their initial configurations and sequences of moves.
Input
Input for your program consists of several puzzles. Each is described by its initial configuration and
the sequence of moves on the puzzle. The first 5 lines of each puzzle description are the starting
configuration. Subsequent lines give the sequence of moves.
The first line of the frame display corresponds to the top line of squares in the puzzle. The other
lines follow in order. The empty position in a frame is indicated by a blank. Each display line contains
exactly 5 characters, beginning with the character on the leftmost square (or a blank if the leftmost
square is actually the empty frame position). The display lines will correspond to a legitimate puzzle.
The sequence of moves is represented by a sequence of As, Bs, Rs, and Ls to denote which square
moves into the empty position. A denotes that the square above the empty position moves; B denotes
that the square below the empty position moves; L denotes that the square to the left of the empty
position moves; R denotes that the square to the right of the empty position moves. It is possible that
there is an illegal move, even when it is represented by one of the 4 move characters. If an illegal move
occurs, the puzzle is considered to have no final configuration. This sequence of moves may be spread
over several lines, but it always ends in the digit 0. The end of data is denoted by the character Z.
Output
Output for each puzzle begins with an appropriately labeled number (Puzzle #1, Puzzle #2, etc.). If
the puzzle has no final configuration, then a message to that effect should follow. Otherwise that final
configuration should be displayed.
Format each line for a final configuration so that there is a single blank character between two
adjacent letters. Treat the empty square the same as a letter. For example, if the blank is an interior
position, then it will appear as a sequence of 3 blanks — one to separate it from the square to the left,
one for the empty position itself, and one to separate it from the square to the right.
Separate output from different puzzle records by one blank line.
Note: The first record of the sample input corresponds to the puzzle illustrated above.
Sample Input
TRGSJ
XDOKI
M VLN
WPABE
UQHCF
ARRBBL0
ABCDE
FGHIJ
KLMNO
PQRS
TUVWX
AAA
LLLL0
ABCDE
FGHIJ
KLMNO
PQRS
TUVWX
AAAAABBRRRLL0
Z
Sample Output
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
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
Puzzle #3:
This puzzle has no final configuration.
还是英文题，这个题主要就是看：
//沙雕题。。。读题读了半天。。好气。。。
//题意：就是输入5行24个字母，然后输入操作的字母语句，A上，B下，L左，R右
//注意，操作语句以O结束,可以是好多行（这里一开始没看懂样例。。。）
//当输入Z，结束此题。。
#include<bits/stdc++.h>
using namespace std;
int hang,lie; //记录空格在哪行哪列
char caozuo; //记录要操作的字符
int kase=0; //第几种情况
char s[8][8];
int flag=0; //0是正常标记，1是越界
void yidong(int h,int l)
{
s[hang][lie]=s[hang+h][lie+l]; //初始空格
s[hang+h][lie+l]=' ';
}
void han(char a)
{
if(a=='A' && s[hang-1][lie]) //向上，行-1
{
yidong(-1,0);
hang--;
}
else if(a=='B' && s[hang+1][lie]) //向下，行+1
{
yidong(1,0);
hang++;
}
else if(a=='R' && s[hang][lie+1]) //向右，列+1
{
yidong(0,1);
lie++;
}
else if(a=='L' && s[hang][lie-1]) //向左，列-1
{
yidong(0,-1);
lie--;
}
else if(a=='\n'){} //就像第二个例子，AAA就换行了。。。。。。沙雕
else flag=1; //越界了。。
}
int main()
{
for(;;)
{
memset(s,0,sizeof(s));
flag=0; //初始标记
//输入：s[1][1] == 'Z'才结束此题
for(int i=1;i<=5;i++)
{
for(int j=1;j<=5;j++)
{
scanf("%c",&s[i][j]);
if(s[i][j]==' ')hang=i,lie=j; //记录空格在哪里
if(s[1][1]=='Z')goto jieshu; //goto语句跳出多重循环
}
getchar(); //吸收换行符
}
while((caozuo=getchar())!='0') //是数字0不是字母O。。。。。。
han(caozuo);
getchar();
if(kase!=0)
printf("\n");//putchar('\n');
kase++;
printf("Puzzle #%d:\n",kase);
//输出两种结果：
if(flag)
printf("This puzzle has no final configuration.\n");
else
{
for(int i=1;i<=5;i++)
{
for(int j=1;j<=5;j++)
{
if(j>1)printf(" ");
printf("%c",s[i][j]);
}
printf("\n");
}
}
}
jieshu:;
return 0;
}