POJ - 2676 Sudoku

最新推荐文章于 2020-03-17 14:37:43 发布

Lngxling

最新推荐文章于 2020-03-17 14:37:43 发布

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分类专栏： DFS

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本文介绍了一种解决数独问题的算法实现，采用深度优先搜索策略，并利用三个二维数组记录行、列及小方格内的已使用数字。通过递归方式填充空缺单元格，确保每行、每列及每个3x3子区域内的数字不重复。

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Sudoku

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 21297		Accepted: 10146		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

题意：给定一个未完成的数独 0是未填入的数据把数独填完

思路：比较笨..开了三个二维数组来记录每一行每一列和每个3*3的小方格内被用过的数字深搜

#include <iostream>
#include <queue>
#include <cstdio>
#include <vector>
#include <cstring>
using namespace std;
bool vis1[10][10],vis2[10][10],vis3[10][10];
int num[10][10],sum=0,flag=0;
struct point
{
	int x,y;
}zz;
vector<point>v;
int cal(struct point p)
{
	int i=p.x;int j=p.y;
	return (((i-1)/3)*3+((j-1)/3)+1);
}
void show()
{
	for(int i=1;i<=9;i++)
	{
		for(int j=1;j<=9;j++)
		{
			cout<<num[i][j];
		}
		cout<<endl;
	}
}
void dfs(int i)
{
	if(i==sum)
	flag=1;
	if(flag==1)
	return;
	int j;
	point z=v[i];
	for(j=1;j<=9;j++)
	{
		if(flag==1)
		break;
		if(vis1[z.x][j]==false&&vis2[z.y][j]==false&&vis3[cal(z)][j]==false)
		{
			num[z.x][z.y]=j;
			vis1[z.x][j]=true;
			vis2[z.y][j]=true;
			vis3[cal(z)][j]=true;
			dfs(i+1);
			vis1[v[i].x][j]=false;
			vis2[v[i].y][j]=false;
			vis3[cal(v[i])][j]=false;
		}
	}

}
int main()
{
	int t;
	cin>>t;
	while(t--)
	{
		memset(vis1,false,sizeof(vis1));
		memset(vis2,false,sizeof(vis2));
		memset(vis3,false,sizeof(vis3));
		sum=0;
		flag=0;
		v.clear();
		int i,j;
		for(i=1;i<=9;i++)
		{
			for(j=1;j<=9;j++)
			{
				scanf(" %1d",&num[i][j]);
				if(num[i][j]==0)
				{
					zz.x=i;
					zz.y=j;
					v.push_back(zz);
					sum++;
				}
				else
				{
					vis1[i][num[i][j]]=true;
					vis2[j][num[i][j]]=true;
					vis3[((i-1)/3)*3+((j-1)/3)+1][num[i][j]]=true;
				}
			}
		}
		dfs(0);
		show();
	}
	
}