POJ 2676 Sudoku

2020-10-19 09:32:39 作者：互联网

POJ 2676 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

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

Sample Output

题解：

暴搜填数独。稍稍有一点进阶的暴搜。思路也很简单：尝试往里填数，然后暴力判断即可。

这个暴力判断可以用位运算优化一下.但是本题我并没有这么干，用了个bool数组。

需要提一嘴的是这个第几个九宫格的处理。式子已经打在代码里了，大家手画模拟一下就能看出来这个东西的正确性。

代码：

#include<cstdio>
#include<cstring>
using namespace std;
int t,cnt;
int mp[11][11];
char mapi[11][11];
bool flag;
bool row[11][11],col[11][11],grid[11][11];
void print()
{
    for(int i=1;i<=9;i++)
    {
        for(int j=1;j<=9;j++)
            printf("%d",mp[i][j]);
        puts("");
    }
}
void dfs(int x,int y)
{
    if(flag)
        return;
    if(mp[x][y])
    {
        if(x==9&&y==9)
        {
            for(int i=1;i<=9;i++)
            {
                for(int j=1;j<=9;j++)
                    printf("%d",mp[i][j]);
                puts("");
            }
            flag=1;
            return;
        }
        if(y==9)
            dfs(x+1,1);
        else
            dfs(x,y+1);
    }
    if(!mp[x][y])
    {
        int k=3*((x-1)/3)+(y-1)/3+1;
        for(int i=1;i<=9;i++)
        {
            if(!row[x][i]&&!col[y][i]&&!grid[k][i])
            {
                row[x][i]=1;
                col[y][i]=1;
                grid[k][i]=1;
                mp[x][y]=i;
                if(x==9&&y==9)
                {
                    for(int i=1;i<=9;i++)
                    {
                        for(int j=1;j<=9;j++)
                            printf("%d",mp[i][j]);
                        puts("");
                    }
                    flag=1;
                    puts("");
                    return;
                }
                if(y==9)
                    dfs(x+1,1);
                else
                    dfs(x,y+1);
                row[x][i]=0;
                col[y][i]=0;
                grid[k][i]=0;
                mp[x][y]=0;
            }
        }
    }
}
int main()
{
    scanf("%d",&t);
    while(t--)
    {
        memset(col,0,sizeof(col));
        memset(row,0,sizeof(row));
        memset(grid,0,sizeof(grid));
        flag=0;
        for(int i=1;i<=9;i++)
            scanf("%s",mapi[i]+1);
        for(int i=1;i<=9;i++)
            for(int j=1;j<=9;j++)
            {
                mp[i][j]=mapi[i][j]-'0';
                if(mp[i][j])
                {
                    int k=3*((i-1)/3)+(j-1)/3+1;
                    row[i][mp[i][j]]=1;
                    col[j][mp[i][j]]=1;
                    grid[k][mp[i][j]]=1;
                }
            }
        dfs(1,1);
    }
    return 0;
}

标签：11,Sudoku,cells,2676,digits,POJ,each,empty
来源： https://www.cnblogs.com/fusiwei/p/13838478.html