POJ 2676 Sudoku(经典DFS)

Sudoku

Time Limit: 2000MS		Memory Limit: 65536K
Total 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

1
103000509
002109400
000704000
300502006
060000050
700803004
000401000
009205800
804000107

Sample Output

143628579
572139468
986754231
391542786
468917352
725863914
237481695
619275843
854396127

题目链接：http://poj.org/problem?id=2676

#include <stdio.h>
#include <string.h>

bool  bResult;
const int MAX_N = 10;
char  sudoku[MAX_N][MAX_N];

void printResult()
{
    for(int i = 0; i < 9; i++)
        printf("%s\n", sudoku[i]);
}

bool JudgeStatue(int curCount, int num)
{
    int row = curCount / 9;
    int col = curCount % 9;
    for(int j = 0; j < 9; j++)     //判断同一行是否有相同的数字
        if(sudoku[row][j] == '0' + num)
            return false;
    for(int i = 0; i < 9; i++)    //判断同一列是否有相同的数字
        if(sudoku[i][col] == '0' + num)
            return false;

                                 //判断在一个小的九宫格里是否有相同的数字
    for(int i = row-row%3; i < row-row%3 + 3; i++)
        for(int j = col-col%3; j < col-col%3 + 3; j++)
            if(sudoku[i][j] == '0' + num)
                return false;
    return true;
}

void dfs(int curCount)        //curCount代表当前下标
{
    if(curCount == -1)
    {
        bResult = true;
        printResult();
        return ;
    }
    if(bResult)return ;

    for(int num = 1; num <= 9; num++)
    {
        if(sudoku[curCount/9][curCount%9]-'0' != 0)
        {
            curCount--;
            if(curCount == -1)
            {
                bResult = true;
                printResult();
                return ;
            }
            num = 0;
            continue;
        }

        if(JudgeStatue(curCount, num))
        {
            sudoku[curCount/9][curCount%9] = num + '0';
            dfs(curCount - 1);
            sudoku[curCount/9][curCount%9] = '0';
        }
    }
}

int main()
{
    int  nCase;

    scanf("%d", &nCase);
    getchar();
    while(nCase--)
    {
        bResult = false;
        memset(sudoku, 0, sizeof(sudoku));
        for(int i = 0; i < 9; i++)
        {
            gets(sudoku[i]);
        }

        dfs(80);               //从最后一个数字开始搜索
    }
    return 0;
}

 题目总结：经典的DFS题目，可以从前往后搜索，也可以倒着搜索，从前往后搜索1704MS，倒着搜索16MS，

               应该是测试数据的原因，导致有这么大的差别。