Plato‘s Blocks
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 734 | Accepted: 296 |
Description
Plato believed what we perceive is but a shadow of reality. Recent archaeological excavations have uncovered evidence that this belief may have been encouraged by Plato‘s youthful amusement with cleverly-designed blocks. The blocks have the curious property that, when held with any face toward a light source, they cast shadows of various letters, numbers, shapes, and patterns. It is possible for three faces incident to a corner to correspond to three different shadow patterns. Opposite faces, of course, cast shadows which are mirror images of one another.
The blocks are formed by gluing together small cubes to form a single, connected object. As an example, the figures below show, layer by layer, the internal structure of a block which can cast shadows of the letters "E", "G", or "B". 
Only a partial set of blocks was discovered, but the curious scientists would like to determine what combinations of shadows are possible. Your program, the solution to this problem, will help them! The program will input groups of shadow patterns, and for each group will report whether or not a solid can be constructed that will cast those three shadows.
Input
The input contains a sequence of data sets, each specifying a dimension and three shadow patterns. The first line of a data set contains a positive integer 1 <= n <= 20 that specifies the dimensions of the input patterns. The remainder of the data set consists of 3n lines, each containing a string of n "X" and "-" characters. Each group of n lines represents a pattern. Where an "X" appears a shadow should be cast by the final solid, and where a "-" appears, light should pass through. For this problem, the input patterns may be assumed to have at least one "X" along each edge of the pattern. The input is terminated by a line containing a single zero in place of a valid dimension.
Output
For each data set in the input, output the data set number and one of the following messages:
Valid set of patterns
Impossible combination
For a set of patterns to be considered valid, it must be possible to construct, by gluing unit cubes together along their faces, a one-piece solid capable of casting the shadow of each of the input patterns.
Sample Input
5 XXXXX X---- X--XX X---X XXXXX XXXXX X---- XXXXX X---- XXXXX XXXXX X---X XXXX- X---X XXXXX 3 X-- -X- --X XX- XXX -XX -XX XXX XX- 0
Sample Output
Data set 1: Valid set of patterns Data set 2: Impossible combination 这个问题刚开始没有思路，看了网上的一些方法，就自己写了一个；
1 //注意每个面都由八种方式，旋转+翻转
2 //先建一个完整的立方块，然后删去中间的空缺部分
3 //最后检查一下（深搜）是不是所有的小立方块都连在一起
4
5 #include<iostream>
6 #include<algorithm>
7 #include<cstring>
8 #include<cstdio>
9
10 using namespace std;
11 char mpr[20][20];
12 int n;
13 char mp[3][8][20][20];
14 char cube[20][20][20];
15 int dx[]={0,0,0,0,1,-1};
16 int dy[]={0,0,1,-1,0,0};
17 int dz[]={1,-1,0,0,0,0};
18 void cs(int t)
19 {
20 for(int i=0;i<n;i++)
21 for(int j=0;j<n;j++)
22 mp[t][0][i][j]=mpr[i][j];
23 for(int i=0;i<n;i++)
