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POJ 3074 Sudoku(数据结构,DLX)

2014-06-27 14:52 417 查看

http://poj.org/problem?id=3074

Sudoku

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8141		Accepted: 2854

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

Source

Stanford Local 2006

数独求解，用一般的搜索要超时，从中间开始搜大概900+ms过。优雅的求解数独（精确覆盖问题）需要用到Dancing Links这个非常高大上的数据结构，《算法竞赛入门经典——训练指南》（刘汝佳）中有对DLX（Dancing Links X算法）的介绍。

DLX模板:

/article/2059866.html

DLX本身并不复杂，写起来可能有点小麻烦，但是求解这类问题最难的是构造出01矩阵。

矩阵的行表示决策（我们要选择行），一共有9*9*9个决策：在r行c列的格子里填上数字v;

矩阵的列表示任务（目的是使得每列有且仅有1个1）,一共有4种任务：

1.a行b列有数字

2.a行有数字b

3.a列有数字b

4.第a个方块要有数字b

所以一共有9*9*4列;

这样9*9的数独问题就转化成9*9*9行，9*9*4列的精确覆盖问题

#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<algorithm>
#include<ctime>
#include<cctype>
#include<cmath>
#include<string>
#include<cstring>
#include<stack>
#include<queue>
#include<list>
#include<vector>
#include<map>
#include<set>
#define sqr(x) ((x)*(x))
#define LL long long
#define itn int
#define INF 0x3f3f3f3f
#define PI 3.1415926535897932384626
#define eps 1e-10
#define mm

using namespace std;

const int maxc=9*9*4+10;
const int maxnode=9*9*9*4+10;
const int maxr=9*9*9+10;

struct DLX
{
    int n,sz;//列数，结点数
    int S[maxc];//各列结点数
    int row[maxnode],col[maxnode];//各点行列编号
    int L[maxnode],R[maxnode],U[maxnode],D[maxnode];//十字链表
    int ansd,ans[maxr];//解

    void init(int n)
    {
        this->n=n;

        //虚拟结点
        for (int i=0;i<=n;i++)
        {
            U[i]=i;D[i]=i;L[i]=i-1;R[i]=i+1;
        }

        R
=0;L[0]=n;
        sz=n+1;
        memset(S,0,sizeof S);
    }

    //插入决策行  _r 行号  columns 记录结点列号·[1,n]
    void add_row(int _r,vector<int> columns)
    {
        int first=sz;
        for (int i=0;i<columns.size();i++)
        {
            int _c=columns[i];
            L[sz]=sz-1;R[sz]=sz+1;
            D[sz]=_c;U[sz]=U[_c];//成环
            D[U[_c]]=sz;U[_c]=sz;
            row[sz]=_r;col[sz]=_c;
            S[_c]++;sz++;
        }
        R[sz-1]=first;L[first]=sz-1;
    }

    //删除一列结点
    void _remove(int _c)
    {
        L[R[_c]]=L[_c];
        R[L[_c]]=R[_c];

        for (int i=D[_c];i!=_c;i=D[i])
            for (int j=R[i];j!=i;j=R[j])
            {
                U[D[j]]=U[j];D[U[j]]=D[j];S[col[j]]--;
            }
    }

    //恢复一列结点，和删除顺序相反
    void _resume(int _c)
    {
        for (int i=U[_c];i!=_c;i=U[i])
            for (int j=L[i];j!=i;j=L[j])
            {
                U[D[j]]=j;D[U[j]]=j;S[col[j]]++;
            }

        L[R[_c]]=_c;
        R[L[_c]]=_c;
    }

    bool dfs(int d)
    {
        if (R[0]==0)
        {
            ansd=d; //记录解的长度
            return true;
        }

        //找最少结点的列删除
        int _c=R[0];
        for (int i=R[0];i!=0;i=R[i])
            if (S[i]<S[_c]) _c=i;

        _remove(_c);
        for (int i=D[_c];i!=_c;i=D[i])
        {
            ans[d]=row[i];
            for (int j=R[i];j!=i;j=R[j])
                _remove(col[j]);

            if (dfs(d+1))   return true;

            for (int j=L[i];j!=i;j=L[j])//反向恢复
                _resume(col[j]);
        }
        _resume(_c);

        return false;
    }

    bool solve(vector<int> &v)
    {
        v.clear();
        if (!dfs(0)) return false;

        for (int i=0;i<ansd;i++)    v.push_back(ans[i]);

        return true;
    }

};

int encode(int a,int b,int c)
{
    return a*81+b*9+c+1;
}

void decode(int code,int &a,int &b,int &c)
{
    code--;

    c=code%9; code/=9;
    b=code%9; code/=9;
    a=code;
}

char sudoku[100];
DLX solver;

int main()
{
    #ifndef ONLINE_JUDGE
        freopen("/home/fcbruce/文档/code/t","r",stdin);
    #endif // ONLINE_JUDGE

    while (scanf("%s",sudoku),strcmp(sudoku,"end"))
    {

        solver.init(9*9*4);
        for (int i=0;i<9;i++)
        {
            for (int j=0;j<9;j++)
            {
                for (int v=0;v<9;v++)
                {
                    if (sudoku[i*9+j]=='.' || sudoku[i*9+j]=='1'+v)
                    {
                        vector<int> columns;
                        columns.push_back(encode(0,i,j));//i行j列要有数字
                        columns.push_back(encode(1,i,v));//i行要有数字v
                        columns.push_back(encode(2,j,v));//j列要有数字v
                        columns.push_back(encode(3,(i/3)*3+j/3,v));//第(i/3)*3+j/3个方块要有数字v
                        solver.add_row(encode(i,j,v),columns);
                    }
                }

            }
        }

        vector<int> ans;
        solver.solve(ans);

        for (int i=0;i<ans.size();i++)
        {
            int r,c,v;
            decode(ans[i],r,c,v);
            sudoku[r*9+c]='1'+v;
        }

        puts(sudoku);
    }

    return 0;
}

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