数独求解问题（DFS+位运算优化）

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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,

.2738..1.
.1...6735
.......29
3.5692.8.
.........
.6.1745.3
64.......
9518...7.
.8..6534.

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;
}

 

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