Problem J. Let Sudoku Rotate

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1363 Accepted Submission(s): 717

Problem Description

Sudoku is a logic-based, combinatorial number-placement puzzle, which is popular around the world.
In this problem, let us focus on puzzles with

Input

The first line of the input contains an integer

Output

For each test case, print a non-negative integer indicating the minimum possible number of operations.

Sample Input

1 681D5A0C9FDBB2F7 0A734B62E167D9E5 5C9B73EF3C208410 F24ED18948A5CA63 39FAED5616400B74 D120C4B7CA3DEF38 7EC829A085BE6D51 B56438F129F79C2A 5C7FBC4E3D08719F AE8B1673BF42A58D 60D3AF25619C30BE 294190D8EA57264C C7D1B35606835EAB AF52A1E019BE4306 8B36DC78D425F7C9 E409492FC7FA18D2

Sample Output

Hint

The original sudoku is same as the example in the statement.

Source

2018 Multi-University Training Contest 4

题意概括：

有一个 16×16 的已经打乱的数独，4×4为一宫，每次可对宫顺时针旋转 90 度。最少要操作多少次可以还原数独。

解题思路：

递归每一宫的左上角坐标 (x, y) ，DFS枚举每一宫的旋转次数，可知这样暴力的方案数为 4^(16) = 4294967296，需要剪枝。

由于数独的特殊性，我们枚举下一个状态前可以先判断上一个状态是否满足条件，即每行每列的数都要单独存在。

对于矩阵旋转的操作，找一下下标的规律：第一行变成第一列，第二行变成第二列，第三行变成第三列.....

AC code：

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<cstring>
 5 #include<vector>
 6 #include<queue>
 7 #include<cmath>
 8 #include<set>
 9 #define INF 0x3f3f3f3f
10 #define LL long long
11 using namespace std;
12 const int MAXN = 20;
13 char str[MAXN][MAXN];
14 int mmp[MAXN][MAXN];
15 int tmp[MAXN][MAXN];
16 bool vis[MAXN];
17 int ans;
18 
19 void rt(int x, int y)
20 {
21     int rx = 4*x, ry = 4*y-3;
22     for(int i = 4*x-3; i <= 4*x; i++){
23         rx = 4*x;
24         for(int k = 4*y-3; k <= 4*y; k++){
25             tmp[i][k] = mmp[rx][ry];
26             rx--;
27         }
28         ry++;
29     }
30 
31     for(int i = 4*x-3; i <= 4*x; i++){
32         for(int j = 4*y-3; j <= 4*y; j++){
33             mmp[i][j] = tmp[i][j];
34         }
35     }
36 
37 }
38 
39 bool check(int x, int y)
40 {
41     for(int i = 4*x-3; i <= 4*x; i++){
42         memset(vis, 0, sizeof(vis));
43         for(int j = 1; j <= 4*y; j++){
44             if(vis[mmp[i][j]]){
45                     return false;
46             }
47             vis[mmp[i][j]] = true;
48         }
49     }
50     for(int i = 4*y-3; i <= 4*y; i++){
51         memset(vis, 0, sizeof(vis));
52         for(int j = 1; j <= 4*x; j++){
53             if(vis[mmp[j][i]]){
54                 return false;
55             }
56             vis[mmp[j][i]] = true;
57         }
58     }
59     return true;
60 }
61 
62 void dfs(int x, int y, int cnt)
63 {
64     if(x > 4){
65         ans = min(ans, cnt);
66         return;
67     }
68     for(int i = 0; i < 4; i++){
69         if(i) rt(x, y);
70         if(check(x, y)){
71             if(y < 4) dfs(x, y+1, cnt+i);
72             else dfs(x+1, 1, cnt+i);
73         }
74     }
75     rt(x, y);
76 }
77 
78 int main()
79 {
80     int T_case;
81     scanf("%d", &T_case);
82     while(T_case--){
83         for(int i = 0; i < 16; i++){
84             scanf("%s", str[i]);
85         }
86         for(int i = 0; i < 16; i++){
87             for(int j = 0; j < 16; j++){
88                 if(str[i][j] >= '0' && str[i][j] <= '9'){
89                     mmp[i+1][j+1] = str[i][j]-'0';
90                 }
91                 else mmp[i+1][j+1] = str[i][j]-'A'+10;
92             }
93         }
94         ans = INF;
95         dfs(1, 1, 0);
96         printf("%d
", ans);
97     }
98     return 0;
99 }

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