POJ-2676 Sudoku （DFS）
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
题目分析：数独题。就这道数独题而言仅仅DFS是可以AC的。先建立一棵搜索树，然后就搜！
代码如下：
1 # include<iostream> 2 # include<cstdio> 3 # include<cstring> 4 # include<algorithm> 5 using namespace std; 6 char p[12][12]; 7 int cnt,a[12][12]; 8 struct Position 9 { 10 int x,y; 11 Position(){} 12 Position(int a,int b):x(a),y(b){} 13 bool operator < (const Position &a) const { 14 if(x==a.x) 15 return y<a.y; 16 return x<a.x; 17 } 18 }; 19 bool yy; 20 Position pos[100]; 21 bool ok(int x,int y,int m) 22 { 23 for(int k=1;k<=9;++k){ 24 if(a[x][k]==m||a[k][y]==m) 25 return false; 26 } 27 int it=x,jt=y; 28 while(it%3!=1) 29 --it; 30 while(jt%3!=1) 31 --jt; 32 for(int i=it;i<=it+2;++i) 33 for(int j=jt;j<=jt+2;++j) 34 if(a[i][j]==m) 35 return false; 36 return true; 37 } 38 void dfs(int p) 39 { 40 if(yy) 41 return ; 42 if(p==cnt){ 43 for(int i=1;i<=9;++i){ 44 for(int j=1;j<=9;++j) 45 printf("%d",a[i][j]); 46 printf("\n"); 47 } 48 yy=true; 49 return ; 50 } 51 int x=pos[p].x,y=pos[p].y; 52 for(int i=1;i<=9;++i){ 53 if(ok(x,y,i)){ 54 a[x][y]=i; 55 dfs(p+1); 56 a[x][y]=0; ///注意要将这个位置重新置为0，否则影响接下来的搜索过程。 57 } 58 if(yy) 59 return ; 60 } 61 } 62 int main() 63 { 64 int T; 65 scanf("%d",&T); 66 while(T--) 67 { 68 cnt=0; 69 for(int i=1;i<=9;++i){ 70 scanf("%s",p[i]+1); 71 for(int j=1;j<=9;++j){ 72 a[i][j]=p[i][j]-'0'; 73 if(a[i][j]==0){ 74 pos[cnt++]=Position(i,j); 75 } 76 } 77 } 78 sort(pos,pos+cnt); 79 yy=false; 80 dfs(0); 81 } 82 return 0; 83 }
