DFS 2676POJSudoku数独

Sudoku
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 16199 Accepted: 7915 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

716k 0ms
dfs枚举待填空格
每个空格 枚举1-9的数
先判断该空格是否可以填入枚举数
如果不可以填，那么返回枚举下一个
如果可以填，那么填入，接着判断下一个格可不可以填
如果下一个也可以填，那么返回true.
如果该空格所有的9个数都无法枚举，则返回false

716k 0ms
dfs枚举待填空格
每个空格 枚举1-9的数
先判断该空格是否可以填入枚举数
如果不可以填，那么返回枚举下一个
如果可以填，那么填入，接着判断下一个格可不可以填
如果下一个也可以填，那么返回true.
如果该空格所有的9个数都无法枚举，则返回false

#include 
#include
#include
using namespace std;
int col[10][10];//col[j][x]表示第j列有数x
bool row[10][10];//row[i][x]表示第i行有数x
bool grid[4][4][10];//grid[i][j][x]表示第i行第j列的中方格有数x
int  map[10][10];
struct Rest
{
    int x,y;
}rest[90];//存放需要填补的空格坐标
bool dfs(int _count){
    if(_count<0) return true;
    int x=rest[_count].x,y=rest[_count].y;
    for(int k=1;k<=9;k++){
        if(col[y][k]||row[x][k]||grid[x/3][y/3][k]) continue;
        map[x][y]=k;
        col[y][k]=row[x][k]=grid[x/3][y/3][k]=true;
        if(dfs(_count-1)) return true;
        col[y][k]=row[x][k]=grid[x/3][y/3][k]=false;//如果该格所有数都不能填，回溯
    }
     return false;
}

int main(){
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    int T,_count;
        scanf("%d",&T);
        while (T--){
            _count=0;
            memset(grid,0,sizeof(grid));
            memset(col,0,sizeof(col));
            memset(row,0,sizeof(row));
            for(int i=0;i<9;i++){
                for(int j=0;j<9;j++){
                  scanf("%1d",&map[i][j]);
                  if(0!=map[i][j]){
                    col[j][map[i][j]]=true;
                    row[i][map[i][j]]=true;
                    grid[i/3][j/3][map[i][j]]=true;
                  }
                  else rest[_count++]=(Rest){i,j};
                }
           }
        dfs(_count-1);
        for(int i=0;i<9;i++){
             for(int j=0;j<9;j++)
                cout<
             cout<
        }
        }
      return 0;
}