ZOJ 2714 Uva LA 2995 Image is everything 机智题

题意：

一个n*n*n的立方体，每个单位格子可能是空的，或是一个纯色方块，给你六个面的视图，能看穿用.表示，否则是一个大写字母代表颜色。问最大可能的体积。

分析：

Final题，如同题目，考察机智程度的题。反正我是不够机智。。想半天不会，看了题解还是不知道怎么写。。最后参考了白书训练指南的代码，做法如下：

建一个3维数组存放整个立方体，首先对于能看穿的，那一溜过去都是空的。接下来就是一个染色判断矛盾的过程，循环所有的面的每个位置去给立方体对应位置染色，如果出现矛盾，这个位置就是空的，一直重复做到没有删除。

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 using namespace std;
 5
 6 char read(){
 7     char ret;
 8     while(1){
 9         ret = getchar();
10         if (ret == ‘.‘ || (ret >= ‘A‘ && ret <= ‘Z‘)) return ret;
11     }
12 }
13 int n;
14 void getp(int id, int r, int c, int d, int &x, int &y, int &z){
15     if (id == 0) x = n-1-d, y = c, z = n-1-r;
16     if (id == 1) x = c, y = d, z = n-1-r;
17     if (id == 2) x = d, y = n-1-c, z = n-1-r;
18     if (id == 3) x = n-1-c, y = n-1-d, z = n-1-r;
19     if (id == 4) x = r, y = c, z = n-1-d;
20     if (id == 5) x = n-1-r, y = c, z = d;
21 }
22 int x, y, z;
23 char pos[15][15][15], view[6][15][15];
24 int main()
25 {
26     while(scanf("%d", &n) && n)
27     {
28         for (int j = 0; j < n; j++)
29             for (int i = 0; i < 6; i++)
30                 for (int k = 0; k < n; k++)
31                     view[i][j][k] = read();
32         for (int i = 0; i < n; i++)
33             for (int j = 0; j < n; j++)
34                 for (int k = 0; k < n; k++)
35                     pos[i][j][k] = ‘#‘;
36         for (int i = 0; i < 6; i++)
37             for (int j = 0; j < n; j++)
38                 for (int k = 0; k < n; k++) if (view[i][j][k] == ‘.‘){
39                     for (int p = 0; p < n; p++){
40                         getp(i, j, k, p, x, y, z);
41                         pos[x][y][z] = ‘.‘;
42                     }
43                 }
44         bool flag = true;
45         while(flag){
46             flag = false;
47             for (int i = 0; i < 6; i++)
48                 for (int j = 0; j < n; j++)
49                     for (int k = 0; k < n; k++) if (view[i][j][k] != ‘.‘){
50                         for (int p = 0; p < n; p++){
51                             getp(i, j, k, p, x, y, z);
52                             if (pos[x][y][z] == ‘.‘) continue;
53                             if (pos[x][y][z] == ‘#‘){
54                                 pos[x][y][z] = view[i][j][k];
55                                 break;
56                             }
57                             if (pos[x][y][z] == view[i][j][k]) break;
58                             pos[x][y][z] = ‘.‘;
59                             flag = true;
60                         }
61                     }
62         }
63         int ans = 0;
64         for (int i = 0; i < n; i++)
65             for (int j = 0; j < n; j++)
66                 for (int k = 0; k < n; k++)
67                     if (pos[i][j][k] != ‘.‘) ans++;
68         printf("Maximum weight: %d gram(s)\n", ans);
69     }
70     return 0;
71 }