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ACM : POJ 2676 SudoKu DFS - 数独
SudoKu
Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu
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
1103000509002109400000704000300502006060000050700803004000401000009205800804000107
Sample Output
143628579572139468986754231391542786468917352725863914237481695619275843854396127
/*/数独AC代码：/*/
#include"iostream"#include"cstdio"#include"cstring"#include"queue"#include"string"using namespace std;int map[11][11],count,n;bool F[11][11];bool H[11][11];bool L[11][11];struct Node { int x,y;} node[100];int MAP(int i,int j) { if(i<=3&&j<=3)return 1; else if(i<=3&&j<=6)return 2; else if(i<=3&&j<=9)return 3; else if(i<=6&&j<=3)return 4; else if(i<=6&&j<=6)return 5; else if(i<=6&&j<=9)return 6; else if(i<=9&&j<=3)return 7; else if(i<=9&&j<=6)return 8; else if(i<=9&&j<=9)return 9;}int DFS(int n) { if(n>count) { return 1; } for(int j=1; j<=9; j++) { if(!H[node[n].x][j]&&!L[node[n].y][j]&&!F[MAP(node[n].x,node[n].y)][j]) { H[node[n].x][j]=L[node[n].y][j]=F[MAP(node[n].x,node[n].y)][j]=1; map[node[n].x][node[n].y]=j; if(DFS(n+1))return 1; H[node[n].x][j]=L[node[n].y][j]=F[MAP(node[n].x,node[n].y)][j]=0; } } return 0;}int main() { int flag=0; while(cin>>n) { for(int i=1; i<=n; i++) { memset(map,0,sizeof(map)); memset(H,0,sizeof(H)); memset(F,0,sizeof(F)); memset(L,0,sizeof(L)); count = 0; for(int j=1; j<=9; j++) for(int k=1; k<=9; k++) { scanf("%1d",&map[j][k]); if(map[j][k]==0) { count++; node[count].x=j; node[count].y=k; } else { H[j][map[j][k]]=1; L[k][map[j][k]]=1; F[MAP(j,k)][map[j][k]]=1; } } DFS(1); for(int j=1; j<=9; j++) { for(int k=1; k<=9; k++) cout<<map[j][k]; cout<<endl; } } } return 0;}
ACM : POJ 2676 SudoKu DFS - 数独
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