数独（DFS）

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本文介绍了一种使用深度优先搜索解决数独问题的算法，并通过C++实现。该算法能够找到有效的解决方案，确保每个单元格内的数字在每行、每列及每个3x3的小方格内都是唯一的。

Link：http://poj.org/problem?id=2676

Sudoku

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 14696		Accepted: 7248		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

Sample Output

Source

Southeastern Europe 2005

AC code：

#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<queue>
#include<cstdio>
#include<queue>
#define LL long long
#define MAXN 1000010
#define EPS 1e-9
using namespace std;
int g[11][11],r[11][11],c[11][11],cb[11][11];
char m[11][11];
int fg;
void dfs(int k)
{
	if(k==81)
	{
		for(int i=0;i<9;i++)
		{
			for(int j=0;j<9;j++)
			{
				printf("%d",g[i][j]);
			}
			printf("\n");
		}
		fg=1;
	}
	int row=k/9;
	int col=k%9;
	if(g[row][col])
	{
		dfs(k+1);
	}
	else
	{
		int num=row/3*3+col/3;
		for(int i=1;i<=9;i++)
		{
			if(!r[row][i]&&!c[col][i]&&!cb[num][i])
			{
				r[row][i]=1;
				c[col][i]=1;
				cb[num][i]=1;
				g[row][col]=i;
				dfs(k+1);
				if(fg)
					return;
				r[row][i]=0;
				c[col][i]=0;
				cb[num][i]=0;
				g[row][col]=0;
			}
		}
	}
}
int main()
{
	
	int t;
	//freopen("in.txt","r",stdin);
	scanf("%d",&t);
	while(t--)
	{
		memset(g,0,sizeof(g));
    	memset(c,0,sizeof(c));
    	memset(r,0,sizeof(r));
    	memset(cb,0,sizeof(cb));
    	for(int i=0;i<9;i++)
    	{
    		cin>>m[i];
    		for(int j=0;j<9;j++)
    		{
    			g[i][j]=m[i][j]-'0';
    			if(g[i][j])
    			{
    				int num=i/3*3+j/3;
    				cb[num][g[i][j]]=1;
    				c[j][g[i][j]]=1;
    				r[i][g[i][j]]=1;
    				
				}
			}
		}
		fg=0;
		dfs(0);
	}
	return 0;
}

注意：以下这种做法会OLE！！！

#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<queue>
#include<cstdio>
#include<queue>
#define LL long long
#define MAXN 1000010
#define EPS 1e-9
using namespace std;
int g[11][11],r[11][11],c[11][11],cb[11][11];
char m[11][11];
void dfs(int k)
{
	if(k==81)
	{
		for(int i=0;i<9;i++)
		{
			for(int j=0;j<9;j++)
			{
				printf("%d",g[i][j]);
			}
			printf("\n");
		}
		return;
	}
	int row=k/9;
	int col=k%9;
	if(g[row][col])
	{
		dfs(k+1);
	}
	else
	{
		int num=row/3*3+col/3;
		for(int i=1;i<=9;i++)
		{
			if(!r[row][i]&&!c[col][i]&&!cb[num][i])
			{
				r[row][i]=1;
				c[col][i]=1;
				cb[num][i]=1;
				g[row][col]=i;
				dfs(k+1);
				r[row][i]=0;
				c[col][i]=0;
				cb[num][i]=0;
				g[row][col]=0;
			}
		}
	}
}
int main()
{
	
	int t;
	//freopen("in.txt","r",stdin);
	scanf("%d",&t);
	while(t--)
	{
		memset(g,0,sizeof(g));
    	memset(c,0,sizeof(c));
    	memset(r,0,sizeof(r));
    	memset(cb,0,sizeof(cb));
    	for(int i=0;i<9;i++)
    	{
    		cin>>m[i];
    		for(int j=0;j<9;j++)
    		{
    			g[i][j]=m[i][j]-'0';
    			if(g[i][j])
    			{
    				int num=i/3*3+j/3;
    				cb[num][g[i][j]]=1;
    				c[j][g[i][j]]=1;
    				r[i][g[i][j]]=1;
    				
				}
			}
		}
		dfs(0);
	}
	return 0;
}