数独求解问题（DFS+位运算优化）

最新推荐文章于 2022-07-22 21:46:30 发布

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本文探讨了数独游戏的求解算法，采用深度优先搜索（DFS）结合位运算优化的方法，成功在700毫秒内解决数独谜题。通过对比优化前后的代码，展示了如何利用位运算提升算法效率。

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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

思路：DFS解这道题有点极限，需要优化的东西比较复杂，参考了网上的位运算优化，终于以700ms的解决了，如果不位运算优化很难去不超时的去解这道问题

附上原超时代码,此代码修改一下应该可以过F题，单纯的求数独的题

代码：


#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<vector>
#include<cmath>
#include<set>
#include<stack>
#include<map> 
#define MAX 10005

typedef long long ll;

using namespace std;
 
int cnt,flag;
int Map[10][10],vistrow[10][10],vistcol[10][10],vistkaui[10][10][10];
struct node
{
	int x,y;
}a[100];
 
void DFS(int x)
{
	if(x == cnt)
	{
		flag = 1;
		return;
	}
	int I = a[x].x;
	int J = a[x].y;
	for(int i = 1; i < 10; i++)
	{
		if(!vistrow[I][i] && !vistcol[J][i] && !vistkaui[I/3][J/3][i])
		{
			Map[I][J] = i;
			vistrow[I][i] = 1;
			vistcol[J][i] = 1;
			vistkaui[I/3][J/3][i] = 1;
			DFS(x+1);
			if(!flag)
			{
				Map[I][J] = '.';
				vistrow[I][i] = 0;
				vistcol[J][i] = 0;
				vistkaui[I/3][J/3][i] = 0;
			}
		}
	}
}
 
int main()
{
	int T;
	
	while(1)
	{
		memset(vistrow,0,sizeof(vistrow));
		memset(vistcol,0,sizeof(vistcol));
		memset(vistkaui,0,sizeof(vistkaui));
 
		cnt = 0;
		flag = 0;
		
		for(int i = 0; i < 9; i++)
		{
			
			
			for(int j = 0; j < 9; j++)
			{
			    scanf("%c",Map[i][j]);
 
				if(Map[i][j] == '.')
				{
					a[cnt].x = i;
					a[cnt++].y = j;
				}
				else
				{
					vistrow[i][Map[i][j]-'] = 1;
					vistcol[j][Map[i][j]] = 1;
					vistkaui[i/3][j/3][Map[i][j]] = 1;
				}
			}
		}
		DFS(0);
		for(int i = 0; i < 9; i++)
			for(int j = 0; j < 9; j++)
			{
				if(j == 8)
					printf("%c\n",Map[i][j]);
				else
					printf("%c",Map[i][j]);
			}
	}
	return 0;
}

AC代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<vector>
#include<cmath>
#include<set>
#include<stack>
#include<map> 
#define MAX 10005

using namespace std;
char Map[10][10];
int vistrow[10],vistcol[10];
int grid[10],rec[512],num[512];

inline int g(int x,int y) 
{
    return (x/3)*3+y/3;
}
inline void flip(int x,int y,int to) 
{
    vistrow[x]^=1<<to;
    vistcol[y]^=1<<to;
    grid[g(x,y)]^=1<<to;
}
bool DFS(int x) 
{
    if(x==0) return 1;
    int minn=10,xx,yy;
    for(int i=0;i<9;i++) {
        for(int j=0;j<9;j++) 
		{
            if(Map[i][j]=='.') 
			{
                int val=vistrow[i]&vistcol[j]&grid[g(i,j)];
                if(!val) return 0;
                if(rec[val]<minn) {
                    minn=rec[val],xx=i,yy=j;
                }
            }
        }
    }
    int val=vistrow[xx]&vistcol[yy]&grid[g(xx,yy)];
    for(;val;val-=val&-val) {
        int to=num[val&-val];
        Map[xx][yy]=to+'1';
        flip(xx,yy,to);
        if(DFS(x-1)) return 1;
        flip(xx,yy,to);
        Map[xx][yy]='.';
    }
    return 0;
}
int main() {
    for(int i=0;i<1<<9;i++) {
        for(int j=i;j;j-=j&-j)
            rec[i]++;
    }
    for(int i=0;i<9;i++) {
        num[1<<i]=i;
    }
    char s[100];
    while(~scanf("%s",s)&&s[0]!='e') {
        for(int i=0;i<9;i++) 
		vistrow[i]=vistcol[i]=grid[i]=(1<<9)-1;
        int tot=0;
        for(int i=0;i<9;i++) {
            for(int j=0;j<9;j++) {
                Map[i][j]=s[i*9+j];
                if(Map[i][j]!='.') flip(i,j,Map[i][j]-'1');
                else ++tot;
            }
        }
        DFS(tot);
        for(int i=0;i<9;i++)
            for(int j=0;j<9;j++)
                printf("%c",Map[i][j]);

        printf("\n");
    }
    return 0;
}