Sudoku
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Submit Status Practice POJ 3074
Appoint description:
System Crawler (2015-04-18)
Description
In the game of Sudoku, you are given a large 9 × 9 grid divided into smaller 3 × 3 subgrids. For example,
| . | 2 | 7 | 3 | 8 | . | . | 1 | . |
| . | 1 | . | . | . | 6 | 7 | 3 | 5 |
| . | . | . | . | . | . | . | 2 | 9 |
| 3 | . | 5 | 6 | 9 | 2 | . | 8 | . |
| . | . | . | . | . | . | . | . | . |
| . | 6 | . | 1 | 7 | 4 | 5 | . | 3 |
| 6 | 4 | . | . | . | . | . | . | . |
| 9 | 5 | 1 | 8 | . | . | . | 7 | . |
| . | 8 | . | . | 6 | 5 | 3 | 4 | . |
Given some of the numbers in the grid, your goal is to determine the remaining numbers such that the numbers 1 through 9 appear exactly once in (1) each of nine 3 × 3 subgrids, (2) each of the nine rows, and (3) each of the nine columns.
Input
The input test file will contain multiple cases. Each test case consists of a single line containing 81 characters, which represent the 81 squares of the Sudoku grid, given one row at a time. Each character is either a digit (from 1 to 9) or a period (used to indicate an unfilled square). You may assume that each puzzle in the input will have exactly one solution. The end-of-file is denoted by a single line containing the word “end”.
Output
For each test case, print a line representing the completed Sudoku puzzle.
Sample Input
.2738..1..1...6735.......293.5692.8...........6.1745.364.......9518...7..8..6534. ......52..8.4......3...9...5.1...6..2..7........3.....6...1..........7.4.......3. end
Sample Output
527389416819426735436751829375692184194538267268174593643217958951843672782965341 416837529982465371735129468571298643293746185864351297647913852359682714128574936 用dancing links解数独，DLX专题的终极目标，手机上的数独玩了两天后终于A了，等下把记录用这个程序刷一遍，想象下小伙伴看到我最高难度下的耗时记录的表情，啊哈哈~~难点在于怎么建立模型，一共有9行，每行有9个数字，所以就是9 * 9，同理，列也是9 * 9，小格子也是有9个，每个也是9个数字，所以也是9 * 9，另外整个图有9 * 9 = 81的格子。所以要覆盖的列就是 9 * 9 + 9 * 9 + 9 * 9 + 81。至于行，一共有81个格子，每个格子有9种取法，所以就有81 * 9行，行和列相乘就是开的数组的大小。还有个剪枝要注意下，如果某个格子的数字已经给出，那么这一行，这一列，这一个小格子，就没必要再填这个数了，不剪枝的话会T。
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <string>
5 #include <cstdlib>
6 #include <cmath>
7 #include <map>
8 #include <cctype>
9 using namespace std;
10
11 const int HEAD = 0;
12 const int SIZE = (81 * 9) * (81 * 4);
13 const int COL = 81 * 4;
14 int U[SIZE],D[SIZE],L[SIZE],R[SIZE],S[SIZE],C[SIZE],N[SIZE],P_H[SIZE],P_C[SIZE];
15 int COUNT;
16 int TEMP[100][100];
17 bool VIS_I[15][15],VIS_J[15][15],VIS_G[15][15];
18 struct Node
19 {
20 int i;
21 int j;
22 int num;
23 }ANS[100];
24
25 void ini(void);
26 void link(int,int,int,int,int,int,int);
27 bool dancing(int);
28 void remove(int);
29 void resume(int);
30 void debug(int);
31 int main(void)
32 {
33 char s[1000];
34 while(scanf(" %s",s + 1) && strcmp(s + 1,"end"))
35 {
36 ini();
37 for(int i = 1;i <= 9;i ++)
38 for(int j = 1;j <= 9;j ++)
39 {
40 int k = s[(i - 1) * 9 + j];
41 int c_1,c_2,c_3,c_4;
42 if(k != ‘.‘)
43 {