24 for(int j=0;j<n;j++)
25 mp[t][1][i][j]=mp[t][0][i][n-1-j];
26 for(int i=0;i<n;i++)
27 for(int j=0;j<n;j++)
28 mp[t][2][i][j]=mp[t][1][j][n-1-i];
29 for(int i=0;i<n;i++)
30 for(int j=0;j<n;j++)
31 mp[t][3][i][j]=mp[t][0][i][n-1-j];
32 for(int i=0;i<n;i++)
33 for(int j=0;j<n;j++)
34 mp[t][4][i][j]=mp[t][3][j][n-1-i];
35 for(int i=0;i<n;i++)
36 for(int j=0;j<n;j++)
37 mp[t][5][i][j]=mp[t][4][i][n-1-j];
38 for(int i=0;i<n;i++)
39 for(int j=0;j<n;j++)
40 mp[t][6][i][j]=mp[t][5][j][n-1-i];
41 for(int i=0;i<n;i++)
42 for(int j=0;j<n;j++)
43 mp[t][7][i][j]=mp[t][6][i][n-1-j];
44 }
45 int checkview1(int a)
46 {
47 for(int i=0;i<n;i++)
48 for(int j=0;j<n;j++)
49 {
50 if(mp[0][a][i][j]==‘X‘)
51 {
52 int flag=0;
53 for(int k=0;k<n;k++)
54 if(cube[i][j][k]==0)
55 flag=1;
56 if(flag==0) return 0;
57 }
58 }
59 return 1;
60 }
61 int checkview2(int a)
62 {
63 for(int i=0;i<n;i++)
64 for(int j=0;j<n;j++)
65 {
66 if(mp[1][a][i][j]==‘X‘)
67 {
68 int flag=0;
69 for(int k=0;k<n;k++)
70 if(cube[i][k][j]==0)
71 flag=1;
72 if(flag==0) return 0;
73 }
74 }
75 return 1;
76 }
77 int checkview3(int a)
78 {
79 for(int i=0;i<n;i++)
80 for(int j=0;j<n;j++)
81 {
82 if(mp[2][a][i][j]==‘X‘)
83 {
84 int flag=0;
85 for(int k=0;k<n;k++)
86 if(cube[k][i][j]==0)
87 flag=1;
88 if(flag==0) return 0;
89 }
90 }
91 return 1;
92 }
93 int check(int x,int y,int z)
94 {
95 if(x<n&&x>=0&&y<n&&y>=0&&z<n&&z>=0) return 1;
96 return 0;
97 }
98 void dfs(int a,int b,int c)
99 {
100 for(int i=0;i<6;i++)
101 {
102 int curx=a+dx[i];
103 int cury=b+dy[i];
104 int curz=c+dz[i];
105 if(check(curx,cury,curz)&&cube[curx][cury][curz]==0)
106 {
107 cube[curx][cury][curz]=1;
108 dfs(curx,cury,curz);
109 }
110 }
111 return ;
112 }
113 int Num()
114 {
115 int num=0;
116 for(int i=0;i<n;i++)
117 for(int j=0;j<n;j++)
118 for(int k=0;k<n;k++)
119 if(cube[i][j][k]==0)
120 {
121 cube[i][j][k]=1;
122 dfs(i,j,k);
123 num++;
124 }
125 if(num>1) return 0;
126 return 1;
127 }
128 int solve(int a,int b,int c)
129 {
130 //建一个完全的立方块
131 memset(cube,0,sizeof(cube));
132
133 //删去其中的空缺部分
134 for(int i=0;i<n;i++)
135 for(int j=0;j<n;j++)
136 if(mp[0][a][i][j]==‘-‘)
137 for(int k=0;k<n;k++)
138 cube[i][j][k]=1;
139 for(int i=0;i<n;i++)
140 for(int j=0;j<n;j++)
141 if(mp[1][b][i][j]==‘-‘)
142 for(int k=0;k<n;k++)
143 cube[i][k][j]=1;
144 for(int i=0;i<n;i++)
145 for(int j=0;j<n;j++)
146 if(mp[2][c][i][j]==‘-‘)
147 for(int k=0;k<n;k++)
148 cube[k][i][j]=1;
149
150
151 //检查三视图是否还是符合的还有立方块是否都连在一起（dfs）
152 if(checkview1(a)&&checkview2(b)&&checkview3(c)&&Num())
153 return 1;
154 return 0;
155 }
156 int main()
157 {
158 int num=1;
159 while(cin >> n&&n)
160 {
161 for(int i=0;i<3;i++)
162 {
163 for(int j=0;j<n;j++)
164 cin >> mpr[j];
165 cs(i);//这里构建八个面，那么8*8*8=512种情况，只要有一种情况符合就行了
166 }
167 int flag=0;
168 for(int i=0;i<8;i++)
169 for(int j=0;j<8;j++)
170 for(int k=0;k<8;k++)
171 if(solve(i,j,k))//检查每种情况是否符合
172 flag=1;
173 if(flag)
174 printf("Data set %d: Valid set of patterns\n",num++);
175 else printf("Data set %d: Impossible combination\n",num++);
176 }
177 return 0;
178 }
POJ 1052 Plato's Blocks