数独（深搜）(poj2726,poj3074)

数独（深搜）数据最弱版本（poj 2676)

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

分析

• 可以从左上角一行一行扫描到右下角，对于每一个块列举每一种可能，然后从每个可能出发继续深度遍历直到发现有一个块没有数字可以填时停止
• 如何储存每一块可以填写的数字？可以利用九位二进制数来表示每一行，每一列，每个九宫格的数字填写情况，然后直接对这三个数字做按位与运算就可以得到某一具体块可以填的数字了。
• 这里直接用bitset，对于每一个结果，直接遍历一下就可以了。

char a[15][15];
int b[11][11];
bitset<10> row[10],col[10],room[10];
inline int g(int i,int j)//计算i,j对应的块的序号
{
	return ((i-1)/3)*3+((j-1)/3)+1;
}
void init()
{
   //把每个九位二进制数都全部置位1，表示可以填
	for(int i=1;i<=9;i++)
	{
		row[i].set();
		col[i].set();
		room[i].set();
	}
	for(int i=1;i<=9;i++)
	{
		for(int j=1;j<=9;j++)
		{
			int x = a[i][j]-'0';
			b[i][j] = x;
			//如果不是0，就要更新该行该列该九宫格的数字填写范围了
			if(x)
			{
				row[i][x] = 0;
				col[j][x] = 0;
				room[g(i,j)][x] = 0;
			}
		}
	}
}
bool dfs(int x,int y)
{
  //如果扫描到了第10行，返回正确
	if(x==10)
		return true;
	bool flag = false;
	if(b[x][y])
	{
		if(y==9)//如果扫描到行末，继续扫描下一行行首
			flag = dfs(x+1,1);
		else
			flag = dfs(x,y+1);
		if(flag)
			return true;
		else
			return false;
	}
	int k = g(x,y);
	bitset<10> tmp = row[x]&col[y]&room[k];//得到该处可填写的数字二进制表示
	//cout<<tmp<<endl;
	for(int i=1;i<=9;i++)
	{
		if(tmp[i]==1)//如果可以填写i
		{
			b[x][y] = i;
			row[x][i] = 0;
			col[y][i] = 0;
			room[k][i] = 0;
			if(y==9)
				flag = dfs(x+1,1);
			else
				flag = dfs(x,y+1);
			if(flag)
				return true;
			row[x][i] = 1;
			col[y][i] = 1;
			room[k][i] = 1;
			b[x][y] = 0;
		}
	}
	return false;
}

int main()
{
	int t;
	cin>>t;
	while(t--)
	{
		for(int i=1;i<=9;i++)
		{
			scanf("%s",a[i]+1);
		}
		init();
		dfs(1,1);
		for(int i=1;i<=9;i++)
		{
			for(int j=1;j<=9;j++)
			{
				cout<<b[i][j];
			}
			cout<<endl;
		}
	}
    return 0;
}

---

数独（深搜，剪枝）数据较强版本(poj 3074)

Description

In the game of Sudoku, you are given a large 9 × 9 grid divided into smaller 3 × 3 subgrids. For example,

.	2	7	3	8	.	.	1	.
.	1	.	.	.	6	7	3	5
.	.	.	.	.	.	.	2	9
3	.	5	6	9	2	.	8	.
.	.	.	.	.	.	.	.	.
.	6	.	1	7	4	5	.	3
6	4	.	.	.	.	.	.	.
9	5	1	8	.	.	.	7	.
.	8	.	.	6	5	3	4	.

Given some of the numbers in the grid, your goal is to determine the remaining numbers such that the numbers 1 through 9 appear exactly once in (1) each of nine 3 × 3 subgrids, (2) each of the nine rows, and (3) each of the nine columns.

Input

The input test file will contain multiple cases. Each test case consists of a single line containing 81 characters, which represent the 81 squares of the Sudoku grid, given one row at a time. Each character is either a digit (from 1 to 9) or a period (used to indicate an unfilled square). You may assume that each puzzle in the input will have exactly one solution. The end-of-file is denoted by a single line containing the word “end”.

