POJ2676 Sudoku

最新推荐文章于 2019-11-15 15:57:17 发布

weixin_30516243

最新推荐文章于 2019-11-15 15:57:17 发布

阅读量102

点赞数

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/huibixiaoxing/p/7354535.html

本文介绍了一种解决Sudoku（数独）问题的算法实现，通过深度优先搜索（DFS）策略来填充空缺的单元格，确保每行、每列及每个3x3的小方格内的数字1到9各出现一次。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Sudoku

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 21055		Accepted: 10043		Special Judge

Description

Sudoku is a very simple task. A square table with 9 rows and 9 columns is divided to 9 smaller squares 3x3 as shown on the Figure. In some of the cells are written decimal digits from 1 to 9. The other cells are empty. The goal is to fill the empty cells with decimal digits from 1 to 9, one digit per cell, in such way that in each row, in each column and in each marked 3x3 subsquare, all the digits from 1 to 9 to appear. Write a program to solve a given Sudoku-task.

Input

The input data will start with the number of the test cases. For each test case, 9 lines follow, corresponding to the rows of the table. On each line a string of exactly 9 decimal digits is given, corresponding to the cells in this line. If a cell is empty it is represented by 0.

Output

For each test case your program should print the solution in the same format as the input data. The empty cells have to be filled according to the rules. If solutions is not unique, then the program may print any one of them.

Sample Input

Sample Output

Source

Southeastern Europe 2005

【题解】

爆搜题，本来我的参数里记录了剩余个数，然后愉快炸掉。。后来去了这个参数，突然飞凯过了这个题。。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cstdlib>
 5 #include <algorithm>
 6 
 7 inline void read(int &x)
 8 {
 9     x = 0;char ch = getchar(),c = ch;
10     while(ch < '0' || ch > '9')  c = ch, ch = getchar();
11     while(ch <= '9' && ch >= '0')x = x * 10 + ch - '0', ch = getchar();
12     if(c == '-')x = -x;
13 }
14 
15 int num[10][10], col[10][10], row[10][10], ge[10][10],ok[10][10],flag;
16 char ch[10];
17 
18 //还剩下cnt个数可以填 
19 int dfs(int x, int y)
20 {
21     register int pp;
22     if(x == 10)return 1;
23     if(num[x][y])
24     {
25         if(y == 9)flag = dfs(x + 1, 1);
26         else flag = dfs(x, y + 1);
27         if(flag)return 1;
28         else return 0;    
29     }
30     int i = x, j = y,p,q;
31     pp = (i - 1)/3 * 3 + (j + 2)/3;
32     for(register int t = 1;t <= 9;++ t)
33     {
34         if(row[i][t] || col[j][t] || ge[pp][t])continue;
35         row[i][t] = col[j][t] = ge[pp][t] = 1;
36         num[i][j] = t;
37         p = i, q = j;
38         if(q == 9)q = 0, ++ p;
39         flag = dfs(p, q + 1);
40         if(flag)return 1;
41         num[i][j] = 0, col[j][t] = row[i][t] = ge[pp][t] = 0;
42     }
43     return 0;
44 }
45 
46 int main()
47 {
48     register int tmp = 0, t;
49     read(t);
50     for(;t;--t)
51     {
52         memset(col, 0, sizeof(col));
53         memset(row, 0, sizeof(row));
54         memset(ge, 0, sizeof(ge));
55         for(int i = 1;i <= 9;++ i)
56         {
57             scanf("%s", ch + 1);
58             for(register int j = 1;j <= 9;++ j)
59             {
60                 num[i][j] = ch[j] - '0';
61                 if(num[i][j])
62                     ++ tmp,row[i][num[i][j]] = col[j][num[i][j]] = ge[(i - 1)/3 * 3 + (j + 2)/3][num[i][j]] = 1;
63             }
64         }
65         dfs(1,1);
66         for(int i = 1;i <= 9;++ i)
67         {
68             for(register int j = 1;j <= 9;++ j)
69                 printf("%d", num[i][j]);
70             putchar('\n');
71         }
72     }
73     return 0;
74 }