【POJ 3074】 Sudoku

【题目链接】

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

【算法】

将数独问题转化为精确覆盖问题，用Dancing Links求解

转化方法如下 ：

我们知道，在一个数独中 ：

1.每个格子填且只填一个数

2.每一行填1-9这九个数

3.每一列填1-9这九个数

4.每个格子填1-9这九个数

对于第一个约束条件，我们用81列，表示是否填入

对于第二个约束条件，我们每一行用9列，表示这一行是否有1-9

第三，四个约束条件的处理方式和第二个类似

【代码】

#include <algorithm>
#include <bitset>
#include <cctype>
#include <cerrno>
#include <clocale>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <limits>
#include <list>
#include <map>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <utility>
#include <vector>
#include <cwchar>
#include <cwctype>
#include <stack>
#include <limits.h>
using namespace std;
#define MAXS 250000

int i,j,cnt;
int mat[][];
char s[];

struct info
{
        int pos,val;
} a[MAXS];

inline int getRow(int pos)
{
        return (pos - ) /  + ;
}
inline int getCol(int pos)
{
        return (pos - ) %  + ;
}
inline int getGrid(int pos)
{
        int x = getRow(pos),y = getCol(pos);
        return (x - ) /  *  + (y - ) /  + ;
}

struct DancingLinks
{
        int n,m,step,size;
        int U[MAXS],D[MAXS],L[MAXS],R[MAXS],Row[MAXS],Col[MAXS];
        int H[MAXS],S[MAXS];
        int ans[MAXS];
        inline void init(int _n,int _m)
        {
                int i;
                n = _n;
                m = _m;
                for (i = ; i <= m; i++)
                {
                        S[i] = ;
                        U[i] = D[i] = i;
                        L[i] = i - ;
                        R[i] = i + ;
                }
                L[] = m; R[m] = ;
                size = m;
                for (i = ; i <= n; i++) H[i] = -;
        }
        inline void link(int r,int c)
        {
                size++;
                Row[size] = r;
                Col[size] = c;
                S[c]++;
                D[size] = D[c];
                U[D[c]] = size;
                U[size] = c;
                D[c] = size;
                if (H[r] < ) L[size] = R[size] = H[r] = size;
                else
                {
                        R[size] = R[H[r]];
                        L[R[H[r]]] = size;
                        L[size] = H[r];
                        R[H[r]] = size;
                }
        }
        inline void Remove(int c)
        {
                int i,j;
                R[L[c]] = R[c];
                L[R[c]] = L[c];
                for (i = D[c]; i != c; i = D[i])
                {
                        for (j = R[i]; j != i; j = R[j])
                        {
                                D[U[j]] = D[j];
                                U[D[j]] = U[j];
                                S[Col[j]]--;
                        }
                }
        }
        inline void Resume(int c)
        {
                int i,j;
                for (i = U[c]; i != c; i = U[i])
                {
                        for (j = L[i]; j != i; j = L[j])
                        {
                                D[U[j]] = j;
                                U[D[j]] = j;
                                S[Col[j]]++;
                        }
                }
                L[R[c]] = c;
                R[L[c]] = c;
        }
        inline bool solve(int dep)
        {
                int i,j,c;
                if (R[] == )
                {
                        step = dep;
                        return true;
                }
                c = R[];
                for (i = R[]; i != ; i = R[i])
                {
                        if (S[i] < S[c])
                                c = i;
                }
                Remove(c);
                for (i = D[c]; i != c; i = D[i])
                {
                        ans[dep] = Row[i];
                        for (j = R[i]; j != i; j = R[j])
                                Remove(Col[j]);
                        if (solve(dep+)) return true;
                        for (j = L[i]; j != i; j = L[j])
                                Resume(Col[j]);
                }
                Resume(c);
                return false;
        }
} DLX;

int main()
{

        while (scanf("%s",s+) && s[] != 'e')
        {
                cnt = ;
                memset(mat,,sizeof(mat));
                for (i = ; i <= ; i++)
                {
                        if (s[i] != '.')
                        {
                                mat[][i] = ;
                                mat[][+(getRow(i)-)*+s[i]-''] = ;
                                mat[][+(getCol(i)-)*+s[i]-''] = ;
                                mat[][+(getGrid(i)-)*+s[i]-''] = ;
                        } else
                        {
                                for (j = ; j <= ; j++)
                                {
                                        cnt++;
                                        mat[cnt][i] = ;
                                        mat[cnt][+(getRow(i)-)*+j] = ;
                                        mat[cnt][+(getCol(i)-)*+j] = ;
                                        mat[cnt][+(getGrid(i)-)*+j] = ;
                                        a[cnt] = (info){i,j};
                                }
                        }
                }
                DLX.init(cnt,);
                for (i = ; i <= cnt; i++)
                {
                        for (j = ; j <= ; j++)
                        {
                                if (mat[i][j])
                                        DLX.link(i,j);
                        }
                }
                DLX.solve();
                for (i = ; i < DLX.step; i++) s[a[DLX.ans[i]].pos] = '' + a[DLX.ans[i]].val;
                for (i = ; i <= ; i++) printf("%c",s[i]);
                printf("\n");
        }

        return ;

}