POJ2676 Sudoku

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 18193		Accepted: 8803		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

依旧是数独题，解法见http://www.cnblogs.com/SilverNebula/p/5857587.html [靶形数独]

  1 /*by SilverN*/
  2 #include<algorithm>
  3 #include<iostream>
  4 #include<cstring>
  5 #include<cstdio>
  6 using namespace std;
  7 int mp[15][15];
  8 int block[15],h[15],r[15];
  9 int in[15][15];
 10 int sx[100],sy[100],cnt=0;
 11 inline int blcnt(int x){
 12     int res=9;
 13     while(x){
 14         res-=(x&1);
 15         x>>=1;
 16     }
 17     return res;
 18 }
 19 void init(){
 20     for(int i=1;i<=9;i++)
 21      for(int j=1;j<=9;j++)
 22         in[i][j]=(i+2)/3*3+(j+2)/3-3;
 23 }
 24 void Print(){
 25     for(int i=1;i<=9;i++){    
 26          for(int j=1;j<=9;j++)
 27              printf("%d",mp[i][j]);
 28          printf("
");
 29     }
 30 }
 31 bool dfs(){
 32     int i,j;
 33     int mini=60,pos=0;
 34     int tmp,now;
 35     int ctmp;
 36     for(i=1;i<=cnt;i++){
 37         if(!mp[sx[i]][sy[i]]){
 38             tmp=h[sx[i]]|r[sy[i]]|block[in[sx[i]][sy[i]]];
 39             now=blcnt(tmp);
 40             if(!now)return 0;
 41             if(now<mini){
 42                 ctmp=tmp;//记录最小值对应的tmp
 43                 //先前由于没有记录ctmp，而在后面计算时用tmp算，出现了bug，费了一个小时查出来错 
 44                 mini=now;
 45                 pos=i;
 46             }
 47         }
 48     }
 49     if(mini==60){
 50         Print();
 51         return 1;
 52     }
 53     i=sx[pos];
 54     j=sy[pos];
 55              for(int k=1;k<=9;k++){
 56                  if(!(ctmp&(1<<k))){
 57                      mp[i][j]=k;
 58                      h[i]^=(1<<k);
 59                      r[j]^=(1<<k);
 60                      block[in[i][j]]^=(1<<k);
 61                      if(dfs())return 1;
 62                      //回溯 
 63                      block[in[i][j]]^=(1<<k);
 64                      h[i]^=(1<<k);
 65                      r[j]^=(1<<k);
 66                      mp[i][j]=0;
 67                  }
 68              }
 69     return 0;
 70 }
 71 
 72 int main(){
 73     int i,j;
 74     int T;
 75     scanf("%d",&T);
 76     init();
 77     while(T--){
 78         memset(h,0,sizeof h);
 79         memset(r,0,sizeof r);
 80         memset(block,0,sizeof block);
 81         cnt=0;
 82 
 83         char ch[20];
 84         for(i=1;i<=9;i++){
 85             scanf("%s",ch);
 86             for(j=1;j<=9;j++){
 87                 mp[i][j]=ch[j-1]-'0';
 88             }
 89         }    
 90         for(i=1;i<=9;i++)//行 
 91          for(j=1;j<=9;j++){//列 
 92             if(mp[i][j]){
 93                  r[j]^=1<<mp[i][j];
 94                  h[i]^=1<<mp[i][j];
 95                  block[in[i][j]]^=1<<mp[i][j];
 96             }
 97             else{
 98                 cnt++;
 99                 sx[cnt]=i;
100                 sy[cnt]=j;
101             }
102         }
103         dfs();
104     }
105     return 0;
106 }