poj 2676 Sudoku

Sudoku

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 16894		Accepted: 8229		Special Judge

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

Source

Southeastern Europe 2005

思路：

DFS不为0的点1-9枚举，为0的跳过

本来没什么好说的，一个教训：循环的判别条件不要出现任何计算

15234481

njczy2010

2676

Accepted

700K

1547MS

G++

2083B

2016-03-06 09:55:21

  1 #include <cstdio>
  2 #include <cstring>
  3 #include <iostream>
  4 #include <algorithm>
  5 #include <stack>
  6 #include <cctype>
  7 #include <vector>
  8 #include <cmath>
  9 #include <map>
 10 #include <queue>
 11 
 12 #define ll long long
 13 
 14 using namespace std;
 15 
 16 int T;
 17 int ans[100][100];
 18 char s[100][100];
 19 int flag;
 20 
 21 struct PP
 22 {
 23     int x;
 24     int y;
 25 };
 26 
 27 int ok(int x,int y,int v)
 28 {
 29     int i,j;
 30     i = x;
 31     for(j = 1;j <= 9;j++){
 32         if(ans[i][j] == v) return 0;
 33     }
 34     j = y;
 35     for(i = 1;i <= 9;i++){
 36         if(ans[i][j] == v) return 0;
 37     }
 38     int rr = (x-1)/3*3;
 39     int cc = (y-1)/3*3;
 40     for(i = rr + 1 ;i <= rr + 3 ;i++  ){
 41         for(j =  cc +1 ;j <= cc + 3 ;j++  ){
 42             if(ans[i][j] == v) return 0;
 43         }
 44     }
 45     return 1;
 46 }
 47 
 48 void dfs(int x,int y)
 49 {
 50     int i;
 51     if(flag == 1){
 52         return;
 53     }
 54     if(x == 10){
 55         flag = 1;
 56         return;
 57     }
 58     int ff;
 59     if(ans[x][y] == 0){
 60         ff = 0;
 61         for(i = 1;i <= 9 ;i++){
 62             if(ok(x,y,i) == 1){
 63                 ans[x][y] = i;
 64                 ff = 1;
 65                 if(y == 9){
 66                     dfs(x+1,1);
 67                 }
 68                 else{
 69                     dfs(x,y+1);
 70                 }
 71                 if(flag == 1) return;
 72                 ans[x][y] = 0;
 73             }
 74         }
 75         if(ff == 0) return;
 76     }
 77     else{
 78         if(y == 9){
 79             dfs(x+1,1);
 80             if(flag == 1) return;
 81         }
 82         else{
 83             dfs(x,y+1);
 84             if(flag == 1) return;
 85         }
 86     }
 87 }
 88 
 89 int main()
 90 {
 91     //freopen("in.txt","r",stdin);
 92     scanf("%d",&T);
 93     for(int ccnt=1;ccnt<=T;ccnt++){
 94     //while(scanf("%d%s%s",&n,a,b)!=EOF){
 95         flag = 0;
 96         int i,j;
 97         for(i = 1;i <= 9;i++){
 98             scanf("%s",s[i] + 1);
 99         }
100         for(i = 1;i <= 9;i++){
101             for(j = 1;j <= 9;j++){
102                 ans[i][j] = s[i][j] - '0';
103             }
104         }
105         dfs(1,1);
106         for(i = 1;i <= 9;i++){
107             for(j = 1;j <= 9;j++){
108                 printf("%d",ans[i][j]);
109             }
110             printf("
");
111         }
112     }
113     return 0;
114 }