Sudoku

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 19230		Accepted: 9241		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

1
103000509
002109400
000704000
300502006
060000050
700803004
000401000
009205800
804000107

Sample Output

143628579
572139468
986754231
391542786
468917352
725863914
237481695
619275843
854396127

數獨問題；注意DFS在回溯時要進行還原，注意標誌位的運用（只要求輸出一個結果），迭代從0到80比較慢，從80到0較快

import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.util.Scanner;

public class Test39_poj2676 {
	static int[][] Maze = new int[9][9];
	static int flsg = 0;// 標誌位

	public static void main(String[] args) throws FileNotFoundException {
		Scanner in = new Scanner(new FileInputStream(
				"E:\\Test_Case\\Test39_poj2676.txt"));
		// Scanner in = new Scanner(System.in);
		int caseNum = in.nextInt();
		for (int i = 0; i < caseNum; i++) {
			for (int j = 0; j < 9; j++) {
				String rowNum = in.next();
				for (int m = 0; m < rowNum.length(); m++) {
					Maze[j][m] = Integer.parseInt(rowNum.substring(m, m + 1));
				}
			}
			flsg = 0;// 初始化標誌位的值
			// 開始迭代，倒著迭代
			DFS(80);
		}
	}

	// 迭代體
	public static void DFS(int step) {
		int row = step / 9;// 點的橫座標
		int column = step % 9;// 點的縱座標
		// 結束條件
		if (step == -1) {
			flsg = 1;// 將標誌位置1
			// 輸出結果
			for (int i = 0; i < 9; i++) {
				for (int j = 0; j < 9; j++) {
					System.out.print(Maze[i][j]);
				}
				System.out.println();
			}
			return;
		}
		if (Maze[row][column] == 0) {// 如果這個位置的值為0
			for (int temp = 1; temp < 10; temp++) {// 從1到9進行遍歷
				// 判定行、列、分割槽是否有重複
				if (isSafe(row, column, temp)) {
					Maze[row][column] = temp;// 賦值
					DFS(step - 1);// 繼續下一次迭代
					Maze[row][column] = 0;// 這個地方要進行還原！
					// 只要求輸出一個結果，所以當標誌位為1（已經輸出一次）時，直接退出
					if (flsg == 1) {
						return;
					}
				}
			}
		} else {// 如果這個位置的值不為0，繼續下個位置
			DFS(step - 1);
		}
	}

	// 判斷在行、列、分割槽是否有重複
	public static boolean isSafe(int row, int column, int temp) {
		boolean isF = true;
		for (int i = 0; i < 9; i++) {
			// 1.行不能有重複
			if (Maze[row][i] == temp) {
				isF = false;
				return isF;
			}
			// 2.列不能有重複
			if (Maze[i][column] == temp) {
				isF = false;
				return isF;
			}
		}
		// 3.確定分割槽，分割槽內也不能有重複
		int[] tempX = partition(row);
		int[] tempY = partition(column);
		// 遍歷此點所在的分割槽
		for (int i = tempX[0]; i <= tempX[1]; i++) {
			for (int j = tempY[0]; j <= tempY[1]; j++) {
				if (Maze[i][j] == temp) {
					isF = false;
					return isF;
				}
			}
		}
		return isF;
	}

	// 確定分割槽的起止座標
	public static int[] partition(int ind) {
		int[] DX = new int[2];
		if (ind < 3) {
			DX[0] = 0;
			DX[1] = 2;
			return DX;
		} else if (ind < 6 && ind > 2) {
			DX[0] = 3;
			DX[1] = 5;
			return DX;
		} else {
			DX[0] = 6;
			DX[1] = 8;
			return DX;
		}
	}
}