POJ 2676-Sudoku（DFS）

Sudoku

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 14120		Accepted: 6976		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

深搜。。

题意：就是传统的数独游戏。。找出一种方案即可。。有spj

对 行，列，以及9个3*3的方块哈希。。然后暴力深搜就行了。。实践证明倒搜比较快

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <string>
#include <cctype>
#include <vector>
#include <cstdio>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#define ll long long
#define maxn 360
#define pp pair<int,int>
#define INF 0x3f3f3f3f
#define max(x,y) ( ((x) > (y)) ? (x) : (y) )
#define min(x,y) ( ((x) > (y)) ? (y) : (x) )
using namespace std;
char ma[11][11];
bool row[11][11],col[11][11],mar[11][11],ok;
int change(int x,int y)
{
	if(x>=1&&x<=3) return y%3==0?y/3:y/3+1;
	else if(x>=4&&x<=6) return 3+(y%3==0?y/3:y/3+1);
	else return 6+(y%3==0?y/3:y/3+1);
}
void dfs(int num)
{
	if(ok) return ;
	if(num==0)
	{
		for(int i=1;i<=9;i++)
		{
			for(int j=1;j<=9;j++)
				putchar(ma[i][j]);
			puts("");
		}
		ok=1;
		return ;
	}
	int x=num%9==0?num/9:num/9+1;
	int y=num%9==0?9:num%9;
	if(ma[x][y]-'0')
		dfs(num-1);
	else
	{
		for(int i=1;i<=9;i++)
		{
			if(!row[x][i]&&!col[y][i]&&!mar[change(x,y)][i])
			{
				ma[x][y]=i+'0';row[x][i]=1;col[y][i]=1;mar[change(x,y)][i]=1;
				dfs(num-1);
				ma[x][y]='0';row[x][i]=0;col[y][i]=0;mar[change(x,y)][i]=0;
			}
		}
	}
}
int main()
{
	int T;
	scanf("%d",&T);
	while(T--)
	{
		memset(row,0,sizeof(row));
		memset(col,0,sizeof(col));
		memset(mar,0,sizeof(mar));
		for(int i=1;i<=9;i++)
			scanf("%s",ma[i]+1);
			for(int i=1;i<=9;i++)
				for(int j=1;j<=9;j++)
				if(ma[i][j]-'0')
				{
					row[i][ma[i][j]-'0']=1;
					col[j][ma[i][j]-'0']=1;
					mar[change(i,j)][ma[i][j]-'0']=1;
				}
		ok=0;dfs(81);
	}
	return 0;
}