HDOJ-三部曲一(搜索、数学)-1013-Sudoku

Sudoku

Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 8   Accepted Submission(s) : 8

Special Judge

Problem 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

 

Source

PKU

 

 

一道DFS题，虽然过了但耗时都在几百ms，我写了3次，前两次耗时都在7、8百ms，最后一次500ms

 

 

#include<iostream>
#include<cstring>
#include<string>
using namespace std;
int Sudoku[9][9];
int Count,c,vacant[81][2];
bool ro[9][10],co[9][10],sq[3][3][10];               //分别标记每行、每列、每个3*3方格各个数字是否出现过

void DFS(int i)
{
	if(c==Count)
		return;

	int k;
	for(k=1;k<10;k++)
	{
		if(!ro[vacant[i][0]][k]&&!co[vacant[i][1]][k]&&!sq[vacant[i][0]/3][vacant[i][1]/3][k])//如果没出现过
		{
			Sudoku[vacant[i][0]][vacant[i][1]]=k;
			c++;
			ro[vacant[i][0]][k]=true;
			co[vacant[i][1]][k]=true;
			sq[vacant[i][0]/3][vacant[i][1]/3][k]=true;
			DFS(i+1);
			if(c==Count)
				return;
			Sudoku[vacant[i][0]][vacant[i][1]]=0;
			c--;
			ro[vacant[i][0]][k]=false;
			co[vacant[i][1]][k]=false;
			sq[vacant[i][0]/3][vacant[i][1]/3][k]=false;
		}
	}
	return;
}

int main()
{
	int T;
	cin>>T;
	while(T--)
	{
		int i,j;
		Count=0;
		c=0;
		string temp[9];
		memset(ro,false,sizeof(ro));
		memset(co,false,sizeof(co));
		memset(sq,false,sizeof(sq));
		memset(vacant,0,sizeof(vacant));
		for(i=0;i<9;i++)
			cin>>temp[i];
		for(i=0;i<9;i++)
		{
			for(j=0;j<9;j++)
			{
				Sudoku[i][j]=temp[i][j]-'0';
				ro[i][Sudoku[i][j]]=true;
				co[j][Sudoku[i][j]]=true;
				sq[i/3][j/3][Sudoku[i][j]]=true;
				if(Sudoku[i][j]==0)
				{
					vacant[Count][0]=i;          //将需要填的空格位置保存
					vacant[Count][1]=j;
					Count++;                     //计算空格个数
				}
			}
		}
		DFS(0);
		for(i=0;i<9;i++)
		{
			for(j=0;j<9;j++)
				cout<<Sudoku[i][j];
			cout<<endl;
		}
	}
}