POJ 2676 Sudoku1 (DFS)

Sudoku

 

Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 19509 Accepted: 9355 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

 

题意

给出一个16*16矩阵的部分格,其中0为空格,要求填充这些空格。

使矩阵满足横竖和九个3*3的方格内的数字都包含1~9这9个数字。

 

思路

很基础的DFS,先存储当前每行每列和每块的数字状态,然后从左上角开始搜索,遇到零的时候向该点填充一个当前行、当前列、当前块都不存在的一个数字,若无法填充,则结束此层DFS,若可行,继续搜索下一层,直到搜索到右下角,标记已经找到答案,此时结束所有DFS,注意在结束的过程中要保留当前所填充的矩阵。

 

AC 代码

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

int mapp[9][9];     //存储当前状态
bool isx[9][10],isy[9][10],iss[3][3][10],flag;  //行、列、块判断

void dfs(int y)
{
    if(y>=9)    //最后一个数是0的情况下若递归到这里,则说明全部填充完毕
    {
        flag=true;
        return;
    }
    for(int i=y; i<9; i++)  //从第y行开始,遍历剩下的点
        for(int j=0; j<9; j++)
        {
            if(mapp[i][j]==0)   //一个需要判断的点
            {
                bool isc=false;     //这一点是否可以被填充
                for(int k=1; k<=9; k++) //1-9
                    if(!isx[j][k]&&!isy[i][k]&&!iss[i/3][j/3][k])   //三种情况都满足
                    {
                        isc=true;   //假设可以填充
                        mapp[i][j]=k;   //填充的数是k
                        isx[j][k]=isy[i][k]=iss[i/3][j/3][k]=true;  //标记该位置
                        dfs(j!=8?y:y+1);
                        if(flag)return;     //找到答案返回,放在这里可以防止在层层结束递归的时候状态被还原
                        else isc=false;     //下层递归失败,标记该点不能填充
                        mapp[i][j]=0;       //还原状态
                        isx[j][k]=isy[i][k]=iss[i/3][j/3][k]=false;
                    }
                if(!isc)return;
            }
            else if(i==8&&j==8)     //最后一个点不是0的情况
            {
                flag=true;
                return;
            }
        }
}

int main()
{
    int n;
    scanf("%d",&n);
    for(int ni=0; ni<n; ni++)
    {
        memset(isx,false,sizeof(isx));
        memset(isy,false,sizeof(isy));
        memset(iss,false,sizeof(iss));
        flag=false;
        for(int i=0; i<9; i++)
            for(int j=0; j<9; j++)
            {
                scanf("%1d",&mapp[i][j]);
                if(mapp[i][j])
                {
                    isx[j][mapp[i][j]]=true;
                    isy[i][mapp[i][j]]=true;
                    iss[i/3][j/3][mapp[i][j]]=true;
                }
            }
        dfs(0);
        for(int i=0; i<9; i++)
            for(int j=0; j<9; j++)
                printf(j!=8?"%d":"%d\n",mapp[i][j]);
    }
    return 0;
}

  • 2 只已被捕捉
    • 邱雷 Chrome | 55.0.2883.87 Windows 10

      66666有时候总会感叹算法的巧妙!!!!!!

      • 千千 Chrome | 58.0.3000.4 Windows 10

        毕竟这也是程序的精髓呀