POJ3074 Sudoku

Sudoku

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10745		Accepted: 3850

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

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题目大意：
　　数独求解

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分析：
　　　　这道题的数据强于POJ2676
　　所以一般的DFS是TLE的
　　但本蒟蒻又不会dancing links
　　所以学习了一种类似于状态压缩的方法来判断每一行，每一列，每一个九宫格的数字使用情况
　　我们可以对于每一行，每一列，每一个九宫格都用一个9位二进制数来存储，再使用bitset来将可以使用的数字取出
　　（对于每一个空位优先搜索可用数字少的，可以加快速度）
　　最后DFS

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代码如下

#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
int heng[10];
int shu[10];
int ge[10];
int cnt[512],num[512],tot;
char ss[10][10];
char s[100];
int ID(int x,int y)
{
    return ((x/3)*3)+(y/3);
}
void visit(int x,int y,int z)
{
    heng[x]^=1<<z;
    shu[y]^=1<<z;
    ge[ID(x,y)]^=1<<z;
}
bool dfs(int now)
{
    if(now==0)return true;
    int temp=10,x,y;
    for(int i=0;i<9;i++)
    {
        for(int j=0;j<9;j++)
        {
            if(ss[i][j]!='.')continue;
            int v=heng[i]&shu[j]&ge[ID(i,j)];
            if(!v)return false;
            if(cnt[v]<temp)
            {
                temp=cnt[v];
                x=i,y=j;
            }
        }
    }//可用数字最少的空位
    int v=heng[x]&shu[y]&ge[ID(x,y)];
    for( ;v ;v-=v&-v)
    {
        int z=num[v&-v];
        ss[x][y]='1'+z;
        visit(x,y,z);
        if(dfs(now-1))return true;
        visit(x,y,z);
        ss[x][y]='.';
    }//取出数字再DFS
    return false;
}
int main()
{
    for(int i=0;i<1<<9;i++)
    {
        for(int j=i;j;j-=j&-j)cnt[i]++;
    }
    for(int i=0;i<9;i++)
    {
        num[1<<i]=i;
    }
    while(scanf("%s",s)&&s[0]!='e')
    {
        for(int i=0;i<9;i++)
        {
            for(int j=0;j<9;j++)
            {
                ss[i][j]=s[i*9+j];
            }
        }
        for(int i=0;i<9;i++)
        {
            heng[i]=shu[i]=ge[i]=(1<<9)-1;
        }
        tot=0;
        for(int i=0;i<9;i++)
        {
            for(int j=0;j<9;j++)
            {
                if(ss[i][j]!='.')visit(i,j,ss[i][j]-'1');
                else tot++;
            }
        }
        dfs(tot);
        for(int i=0;i<9;i++)
        {
            for(int j=0;j<9;j++)
            {
                s[i*9+j]=ss[i][j];
            }
        }
        printf("%s
",s);
    }
    return 0;
}