UVA227 Puzzle谜题

Puzzle谜题

题目描述

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:

1. The square above the empty position moves.
2. The square to the right of the empty position moves.
3. The square to the right of the empty position moves.
4. The square below the empty position moves.
5. The square below the empty position moves.
6. 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.

解题

本题难度不大，只是普通模拟，主要注意的是输出格式还有读取字符时的方法。
读取字符：使用getchar()才能正确读取空格。注意处理输入时的换行符。输入的操作指示可能不在同一行。
输出格式：每一组输出之间的换行。每一行后面不能有多余空格。

代码

#include<stdio.h>
#include<ctype.h>
#include<cstring>
#include<stdlib.h>
#include<iostream>
using namespace std;

char m[6][6];	//储存Puzzle
bool flag;	//记录是否有解
int n=0;	//组数
int p,q;	//空格位置
char opera;	//操作指令
void moving(char c);	//移动

int main(){
	while(true){
		//数据初始化
		flag=false;
		n++;	p=0;	q=0;
		//Puzzle输入
		for(int i=0;i<5;i++){
			for(int j=0;j<5;j++){
				m[i][j]=getchar();
				if(m[i][j]=='Z'){	//判断输入结束
					return 0;
				}
				if(m[i][j]==' '){	//记录空格
					p=i;	q=j;
				}
			}
			getchar();	//处理换行符 
		}
		//读取、处理操作指令
		for(;;){
			opera=getchar();
			if(opera=='0'){		//标志操作指令结束
				break;
			}else{
				moving(opera);
			}
		}
		getchar();	//处理换行符

		//输出
		if(n!=1)
			cout<<endl;
		cout<<"Puzzle #"<<n<<':'<<endl;
		if(flag){
			cout<<"This puzzle has no final configuration."<<endl;
		}else{
			for(int i=0;i<5;i++){
				for(int j=0;j<5;j++){
					cout<<m[i][j];
					if(j!=4){
						cout<<' ';
					}
				}
				cout<<endl;
			}
		}
	}
	return 0;
}

void moving(char c){
	switch (c){
		case 'A':
			if(p==0){
				flag=true;
			}else{
				char t=m[p-1][q];
				m[p-1][q]=m[p][q];
				m[p][q]=t;
				p--;
			}
			break;
		case 'B':
			if(p==4){
				flag=true;
			}else{
				char t=m[p+1][q];
				m[p+1][q]=m[p][q];
				m[p][q]=t;
				p++;
			}
			break;
		case 'L':
			if(q==0){
				flag=true;
			}else{
				char t=m[p][q-1];
				m[p][q-1]=m[p][q];
				m[p][q]=t;
				q--;
			}
			break;
		case 'R':
			if(q==4){
				flag=true;
			}else{
				char t=m[p][q+1];
				m[p][q+1]=m[p][q];
				m[p][q]=t;
				q++;
			}
			break;
		default:break;
	}
	return;
}