poj2676Sudoku

最新推荐文章于 2020-01-17 23:59:14 发布

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本文介绍了一种使用深度优先搜索(DFS)解决Sudoku难题的方法。通过递归地尝试每一种可能的数字填充空格，并检查是否违反Sudoku规则来找到解决方案。如果当前的尝试导致了冲突，则回溯到上一步并尝试下一个可能的数字。

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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

Sample Output

#include<iostream>
#include<string.h>
using namespace std;

int map[10][10]; //九宫格

bool row[10][10];    //row[i][x]  标记在第i行中数字x是否出现了
bool col[10][10];    //col[j][y]  标记在第j列中数字y是否出现了
bool grid[10][10];   //grid[k][x] 标记在第k个3*3子格中数字z是否出现了
bool DFS(int x,int y)
{
    if(x==10)
        return true;
    bool flag=false;
    if(map[x][y])
    {
        if(y==9)
            flag=DFS(x+1,1);
        else
            flag=DFS(x,y+1);

        if(flag)  //回溯
            return true;
        else
            return false;
    }
    else
    {
        int k=3*((x-1)/3)+(y-1)/3+1;
        for(int i=1;i<=9;i++)   //枚举数字1~9填空
            if(!row[x][i] && !col[y][i] && !grid[k][i])
            {
                map[x][y]=i;
                row[x][i]=true;
                col[y][i]=true;
                grid[k][i]=true;
                if(y==9)
                    flag=DFS(x+1,1);
                else
                    flag=DFS(x,y+1);
                if(!flag)   //回溯，继续枚举
                {
                    map[x][y]=0;
                    row[x][i]=false;
                    col[y][i]=false;
                    grid[k][i]=false;
                }
                else
                    return true;
            }
    }
    return false;
}

int main(int i,int j)
{
    int test;
    cin>>test;
    while(test--)
    {
        memset(row,false,sizeof(row));
        memset(col,false,sizeof(col));
        memset(grid,false,sizeof(grid));
        char MAP[10][10];
        for(i=1;i<=9;i++)
            for(j=1;j<=9;j++)
            {
                cin>>MAP[i][j];
                map[i][j]=MAP[i][j]-'0';

                if(map[i][j])
                {
                    int k=3*((i-1)/3)+(j-1)/3+1;
                    row[i][ map[i][j] ]=true;
                    col[j][ map[i][j] ]=true;
                    grid[k][ map[i][j] ]=true;
                }
            }
        DFS(1,1);

        for(i=1;i<=9;i++)
        {
            for(j=1;j<=9;j++)
                cout<<map[i][j];
            cout<<endl;
        }
    }
    return 0;
}