poj 2676 -- Sudoku

最新推荐文章于 2021-01-28 11:09:53 发布

weixin_34146986

最新推荐文章于 2021-01-28 11:09:53 发布

阅读量73

点赞数

原文链接：http://www.cnblogs.com/ubuntu-kevin/p/3895603.html

版权

本文介绍了一种解决数独游戏的方法，通过逆向搜索算法实现了数独的自动求解。详细阐述了输入数据格式、输出结果形式以及解题过程中的关键步骤。

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: 13723		Accepted: 6791		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

思路：这个题正着搜超时，倒着搜16ms

 1 /*======================================================================
 2  *           Author :   kevin
 3  *         Filename :   suduku.cpp
 4  *       Creat time :   2014-08-06 15:51
 5  *      Description :
 6  ========================================================================*/
 7 #include <iostream>
 8 #include <algorithm>
 9 #include <cstdio>
10 #include <cstring>
11 #include <queue>
12 #include <cmath>
13 #define clr(a,b) memset(a,b,sizeof(a))
14 #define M 20
15 using namespace std;
16 int s[M][M],row[M][M],col[M][M];
17 /*-------判断是否有3×3格子里是否有b-------*/
18 bool judge(int x,int y,int b)
19 {
20     int i,j;
21     for(i = (x-1)/3*3+1; i <= (x-1)/3*3+3; i++){
22         for(j = (y-1)/3*3+1; j <= (y-1)/3*3+3; j++){
23             if(s[i][j] == b) return false;
24         }
25     }
26     return true;
27 }
28 /*----------------end----------------*/
29 bool DFS(int x,int y) // x代表列，y代表行
30 {
31     if(x == 9 && y == 0){
32         return true;
33     }
34     if(s[y][x]){
35         if(x > 1){
36             if(DFS(x-1,y))
37                 return true;
38         }
39         if(x == 1){
40             if(DFS(9,y-1))
41                 return true;
42         }
43     }
44     else{
45         for(int i = 1; i <= 9; i++){      //枚举1到9
46             if(!row[y][i] && !col[x][i]){ //当前行当前列是否有i
47                 if(judge(y,x,i)){         //判断能否放在3×3方格里
48                     row[y][i] = 1; col[x][i] = 1;
49                     s[y][x] = i;
50                     if(x > 1){
51                         if(DFS(x-1,y)){
52                             return true;
53                         }
54                     }
55                     if(x == 1){
56                         if(DFS(9,y-1)){
57                             return true;
58                         }
59                     }
60                     row[y][i] = 0; col[x][i] = 0;
61                     s[y][x] = 0;
62                 }
63             }
64         }
65     }
66     return false;
67 }
68 int main(int argc,char *argv[])
69 {
70     int n;
71     scanf("%d",&n);
72     getchar();
73     while(n--){
74         clr(s,0);
75         clr(row,0);
76         clr(col,0);
77         char c;
78         for(int i  = 1; i <= 9; i++){
79             for(int j = 1; j <= 9; j++){
80                 scanf("%c",&c);
81                 s[i][j] = c-'0';
82                 row[i][s[i][j]] = 1;   //标记行里出现的数
83                 col[j][s[i][j]] = 1;   //标记列里出现的数
84             }
85             getchar();
86         }
87         DFS(9,9);          //倒着搜
88         for(int i = 1; i <= 9; i++){
89             for(int j = 1; j <= 9; j++){
90                 printf("%d",s[i][j]);
91             }
92             printf("\n");
93         }
94     }
95     return 0;
96 }