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

1
103000509
002109400
000704000
300502006
060000050
700803004
000401000
009205800
804000107

Sample Output

143628579
572139468
986754231
391542786
468917352
725863914
237481695
619275843
854396127


  1 #include <iostream>
  2 #include <stdio.h>
  3 #include <string.h>
  4 using namespace std;
  5 int num[10][10];
  6 struct node
  7 {
  8     int x,y;
  9 }dd[85];
 10 int cnt;
 11 bool mark;
 12 bool judge1(int row,int col,int value)//判断一行中是否存在矛盾
 13 {
 14     int i;
 15     for(i=0;i<9;i++)
 16     {
 17         if(num[row][i]==value && i!=col)
 18         {
 19             return false;
 20         }
 21     }
 22     return true;
 23 }
 24 
 25 bool judge2(int row,int col,int value)//判断一列中是否存在矛盾
 26 {
 27     int i;
 28     for(i=0;i<9;i++)
 29     {
 30         if(num[i][col]==value && i!=row)
 31         {
 32             return false;
 33         }
 34     }
 35     return true;
 36 }
 37 
 38 bool judge3(int row,int col,int value)
 39 {
 40     int rr=row/3*3;
 41     int cc=col/3*3;
 42     int i,j;
 43     for(i=rr;i<rr+3;i++)
 44     {
 45         for(j=cc;j<cc+3;j++)
 46         {
 47             if(num[i][j]==value && !(row==i && col==j))
 48             {
 49                 return false;
 50             }
 51         }
 52     }
 53     return true;
 54 }
 55 
 56 void dfs(int step)
 57 {
 58     if(step==cnt)
 59     {
 60         int i,j;
 61         for(i=0;i<9;i++)
 62         {
 63             for(j=0;j<9;j++)
 64             {
 65                 printf("%d",num[i][j]);
 66             }
 67             printf("\n");
 68         }
 69         mark=true;
 70         return;
 71     }
 72     if(mark)
 73     {
 74         return;
 75     }
 76     int i;
 77     for(i=1;i<=9;i++)
 78     {
 79         num[dd[step].x][dd[step].y]=i;
 80         if(!judge1(dd[step].x,dd[step].y,i) || !judge2(dd[step].x,dd[step].y,i) || !judge3(dd[step].x,dd[step].y,i))
 81         {
 82             num[dd[step].x][dd[step].y]=0;
 83             continue;
 84         }
 85         dfs(step+1);
 86         if(mark)
 87         {
 88             return;
 89         }
 90         num[dd[step].x][dd[step].y]=0;
 91     }
 92 }
 93 int main()
 94 {
 95     int t;
 96     scanf("%d",&t);
 97     while(t--)
 98     {
 99         char str[10];
100         int i,j;
101         cnt=0;
102         for(i=0;i<9;i++)
103         {
104             scanf("%s",str);
105             for(j=0;j<9;j++)
106             {
107                 if(str[j]=='0')
108                 {
109                     num[i][j]=0;
110                     dd[cnt].x=i;
111                     dd[cnt].y=j;
112                     cnt++;
113                 }
114                 else
115                 {
116                     num[i][j]=str[j]-'0';
117                 }
118             }
119         }
120         mark=false;
121         dfs(0);
122     }
123     return 0;
124 }
View Code

数独游戏。

暴力深搜；

注意写好三个判断函数judge1（），judge2()，judge3()其他的都是简单深搜的思想

posted @ 2013-06-04 19:28  欧阳生朵  阅读(165)  评论(0)  编辑  收藏  举报