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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
1103000509002109400000704000300502006060000050700803004000401000009205800804000107Sample Output
143628579572139468986754231391542786468917352725863914237481695619275843854396127Source
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("\n"); 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 }
POJ2676 Sudoku
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