sudoku-------九宫格

最新推荐文章于 2020-12-29 21:19:44 发布

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最新推荐文章于 2020-12-29 21:19:44 发布

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分类专栏： DFS

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本文介绍了一种使用回溯法解决数独问题的算法实现，通过判断每一个空位填入数字的可能性来逐步填充完整的数独表格。

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Sudoku

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

Total Submission(s) : 88 Accepted Submission(s) : 28

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

Sample Output

下面是我的代码：

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
int flag;
int map[9][9];

int panduan(int x,int y,int z)
{
    for(int j=0;j<9;j++)
    {
        if(map[x][j]==z&&j!=y)
        {
            return 0;
        }
        if(map[j][y]==z&&j!=x)
        {
            return 0;
        }
    }
    int dx=x/3*3;
    int dy=y/3*3;
    for(int i=dx;i<dx+3;i++)
    {
        for(int j=dy;j<dy+3;j++)
        {
            if(map[i][j]==z&&i!=x&&j!=y)
            return 0;
        }
    }
    return 1;

}
void dfs(int k)
{
    if(k==81)
    {
        for(int i=0;i<9;i++)
        {
            for(int j=0;j<9;j++)
            {
                 printf("%d",map[i][j]);
            }
            cout<<endl;
        }
        flag=1;
    return ;
    }
    if(flag==1)
    return ;

    int x=k/9;
    int y=k%9;
    if(map[x][y]==0)
    {
        for(int i=1;i<=9;i++)
        {
            map[x][y]=i;
            if(panduan(x,y,map[x][y])==1)
            dfs(k+1);
        }
        map[x][y]=0;
    }
    else
    dfs(k+1);
}
int main()
{
    int tt;

    string c;
    cin>>tt;
    //scanf("%s",&tt);
    c.resize(9);
    while(tt--)
    {
        memset(map,0,sizeof(map));
        flag=0;
        for(int i=0;i<9;i++)
        {
            scanf("%s",&c[0]);
            for(int j=0;j<9;j++)
            {
                //scanf("%s",&c[0]);
                map[i][j]=c[j]-'0';
            }
        }
        dfs(0);
    }
    return 0;
}