Output

For each test case, print a line representing the completed Sudoku puzzle.

Sample Input

.2738..1..1...6735.......293.5692.8...........6.1745.364.......9518...7..8..6534.
......52..8.4......3...9...5.1...6..2..7........3.....6...1..........7.4.......3.
end

Sample Output

527389416819426735436751829375692184194538267268174593643217958951843672782965341
416837529982465371735129468571298643293746185864351297647913852359682714128574936

分析：

• 如果我们依然像上道题一样从左上角找到右下角，这道题会很容易Tl的，所以我们要变得更加聪明一点，正如我们做数独的话，肯定是先找最容易确定的位置去填，所以我们可以在搜索时先找到81个空中可填数最少的空
• 深搜时，我们需要的状态就是还没有填的数，这是我们所关心的，如果没有要填的空，就返回true，如果在找最小可填数时有一个空发现没有可以填的数字，则直接返回false
• 另外我们可以直接使用二进制运算lowbit，再加上一个数组num[]，就可以直接查找到对于每一种可能的具体数字。
• 因为我们是用二进制储存可填数的，所以在更新这些数字时，只是单纯的把某一位进行取反，所以我们可以直接用一个flip函数，巧妙地减少了代码量。

#include <stdio.h>
#include <iostream>
#include <cstdlib>
#include <cmath>
#include <cctype>
#include <string>
#include <cstring>
#include <algorithm>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <ctime>
#include <vector>
#include <fstream>
#include <list>
#include <iomanip>
#include <numeric>
#include <bitset>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define ms(s) memset(s, 0, sizeof(s))
const int inf = 0x3f3f3f3f;
#define LOCAL
char a[100];
char b[11][11];
int row[10], col[10], room[10], cnt[513], num[512];
inline int g(int i,int j)
{
	return ((i-1)/3)*3+((j-1)/3)+1;
}
inline void flip(int x, int y, int z) {
	row[x] ^= 1 << z;
	col[y] ^= 1 << z;
	room[g(x, y)] ^= 1 << z;
}
bool dfs(int tot)
{
	if(tot==0)
		return true;
	int tmp = 10,x,y;
	for(int i=1;i<=9;i++)
		for(int j=1;j<=9;j++)
		{
			if(b[i][j] != '.')continue;
			int val = col[j]&row[i]&room[g(i,j)];
			if(!val)return false;//重点
			if (cnt[val]< tmp)
			{
				tmp = cnt[val];
				x = i, y = j;
			}
		}
	int  val = row[x]&col[y]&room[g(x,y)];
	for(;val;val -= val&-val)
	{
		int z = num[val&-val];
		b[x][y] = z + '1';
		flip(x,y,z);
		if(dfs(tot-1))
		{
			return true;
		}
		flip(x,y,z);
		b[x][y] = '.';
	}
	return false;
}
void init()
{
	for(int i=1;i<=9;i++)
		row[i]=col[i]=room[i]=(1<<9)-1;//全部赋值成1
	int tot = 0;
	for(int i=1;i<=9;i++)
		for(int j=1;j<=9;j++)
		{
			b[i][j] = a[(i-1)*9+j-1];
			if(b[i][j]!='.')
				flip(i,j,b[i][j]-'1');//因为二进制的第1位表示1是否可填，为了配合filp函数，这里传参b[i][j]-'1'
			else
				tot++;
		}
	dfs(tot);
	for(int i=1;i<=9;i++)
		for(int j=1;j<=9;j++)
			a[(i-1)*9+j-1] = b[i][j];
}


int main() 
{
	for (int i = 0; i < 1 << 9; i++)
	    for (int j = i; j; j -= j&-j) 
	        cnt[i]++;//对于每一个九位二进制数，储存它二进制表示中1的个数
	for (int i = 0; i <= 9; i++)
		num[1 << i] = i;//储存二进制中每一位1的权值
	while(scanf("%s",a))
	{
		if(a[0]=='e')
			break;
		init();
		puts(a);
	}
    return 0;
}

poj3076是一个16宫格的数独，日后再战。