44 VIS_I[i][k - ‘0‘] = VIS_J[j][k - ‘0‘] = true;
45 VIS_G[(i - 1) / 3 * 3 + (j - 1) / 3 + 1][k - ‘0‘] = true;
46 c_1 = 81 * 0 + (i - 1) * 9 + k - ‘0‘;
47 c_2 = 81 * 1 + (j - 1) * 9 + k - ‘0‘;
48 c_3 = 81 * 2 + ((i - 1) / 3 * 3 + (j - 1) / 3) * 9 + k - ‘0‘;
49 c_4 = 81 * 3 + (i - 1) * 9 + j;
50 link(c_1,c_2,c_3,c_4,k - ‘0‘,i,j);
51 }
52 }
53
54 for(int i = 1;i <= 9;i ++)
55 for(int j = 1;j <= 9;j ++)
56 {
57 if(s[(i - 1) * 9 + j] != ‘.‘)
58 continue;
59 int c_1,c_2,c_3,c_4;
60 for(int k = 1;k <= 9;k ++)
61 {
62 if(VIS_I[i][k] || VIS_J[j][k] ||
63 VIS_G[(i - 1) / 3 * 3 + (j - 1) / 3 + 1][k])
64 continue;
65 c_1 = 81 * 0 + (i - 1) * 9 + k;
66 c_2 = 81 * 1 + (j - 1) * 9 + k;
67 c_3 = 81 * 2 + ((i - 1) / 3 * 3 + (j - 1) / 3) * 9 + k;
68 c_4 = 81 * 3 + (i - 1) * 9 + j;
69 link(c_1,c_2,c_3,c_4,k,i,j);
70 }
71 }
72 dancing(0);
73 }
74
75 return 0;
76 }
77
78 void ini(void)
79 {
80 L[HEAD] = COL;
81 R[HEAD] = 1;
82 for(int i = 1;i <= COL;i ++)
83 {
84 L[i] = i - 1;
85 R[i] = i + 1;
86 U[i] = D[i] = C[i] = i;
87 S[i] = 0;
88 }
89 R[COL] = HEAD;
90
91 fill(&VIS_I[0][0],&VIS_I[12][12],false);
92 fill(&VIS_J[0][0],&VIS_J[12][12],false);
93 fill(&VIS_G[0][0],&VIS_G[12][12],false);
94 COUNT = COL + 1;
95 }
96
97 void link(int c_1,int c_2,int c_3,int c_4,int num,int r,int c)
98 {
99 int first = COUNT;
100 int col;
101
102 for(int i = 0;i < 4;i ++)
103 {
104 switch(i)
105 {
106 case 0:col = c_1;break;
107 case 1:col = c_2;break;
108 case 2:col = c_3;break;
109 case 3:col = c_4;break;
110 }
111
112 L[COUNT] = COUNT - 1;
113 R[COUNT] = COUNT + 1;
114 U[COUNT] = U[col];
115 D[COUNT] = col;
116
117 D[U[col]] = COUNT;
118 U[col] = COUNT;
119 C[COUNT] = col;
120 N[COUNT] = num;
121 P_H[COUNT] = r;
122 P_C[COUNT] = c;
123 S[col] ++;
124 COUNT ++;
125 }
126 L[first] = COUNT - 1;
127 R[COUNT - 1] = first;
128 }
129
130 bool dancing(int k)
131 {
132 if(R[HEAD] == HEAD)
133 {
134 for(int i = 0;i < k;i ++)
135 TEMP[ANS[i].i][ANS[i].j] = ANS[i].num;
136 int count = 0;
137 for(int i = 1;i <= 9;i ++)
138 for(int j = 1;j <= 9;j ++)
139 printf("%d",TEMP[i][j]);
140 puts("");
141 return true;
142 }
143
144 int c = R[HEAD];
145 for(int i = R[HEAD];i != HEAD;i = R[i])
146 if(S[c] > S[i])
147 c = i;
148
149 remove(c);
150 for(int i = D[c];i != c;i = D[i])
151 {
152 ANS[k].i = P_H[i];
153 ANS[k].j = P_C[i];
154 ANS[k].num = N[i];
155 for(int j = R[i];j != i;j = R[j])
156 remove(C[j]);
157 if(dancing(k + 1))
158 return true;
159 for(int j = L[i];j != i;j = L[j])
160 resume(C[j]);
161 }
162 resume(c);
163
164 return false;
165 }
166
167 void remove(int c)
168 {
169 L[R[c]] = L[c];
170 R[L[c]] = R[c];
171 for(int i = D[c];i != c;i = D[i])
172 for(int j = R[i];j != i;j = R[j])
173 {
174 D[U[j]] = D[j];
175 U[D[j]] = U[j];
176 S[C[j]] --;
177 }
178 }
179
180 void resume(int c)
181 {
182 L[R[c]] = c;
183 R[L[c]] = c;
184 for(int i = D[c];i != c;i = D[i])
185 for(int j = L[i];j != i;j = L[j])
186 {
187 D[U[j]] = j;
188 U[D[j]] = j;
189 S[C[j]] ++;
190 }
191 }
时间： 2024-11-08 03:44:56