poj2676 Sudoku

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

Output

Sample Input

1
103000509
002109400
000704000
300502006
060000050
700803004
000401000
009205800
804000107

Sample Output

143628579
572139468
986754231
391542786
468917352
725863914
237481695
619275843
854396127

题意：数独：玩家须要依据9×9盘面上的已知数字，推理出全部剩余空格的数字，并满足每一行、每一列、每个粗线宫内的数字均含1-9，不反复。

思路： 爆搜，枚举，设置一个二维数组x[i][k]表示第i行有木有出现过数字k，y[j][k]表示第j列有木有出现过数字k，xg[(i/3)*3+j/3][k]表示第(i/3)*3+j/3个宫格有木有出现过 数字k。

#include<iostream>

#include<cstdio>

#include<cstdlib>
#include<string>
#include<cstring>
#include<list>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<algorithm>
#include<cmath>
#include<bitset>
#include<climits>
#define mem(a) memset(a,0,sizeof(a))
#define MAXN 100000
using namespace std;
char vw[9][9];
int x[9][9],y[9][9],xg[9][9];
bool vis;
int flag=0;
void dfs(int a,int b)
{
    int u=a*9+b+1;//这个公式相应以下的<span style="font-family: arial, 宋体, sans-serif;"> dfs(u/9,u%9)，非常静止，自己想想为啥</span>
    if(a==9)
    {
        vis=true;//填好就标记直接返回
        for(int i=0; i<9; i++)
        {
            for(int j=0; j<9; j++)
                printf("%d",vw[i][j]+1);
            printf("\n");
        }
    }
    if(vis)<span style="font-family: arial, 宋体, sans-serif;">//填好就直接返回</span>
        return ;
    if(vw[a][b]!=-1)
    {
        dfs(u/9,u%9);
        return ;
    }
    for(int i=0; i<9&&!vis; i++)
    {
        if(!x[a][i]&&!y[b][i]&&!xg[(a/3)*3+b/3][i])
        {
            x[a][i]=y[b][i]=xg[(a/3)*3+b/3][i]=1;
            vw[a][b]=i;
            dfs(u/9,u%9);
            x[a][i]=y[b][i]=xg[(a/3)*3+b/3][i]=0;
            vw[a][b]=-1;
        }
    }
}
int main()
{
    int i,j,k,t;
    char ch;
    cin>>t;
    while(t--)
    {
        mem(x);
        mem(y);
        mem(xg);
        mem(vw);
        for(i=0; i<9; i++)
            for(j=0; j<9; j++)
            {
                cin>>ch;
                k=vw[i][j]=ch-‘1‘;
                //printf("%d %d",k,vw[i][j]);
                if(ch-‘0‘)
                {
                    x[i][k]=y[j][k]=xg[(i/3)*3+j/3][k]=1;
                }
            }
        vis=false;//标记填没填好
        dfs(0,0);
    }
    return 0;
}