Uva 227-Puzzle 解题报告

Posted 2020-07-02 甘乐

 

题目链接：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=163

Puzzle

　　　　　　　　　　　　　　Time limit: 3.000 seconds

 

A children‘s puzzle that was popular 30 years ago consisted of a 5x5 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.

想要解决这道题，有几点需要注意
1.关于puzzle中字母的数据输入保存问题，由于每行末尾有enter，考虑到最后数据以’Z‘作为结束字符，（尤其注意最后程序要正常退出的问题）用gets()函数较好，gets读取换行符‘\n‘之前的所有字符，并在其末尾加上’\0‘后将其返回给调用它的函数，接着读取enter字符并将其丢弃。

2.其次关于调动序列的数据输入问题，这里需要读取’0‘之前的所有序列，用scanf（“%c”，str[i++])即可，但考虑到换行等空白字符，亦可用getchar()函数读取’0‘之前所有非空白字符（利用isspace()）。

3.千万注意在保存调用序列后要用getchar吃掉空白字符，最后注意一下输出格式就OK了。


AC代码如下：

#include <stdio.h>
#include <string.h>
#include <ctype.h>
//2.#include<stdlib.h>
char str[100];
int main()
{
//    freopen("C:\\Users\\Aliez\\Desktop\\input.txt","r",stdin);
//    freopen("C:\\Users\\Aliez\\Desktop\\output.txt","w",stdout);
    char puz[5][5],temp;
    int ch,c,len,m,n,T = 0,real,k;
    while(1){
        real = 1;
        k = 0;
        memset(str,‘\0‘,sizeof(str));
        memset(puz,‘ ‘,sizeof(puz));
        for(int i = 0;i < 5;i++){
            gets(puz[i]);
            if(puz[0][0] == ‘Z‘)
                return 0;
                //1.break;
                //2.exit(0);
            if(puz[i][4] == ‘\0‘)
                puz[i][4] = ‘ ‘;
        }
        /*1.if(puz[0][0] == ‘Z‘)
            break;*/
        for(int i = 0;i < 5;i++){
            for(int j = 0;j < 5;j++){
                if(puz[i][j] == ‘ ‘){
                    m = i;
                    n = j;
                }
            }
        }

        while((c = getchar())!=‘0‘){
            if(!isspace(c))
                str[k++] = c;
        }
        while(1){
            c = getchar();
            if(isspace(c))
                break;
        }
        //以下注释部分为测试代码
//        puts(puz[0]);
//        puts(str);
//        printf("m=%d,n=%d\n",m,n);
        len = strlen(str);
        int x = m, y = n;
        for(int i = 0; i < len;i++){
            switch(str[i])
            {
                case ‘A‘:
                        x--;
                    break;
                case ‘B‘:
                        x++;
                    break;
                case ‘L‘:
                        y--;
                    break;
                case ‘R‘:
                        y++;
                    break;
                default:
                    real = 0;
                    break;
            }
            if(x < 0||x >4||y < 0||y > 4){
                real = 0;
                break;
            }
            else{
                puz[m][n] = puz[x][y];
                puz[x][y] = ‘ ‘;
                m = x;
                n = y;
            }
        }
        if(T++)
            printf("\n");
        printf("Puzzle #%d:\n",T);

        if(real){
            for(int i = 0;i < 5;i++){
                printf("%c",puz[i][0]);
                for(int j = 1;j < 5;j++)
                    printf(" %c",puz[i][j]);
                printf("\n");
            }
        }
        else
            printf("This puzzle has no final configuration.\n");
    }
    return 0;
}