POJ 2676 - Sudoku - [蓝桥杯 数独][DFS]

题目链接：http://poj.org/problem?id=2676

Time Limit: 2000MS Memory Limit: 65536K

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

 

题意：

你一定听说过“数独”游戏。
如【图1.png】，玩家需要根据9×9盘面上的已知数字，推理出所有剩余空格的数字，并满足每一行、每一列、每一个同色九宫内的数字均含1-9，不重复。

数独的答案都是唯一的，所以，多个解也称为无解。

本图的数字据说是芬兰数学家花了3个月的时间设计出来的较难的题目。但对会使用计算机编程的你来说，恐怕易如反掌了。

本题的要求就是输入数独题目，程序输出数独的唯一解。我们保证所有已知数据的格式都是合法的，并且题目有唯一的解。

格式要求：
输入9行，每行9个数字，0代表未知，其它数字为已知。
输出9行，每行9个数字表示数独的解。

 

题解：

从[1,1]到[9,9]地进行DFS

 

AC代码：

#include<cstdio>
#include<cstring>
using namespace std;

int num[10][10];
bool rowVis[10][10],colVis[10][10],matVis[10][10];

struct Pos{
    int row,col;
    Pos nextpos()
    {
        if(col==9) return (Pos){row+1,1};
        else return (Pos){row,col+1};
    }
    int MatID(){return (row-1)/3*3 + (col-1)/3+1;}
    bool ok(int x)
    {
        if(!rowVis[row][x] && !colVis[col][x] && !matVis[MatID()][x]) return 1;
        else return 0;
    }
};

void mark(Pos pos,int x,bool val)
{
    rowVis[pos.row][x] = val;
    colVis[pos.col][x] = val;
    matVis[pos.MatID()][x] = val;
}

bool dfs(Pos pos)
{
    int row=pos.row,col=pos.col;

    if(row==10) return 1;
    if(num[row][col]) return dfs(pos.nextpos());

    for(int i=1;i<=9;i++)
    {
        if(pos.ok(i))
        {
            num[row][col]=i; mark(pos,i,1);
            if(dfs(pos.nextpos())) return 1;
            num[row][col]=0; mark(pos,i,0);
        }
    }
    return 0;
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        memset(rowVis,0,sizeof(rowVis));
        memset(colVis,0,sizeof(colVis));
        memset(matVis,0,sizeof(matVis));
        for(int row=1;row<=9;row++)
        {
            char tmp[10]; scanf("%s",tmp);
            for(int col=1;col<=9;col++)
            {
                num[row][col]=tmp[col-1]-'0';
                if(num[row][col]) mark((Pos){row,col},num[row][col],1);
            }
        }

        dfs((Pos){1,1});

        for(int row=1;row<=9;row++)
        {
            for(int col=1;col<=9;col++) printf("%d",num[row][col]);
            printf("\n");
        }
    }
}