首页 > 代码库 > poj 2676
poj 2676
Sudoku
Time Limit: 2000MS | Memory Limit: 65536K | |||
Total Submissions: 13665 | Accepted: 6767 | 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
1 103000509 002109400 000704000 300502006 060000050 700803004 000401000 009205800 804000107
Sample Output
143628579 572139468 986754231 391542786 468917352 725863914 237481695 619275843 854396127
Source
Southeastern Europe 2005
AC代码:
#include<iostream> using namespace std; char G[15][15]; char ch[15]={'0','1','2','3','4','5','6','7','8','9'}; int a[15]={0,1,1,1,4,4,4,7,7,7}; int sucess; int judge(int x,int y,int k){ for(int i=1;i<=9;i++){ if(G[x][i]==ch[k]) return 0; } for(int i=1;i<=9;i++){ if(G[i][y]==ch[k]) return 0; } for(int i=a[x];i<=a[x]+2;i++) for(int j=a[y];j<=a[y]+2;j++) if(G[i][j]==ch[k]) return 0; return 1; } void dfs(int x,int y){ if(sucess) return ; if(x*y>81){ for(int i=1;i<=9;i++){ for(int j=1;j<=9;j++) cout<<G[i][j]; cout<<endl; } sucess=1; return ; } if(G[x][y]!='0'){ if(y==9) dfs(x+1,1); else dfs(x,y+1); } else{ for(int k=1;k<=9;k++){ if(judge(x,y,k)){ G[x][y]=ch[k]; if(y==9) dfs(x+1,1); else dfs(x,y+1); G[x][y]='0'; } } } } int main(){ int T; cin>>T; while(T--){ for(int i=1;i<=9;i++) cin>>G[i]+1; sucess=0; dfs(1,1); } return 0; }
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。