POJ2676 Sudoku(dfs)

Sudoku

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 16161 Accepted: 7894 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

1103000509002109400000704000300502006060000050700803004000401000009205800804000107

Sample Output

143628579572139468986754231391542786468917352725863914237481695619275843854396127

数独游戏，0代表空格，让你填入数字，使得每行每列都是1－9。

    通过i，j，与网格关系可得出关系式：k = 3 * ((i - 1) / 3) + (j - 1) / 3 + 1，用来标记网格vis。每组数据从(1, 1)开始dfs，dfs内首先判断当前数字是0还是非0，若为非0则判断是否已经到了第九列，是的话就dfs下一行，否则dfs下一列(也就是遍历所有)。若当前数字为0则从1到9扫一遍，若扫到当前行当前列且网格的第i个未访问，就标记，而后再判断是否是第9列，最后判断是否符合题意，不符合则当前网格修改回去。

AC代码：

#include "iostream"#include "cstdio"#include "cstring"#include "algorithm"using namespace std;const int MAXN = 10;int map[MAXN][MAXN];bool row[MAXN][MAXN], col[MAXN][MAXN], vis[MAXN][MAXN];bool dfs(int x, int y){if(x == 10) return true;bool flag = false;if(map[x][y]) {if(y == 9) flag = dfs(x + 1, 1);else flag = dfs(x, y + 1);if(flag) return true;else return false;}else {int k = 3 * ((x - 1) / 3) + (y - 1) / 3 + 1;for(int i = 1; i <= 9; ++i)if(!row[x][i] && !col[y][i] && !vis[k][i]) {map[x][y] = i;row[x][i] = true;col[y][i] = true;vis[k][i] = true;if(y == 9) flag = dfs(x + 1, 1);else flag = dfs(x, y + 1);if(!flag) {map[x][y] = 0;row[x][i] = false;col[y][i] = false;vis[k][i] = false;}else return true;}}return false;}int main(int argc, char const *argv[]){int t;cin >> t;while(t--) {memset(row, false, sizeof(row));memset(col, false, sizeof(col));memset(vis, false, sizeof(vis));char s[MAXN][MAXN];for(int i = 1; i <= 9; ++i)for(int j = 1; j <= 9; ++j) {cin >> s[i][j];map[i][j] = s[i][j] - '0';if(map[i][j]) {int k = 3 * ((i - 1) / 3) + (j - 1) / 3 + 1;row[i][map[i][j]] = true;col[j][map[i][j]] = true;vis[k][map[i][j]] = true;}}dfs(1, 1);for(int i = 1; i <= 9; ++i) {for(int j = 1; j <= 9; ++j)cout << map[i][j];cout << endl;}}return 0;}