HDU 3111 Sudoku(数独,还是深搜)

Sudoku

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 464    Accepted Submission(s): 302

Problem Description

A Sudoku puzzle, once solved, is a 9x9 grid of digits organized as a 3x3 grid of smaller 3x3 units. Each of the nine rows must contain every positive digits exactly once, as do each column and also each 3x3 unit. The puzzle is to start from a partially filled 9x9 grid and to fill in the remaining cells using only logic. The puzzle maker usually makes sure that the solution will be unique and that it can be reached using deduction only, without guessing.

This number placing game is gaining popularity in the west, and every second newspaper publishes weekly instances of the puzzle. Somewhere at the head of one such newspaper, someone decided that buying individual instances from a puzzle maker would be too expansive, and instead decided to steal puzzles from other newspapers and also to print randomly generated Sudoku-like grids.

One week later, his assistant gets stuck with the job of printing the solution to the Sudoku puzzles his boss previously published. Unfortunately, his boss doesn’t have those solutions, the randomly generated problems don’t have any solution, and he doesn’t even remember which is which. In despair, the assistant calls for your help.

 

Input

The first line of the input will contain the number of test cases. Each test case will consist of a 9 by 9 grid of characters, where each character will either be ‘?’ or a digit between 1 and 9 inclusively.

 

Output

For each test case, you must print back the grid to the standard input, replacing each question mark with an appropriate digit to solve the Sudoku. If a test case does not allow any solution, output ”impossible” instead of a completed grid. If a test case do allow a solution, you can assume that the solution will be unique, and that theoretically it could be reached without guessing.

Test cases are separated by “---” both in the input and in the output.

 

Sample Input

   
   
   
   

    
    
    
    3
??4??9??8
?3??5??1?
7??4??2??
3??8??1??
?5?????9?
??6??1??2
??8??3??1
?2??4??5?
6??1??7??
---
??4??9??8
?3??5??1?
7??4??2??
3??8??1??
?5?????9?
??6??1??2
??8??3??1
62??4??5?
6??1??7??
---
3?1????76
7??9?????
2?5?3????
?????64?1
???2?1???
1?25?????
????8?9?3
?????9??4
51????7?8
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    264319578
839257416
715486239
372894165
451632897
986571342
548763921
127948653
693125784
---
impossible
---
391452876
764918532
285637149
953876421
678241395
142593687
426785913
837169254
519324768
   
   
   
   

 

Source

IPCP 2005 Northern Preliminary for Northeast North-America

原题链接:http://acm.hdu.edu.cn/showproblem.php?pid=3111

还是数独,和以前一题差不多,也是数独,链接http://blog.csdn.net/hurmishine/article/details/51333913

但这题更加容易一点了

AC代码:

#include <iostream>
#include <cstring>
using namespace std;
int a[10][10];
int sum;
typedef struct Node
{
    int x,y;
};
Node node[100];
bool flag;
bool OK(int p,int x)//在第p个位置放x
{
    for(int i=0; i<9; i++)
    {
        if(a[node[p].x][i]==x||a[i][node[p].y]==x)
            return false;
    }
    int xx=node[p].x/3*3;
    int yy=node[p].y/3*3;
    for(int i=xx; i<xx+3; i++)
        for(int j=yy; j<yy+3; j++)
            if(a[i][j]==x)
                return false;
    return true;
}
void DFS(int step)
{
    if(step==sum)
    {
        for(int i=0; i<9; i++)
        {
            for(int j=0; j<9; j++)
                cout<<a[i][j];
            cout<<endl;
        }
        flag=true;
    }
    if(flag)
        return;
    for(int i=1; i<=9; i++)
    {
        if(OK(step,i))
        {
            a[node[step].x][node[step].y]=i;
            DFS(step+1);
            a[node[step].x][node[step].y]=0;
        }
    }
}
int main()
{
    int n;
    cin>>n;
    char c,ch[5];
    sum=0;
    for(int ii=1; ii<=n; ii++)
    {
        sum=0;
        for(int i=0; i<9; i++)
        {
            for(int j=0; j<9; j++)
            {
                cin>>c;
                if(c=='?')
                {
                    a[i][j]=0;
                    node[sum].x=i;
                    node[sum].y=j;
                    sum++;
                }
                else
                    a[i][j]=c-'0';
            }
        }
        if(ii<n)
            cin>>ch;
        flag=false;
        DFS(0);
        if(!flag)
            cout<<"impossible"<<endl;
        if(ii<n)
            cout<<"---"<<endl;
    }
    return 0;
}