【HDOJ】3220 Alice’s Cube

状态压缩+逆向BFS。
方向数组就是任意相邻的两点（初始化时减1），每个顶点均有4个相邻点。因此，共有16*4/2=32个方向。按序排列即可找到。

 1 /* 3220 */
 2 #include <iostream>
 3 #include <queue>
 4 #include <algorithm>
 5 #include <cstdio>
 6 #include <cstring>
 7 using namespace std;
 8
 9
10 char visit[1<<17];
11 const int end = 0x0ff;
12 int beg;
13 int dir[32][2] = {
14     1,2, 1,3, 1,5, 1,9,
15     2,4, 2,6, 2,10,
16     3,4, 3,7, 3,11,
17     4,8, 4,12,
18     5,6, 5,7, 5,13,
19     6,8, 6,14,
20     7,8, 7,15,
21     8,16,
22     9,10, 9,11, 9,13,
23     10,12, 10,14,
24     11,12, 11,15,
25     12,16,
26     13,14, 13,15,
27     14,16,
28     15,16
29 };
30
31 void bfs() {
32     int i, j, k, tmp;
33     queue<int> Q;
34     int a, b, s;
35     char t;
36
37     memset(visit, -1, sizeof(visit));
38     visit[end] = 0;
39     Q.push(end);
40
41     while (!Q.empty()) {
42         s = Q.front();
43         Q.pop();
44         t = visit[s];
45         if (t >= 3)
46             continue;
47         ++t;
48         for (i=0; i<32; ++i) {
49             a = s & (1<<dir[i][0]);
50             b = s & (1<<dir[i][1]);
51             if (a ^ b) {
52                 tmp = (s ^ (1<<dir[i][0])) ^ (1<<dir[i][1]);
53                 if (visit[tmp] < 0) {
54                     visit[tmp] = t;
55                     Q.push(tmp);
56                 }
57             }
58         }
59     }
60 }
61
62 void init() {
63     int i;
64
65     for (i=0; i<32; ++i) {
66         --dir[i][0];
67         --dir[i][1];
68     }
69
70     bfs();
71 }
72
73 int main() {
74     int t, tt;
75     int i, j, k;
76
77     #ifndef ONLINE_JUDGE
78         freopen("data.in", "r", stdin);
79         freopen("data.out", "w", stdout);
80     #endif
81
82     init();
83     scanf("%d", &tt);
84     for (t=1; t<=tt; ++t) {
85         beg = 0;
86         for (i=15; i>=0; --i) {
87             scanf("%d", &j);
88             if (j)
89                 beg |= (1<<i);
90         }
91         k = visit[beg];
92         if (k < 0)
93             printf("Case #%d: more\n", t);
94         else
95             printf("Case #%d: %d\n", t, k);
96     }
97
98     return 0;
99 }