UVa1603 The Rotation Game (IDA*)

Posted 2020-08-06

链接：http://acm.hust.edu.cn/vjudge/problem/36627
分析：感觉这题前面的处理比较麻烦。首先把图中的位置从左到右从上到下编号，然后用一个二维数组记录8个方向上各个位置的编号，枚举最大深度maxd，乐观估计函数为至少需要移动次数，因为每次移动最多改变一个位置上的数字，所以中间8个位置除了出现次数最多的数字外的数字个数为x的话那么就至少要移动x次，枚举move8个方向再递归如果check为true则成功找到解返回，否则将a复位继续枚举，枚举完8个方向找不到解则失败返回false。

 1 #include <cstdio>
 2 #include <algorithm>
 3 using namespace std;
 4 
 5 /*
 6       00    01
 7       02    03
 8 04 05 06 07 08 09 10
 9       11    12
10 13 14 15 16 17 18 19
11       20    21
12       22    23
13 */
14 
15 const int rev[8] = {5, 4, 7, 6, 1, 0, 3, 2};
16 const int center[8] = {6, 7, 8, 11, 12, 15, 16, 17};
17 int line[8][7] = {
18     {0, 2, 6, 11, 15, 20, 22},
19     {1, 3, 8, 12, 17, 21, 23},
20     {10, 9, 8, 7, 6, 5, 4},
21     {19, 18, 17, 16, 15, 14, 13},
22 };
23 
24 int a[24];
25 char ans[1000];
26 
27 int diff(int target) {
28     int cnt = 0;
29     for (int i = 0; i < 8; i++)
30         if (a[center[i]] != target) cnt++;
31     return cnt;
32 }
33 
34 inline int h() {
35     return min(min(diff(1), diff(2)), diff(3));
36 }
37 
38 bool check() {
39     for (int i = 0; i < 8; i++)
40         if (a[center[i]] != a[center[0]]) return false;
41     return true;
42 }
43 
44 void move(int i) {
45     int tmp = a[line[i][0]];
46     for (int j = 0; j < 6; j++) a[line[i][j]] = a[line[i][j + 1]];
47     a[line[i][6]] = tmp;
48 }
49 
50 bool dfs(int d, int maxd) {
51     if (check()) {
52         ans[d] = ‘\0‘;
53         printf("%s\n", ans);
54         return true;
55     }
56     if (maxd - d < h()) return false;
57     for (int i = 0; i < 8; i++) {
58         ans[d] = ‘A‘ + i;
59         move(i);
60         if (dfs(d + 1, maxd)) return true;
61         move(rev[i]);
62     }
63     return false;
64 }
65 
66 int main() {
67     for (int i = 4; i < 8; i++)
68         for (int j = 0; j < 7; j++)
69             line[i][j] = line[rev[i]][6 - j];
70     while (scanf("%d", &a[0]) == 1 && a[0]) {
71         for (int i = 1; i < 24; i++) scanf("%d", &a[i]);
72         for (int i = 0; i < 24; i++) if (!a[i]) return 0;
73         if (check()) printf("No moves needed\n");
74         else for (int maxd = 1; ; maxd++)
75             if (dfs(0, maxd)) break;
76         printf("%d\n", a[6]);
77     }
78     return 0;
79 }