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  • POJ 2676 Sudoku(深搜)

    Sudoku

    Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)
    Total Submission(s) : 15   Accepted Submission(s) : 12
    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
    1
    103000509
    002109400
    000704000
    300502006
    060000050
    700803004
    000401000
    009205800
    804000107
     
    Sample Output
    143628579
    572139468
    986754231
    391542786
    468917352
    725863914
    237481695
    619275843
    854396127
     
    Source
    PKU
     
     
    题意:

    九宫格问题,也有人叫数独问题

    把一个9行9列的网格,再细分为9个3*3的子网格

    如图

    要求每行、每列、每个子网格内都只能使用一次1~9中的一个数字,即每行、每列、每个子网格内都不允许出现相同的数字。

    0是待填位置,其他均为已填入的数字。

    要求填完九宫格并输出(如果有多种结果,则只需输出其中一种)

    如果给定的九宫格无法按要求填出来,则输出原来所输入的未填的九宫格

    思路:

    判断某一个不为零的格子,看看它的本行本列以及相联系的一个宫已有那些数字,然后再把

    有数字依次往里面填,填好这个方格后继续填写下一个方格,如果走到后面走不通了,就应该返回来,把

    刚刚它的那个数字还原为零,这样好下一次再继续填写另外的数,如果所有的方格都填完了,整个回溯也就结束了。

    AC代码:

     1 #include<cstdio>
     2 #include<cstdlib>
     3 #include<iostream>
     4 #include<algorithm>
     5 #include<cstring>
     6 #include<string>
     7 #include<cmath>
     8 
     9 using namespace std;
    10 
    11 
    12 int s[12][12]={0};//存储矩阵
    13 bool row[12][12];//row[i][j]表示第i行有没有j这给数字
    14 bool col[12][12];//col[i][j]表示第i列有没有j这给数字
    15 bool grid[12][12];//grid[k][j]表示第k个九宫格有没有j这给数字
    16 
    17 
    18 
    19 bool DFS(int x,int y)
    20 {
    21     if(x==10)
    22         return true;
    23     bool sgin=false;
    24     if(s[x][y]!=0){
    25         if(y==9)
    26             sgin=DFS(x+1,1);
    27         else
    28             sgin=DFS(x,y+1);
    29         if(sgin)
    30             return true;
    31         else
    32             return false;
    33     }
    34     else{
    35         int k=3*((x-1)/3)+(y-1)/3+1;
    36         for(int i=1;i<=9;i++)
    37             if(!row[x][i]&&!col[y][i]&&!grid[k][i]){//确定放入第x行,第y列,第k个九宫格都可以数
    38                 s[x][y]=i;
    39                 row[x][i]=true;
    40                 col[y][i]=true;
    41                 grid[k][i]=true;
    42                 if(y==9)
    43                     sgin=DFS(x+1,1);
    44                 else
    45                     sgin=DFS(x,y+1);
    46                 if(sgin)
    47                     return true;
    48                 else {//这个放入不可以,则返回上一层;
    49                     row[x][i]=false;
    50                     col[y][i]=false;
    51                     grid[k][i]=false;
    52                     s[x][y]=0;
    53                 }
    54             }
    55     }
    56     return false;
    57 }
    58 
    59 
    60 
    61 
    62 int main()
    63 {
    64 //    freopen("1.txt","r",stdin);
    65     int test;
    66     cin>>test;
    67     char a;
    68     while(test){
    69         memset(s,0,sizeof(s));
    70         memset(row,false,sizeof(row));
    71         memset(col,false,sizeof(col));
    72         memset(grid,false,sizeof(grid));
    73         for(int i=1;i<=9;i++){
    74             for(int j=1;j<=9;j++){
    75                 cin>>a;
    76                 s[i][j]=a-'0';
    77                 if(s[i][j]){//本来就不是0,进行排除
    78                 int k=((i-1)/3)*3+(j-1)/3+1;
    79                 row[i][s[i][j]]=true;
    80                 col[j][s[i][j]]=true;
    81                 grid[k][s[i][j]]=true;
    82                 }
    83             }
    84         }
    85         DFS(1,1);
    86         for(int i=1;i<=9;i++){
    87             for(int j=1;j<=9;j++){
    88                 cout<<s[i][j];
    89             }
    90             cout<<endl;
    91         }
    92         test--;
    93     }
    94     return 0;
    95 }
    View Code
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  • 原文地址:https://www.cnblogs.com/zhangchengbing/p/3360188.html
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