POJ 2676 Sudoku(经典DFS）

Sudoku

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 11575 Accepted: 5748 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

1103000509002109400000704000300502006060000050700803004000401000009205800804000107

Sample Output

143628579572139468986754231391542786468917352725863914237481695619275843
854396127
思路 ：暴力深搜加特殊标记，标记每行每列每个小的九方格，一个mk[10][10][10]搞定，运行时间865ms
#include <cstdio>#include <queue>#include <map>#include <stack>#include <cstring>#include <algorithm>#include <iostream>using namespace std;int mp[15][15];int mk[10][10][10];//标记bool yes;void dfs(){    for(int i = 1; i < 10; i++)    for(int j = 1; j < 10; j++)    {        if(mp[i][j] == 0)        {            for(int k = 1; k <= 9; k++)            {                if(mk[i][0][k]||mk[0][j][k]||mk[(i-1)/3+1][(j-1)/3+1][k])                    continue;                mp[i][j] = k;                mk[i][0][k] = 1;  // 表示第 i 行 的标记，                mk[0][j][k] = 1; // 标记第j列                mk[(i-1)/3+1][(j-1)/3+1][k] = 1;//标记9个小九方格。每个九方格有一个固定数组。                dfs();                if(yes) return;//防止修改mp[i][j]                mp[i][j] = 0;                mk[i][0][k] = 0;                mk[0][j][k] = 0;                mk[(i-1)/3+1][(j-1)/3+1][k] = 0;            }            if(mp[i][j] == 0) return;//如果没有合适的不要继续，应该返回。        }    }    yes = 1;}int main(){    int t;    cin >> t;    while(t--)    {        char s[15];        yes = 0;        memset(mk,0,sizeof(mk));        for(int i = 1;i < 10; i++)        {            cin >>s;            for(int j = 1; j < 10; j++)            {                int k = s[j-1] - '0';                mp[i][j] = k;                mk[i][0][k] = 1;                mk[0][j][k] = 1;                mk[(i-1)/3+1][(j-1)/3+1][k] = 1;            }        }        dfs();        for(int i = 1; i <= 9; i++)        {            for(int j = 1; j <= 9; j++)                printf("%d",mp[i][j]);            printf("\n");        }    }    return 0;}