UVa227_puzzle

 1 //按要求来就可以了　　#include<iostream>
 2 #include<stdio.h>
 3 #include<cstring>
 4 using namespace std;
 5 const int M = 5, N = 5, maxn = 1010;
 6
 7 void output(char (*p)[N])
 8 {
 9     for (int i2 = 0; i2 < M; i2++)
10     {
11       printf ("%c %c %c %c %c\n", p[i2][0], p[i2][1],p[i2][2], p[i2][3], p[i2][4]);
12     }
13 }
14
15 int main(void)
16 {
17     char p[M][N]= {0}, c[maxn] = {0};
18     int count = 0;
19     while (cin.getline(p[0],M+1))
20     {
21         if (strcmp(p[0],"Z") == 0)
22             return 0;
23         cin.getline(p[1],M+1);         //这样比循环的效率高.
24         cin.getline(p[2],M+1);
25         cin.getline(p[3],M+1);
26         cin.getline(p[4],M+1);
27         int b_x = 0, b_y = 0;
28         for (int i = 0; i < M; i++)
29             for (int j = 0; j < N; j++)
30             {
31                 if (p[i][j] == ‘ ‘) {
32                     b_x = i; b_y = j;       //b_x,b_y代表空白位置
33                     break;
34                 }
35         }
36         int k = 0;
37         while (cin >> c[k])
38         {
39             if (c[k] != ‘0‘)
40                 k++;
41             else
42                 break;
43         }
44         c[k] = 0;         //因为每次在k的位置数是不定的，k不定！！所以要将第k位初始化等下面条件
45                           //控制到c[i]!=0
46         int flag = 1, x = b_x, y = b_y;
47         for (int i = 0; c[i]; i++)
48         {
49             switch(c[i])
50             {
51                 case ‘A‘ : x = x-1; y = b_y;break;
52                 case ‘B‘ : x = x+1; y = b_y;break;
53                 case ‘L‘ : x = b_x; y = y-1; break;
54                 case ‘R‘ : x = b_x; y = y+1; break;
55                 default : flag = 0; break;
56             }
57             if (x < 0 || x > 4 || y < 0 || y > 4)
58             {
59                 flag = 0; break ;
60             }
61             else {
62                 p[b_x][b_y] = p[x][y];
63                 p[x][y] = ‘ ‘;
64                 b_x = x; b_y = y;             //移动的位置变成了空白位置！！注意！！
65             }
66         }
67             if (count)
68                 printf("\n");
69              printf ("Puzzle #%d:\n", ++count);
70             if (flag)
71             {
72                 output(p);
73             }
74             else {
75                 printf ("This puzzle has no final configuration.\n");
76             }
77             getchar();         //cin.ignore(1,‘\n‘); 吃掉cin >> c[k]后面的‘\n‘
78         }
79     return 0;
80 }