Poj Sudoku (Sudoku) DFS

Sudoku

Time limit:2000 ms		Memory limit:65536 K
Total submissions:13665		Accepted:6767		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 shocould 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

1103000509002109400000704000300502006060000050700803004000401000009205800804000107

Sample output

143628579572139468986754231391542786468917352725863914237481695619275843854396127
Explanation: games have been played, but they have never been made. Haha, haha, and interesting code. I know the meaning of the question. The key is how to search, in fact, it is constantly enumerating and jumping out if all the conditions are met;
AC code:

1 # include <iostream> 2 # include <algorithm> 3 # include <cstdio> 4 # include <cstring> 5 using namespace STD; 6 const int n = 11; 7 int map [N] [N], flag; 8 char map [N] [N]; 9 int check (INT ans, int key) 10 {11 int x = (key-1)/9 + 1; 12 INT y = key-(x-1) * 9; 13 for (INT I = 1; I <= 9; I ++) // does the row conflict? 14 if (Map [x] [I] = ans) 15 return 0; 16 for (INT I = 1; I <= 9; I ++) // whether the column conflicts 17 if (Map [I] [Y] = ans) 18 return 0; 19 if (x <= 3) X = 1 ;// The starting point of X is 20 if (x> 3 & x <7) x = 4; 21 if (x> = 7) x = 7; 22 if (Y <= 3) y = 1; // the starting point of the small nine-grid where Y is located. 23 if (Y> 3 & Y <7) y = 4; 24 if (Y> = 7) y = 7; 25 for (INT I = 0; I <3; I ++) // whether the small nine cells CONFLICT 26 for (Int J = 0; j <3; j ++) 27 if (Map [x + I] [Y + J] = ans) 28 return 0; 29 return 1; // If none of the preceding conditions meet, return 1; 30} 31 int DFS (INT key) 32 {33 int x = (key-1)/9 + 1; 34 int y = key-(x-1) * 9; 35 if (Key> 81) 36 {37 flag = 1; 38 return 0; 39} 40 if (Map [x] [Y]! = 0) // if it is not 0, Jump directly to 41 {42 DFS (Key + 1); 43} 44 else // No. In this position, nine numbers are enumerated, perform recursive search for 45 {46 for (int K = 1; k <= 9; k ++) 47 If (check (K, key )) 48 {49 map [x] [Y] = K; 50 DFS (Key + 1); 51 if (flag = 1) return 0; 52 map [x] [Y] = 0; 53} 54} 55 return 0; 56} 57 int main () 58 {59 int T, I, J; 60 // freopen ("in2.txt", "r", stdin); 61 // freopen ("out2.txt", "W", stdout); 62 scanf ("% d ", & T); 63 while (t --) 64 {65 memset (MAP, 0, sizeof (MAP); 66 for (I = 1; I <= 9; I ++) 67 for (j = 1; j <= 9; j ++) 68 {69 CIN> map [I] [J]; 70 map [I] [J] = map [I] [J]-'0'; 71} 72 flag = 0; 73 DFS (1 ); 74 for (I = 1; I <= 9; I ++) 75 {76 for (j = 1; j <= 9; j ++) 77 printf ("% d", map [I] [J]); 78 printf ("\ n"); 79} 80} 81 return 0; 82}