POJ 3074 Sudoku (DLX)

Sudoku

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 3074

Appoint description: 
System Crawler  (2015-04-18)

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

用dancing links解数独，DLX专题的终极目标，手机上的数独玩了两天后终于A了，等下把记录用这个程序刷一遍，想象下小伙伴看到我最高难度下的耗时记录的表情，啊哈哈~~
难点在于怎么建立模型，一共有9行，每行有9个数字，所以就是9 * 9，同理，列也是9 * 9，小格子也是有9个，每个也是9个数字，所以也是9 * 9，另外整个图有9 * 9 = 81的格子。所以要覆盖的列就是 9 * 9 + 9 * 9 + 9 * 9 + 81。至于行，一共有81个格子，每个格子有9种取法，所以就有81 * 9行，行和列相乘就是开的数组的大小。还有个剪枝要注意下，如果某个格子的数字已经给出，那么这一行，这一列，这一个小格子，就没必要再填这个数了，不剪枝的话会T。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <string>
 #include <cstdlib>
 #include <cmath>
 #include <map>
 #include <cctype>
 using    namespace    std;

 const    int    HEAD = ;
 const    int    SIZE = ( * ) * ( * );
 const    int    COL =  * ;
 int    U[SIZE],D[SIZE],L[SIZE],R[SIZE],S[SIZE],C[SIZE],N[SIZE],P_H[SIZE],P_C[SIZE];
 int    COUNT;
 int    TEMP[][];
 bool    VIS_I[][],VIS_J[][],VIS_G[][];
 struct    Node
 {
     int    i;
     int    j;
     int    num;
 }ANS[];

 void    ini(void);
 void    link(int,int,int,int,int,int,int);
 bool    dancing(int);
 void    remove(int);
 void    resume(int);
 void    debug(int);
 int    main(void)
 {
     char    s[];
     while(scanf(" %s",s + ) && strcmp(s + ,"end"))
     {
         ini();
         for(int i = ;i <= ;i ++)
             for(int j = ;j <= ;j ++)
             {
                 int    k = s[(i - ) *  + j];
                 int    c_1,c_2,c_3,c_4;
                 if(k != '.')
                 {
                     VIS_I[i][k - ''] = VIS_J[j][k - ''] = true;
                     VIS_G[(i - ) /  *  + (j - ) /  + ][k - ''] = true;
                     c_1 =  *  + (i - ) *  + k - '';
                     c_2 =  *  + (j - ) *  + k - '';
                     c_3 =  *  + ((i - ) /  *  + (j - ) / ) *  + k - '';
                     c_4 =  *  + (i - ) *  + j;
                     link(c_1,c_2,c_3,c_4,k - '',i,j);
                 }
             }

         for(int i = ;i <= ;i ++)
             for(int j = ;j <= ;j ++)
             {
                 if(s[(i - ) *  + j] != '.')
                     continue;
                 int    c_1,c_2,c_3,c_4;
                 for(int k = ;k <= ;k ++)
                 {
                     if(VIS_I[i][k] || VIS_J[j][k] ||
                             VIS_G[(i - ) /  *  + (j - ) /  + ][k])
                         continue;
                     c_1 =  *  + (i - ) *  + k;
                     c_2 =  *  + (j - ) *  + k;
                     c_3 =  *  + ((i - ) /  *  + (j - ) / ) *  + k;
                     c_4 =  *  + (i - ) *  + j;
                     link(c_1,c_2,c_3,c_4,k,i,j);
                 }
             }
         dancing();
     }

     return    ;
 }

 void    ini(void)
 {
     L[HEAD] = COL;
     R[HEAD] = ;
     for(int    i = ;i <= COL;i ++)
     {
         L[i] = i - ;
         R[i] = i + ;
         U[i] = D[i] = C[i] = i;
         S[i] = ;
     }
     R[COL] = HEAD;

     fill(&VIS_I[][],&VIS_I[][],false);
     fill(&VIS_J[][],&VIS_J[][],false);
     fill(&VIS_G[][],&VIS_G[][],false);
     COUNT = COL + ;
 }

 void    link(int c_1,int c_2,int c_3,int c_4,int num,int r,int c)
 {
     int    first = COUNT;
     int    col;

     for(int i = ;i < ;i ++)
     {
         switch(i)
         {
             case    :col = c_1;break;
             case    :col = c_2;break;
             case    :col = c_3;break;
             case    :col = c_4;break;
         }

         L[COUNT] = COUNT - ;
         R[COUNT] = COUNT + ;
         U[COUNT] = U[col];
         D[COUNT] = col;

         D[U[col]] = COUNT;
         U[col] = COUNT;
         C[COUNT] = col;
         N[COUNT] = num;
         P_H[COUNT] = r;
         P_C[COUNT] = c;
         S[col] ++;
         COUNT ++;
     }
     L[first] = COUNT - ;
     R[COUNT - ] = first;
 }

 bool    dancing(int k)
 {
     if(R[HEAD] == HEAD)
     {
         for(int i = ;i < k;i ++)
             TEMP[ANS[i].i][ANS[i].j] = ANS[i].num;
         int    count = ;
         for(int i = ;i <= ;i ++)
             for(int j = ;j <= ;j ++)
                 printf("%d",TEMP[i][j]);
         puts("");
         return    true;
     }

     int    c = R[HEAD];
     for(int i = R[HEAD];i != HEAD;i = R[i])
         if(S[c] > S[i])
             c = i;

     remove(c);
     for(int i = D[c];i != c;i = D[i])
     {
         ANS[k].i = P_H[i];
         ANS[k].j = P_C[i];
         ANS[k].num = N[i];
         for(int j = R[i];j != i;j = R[j])
             remove(C[j]);
         if(dancing(k + ))
             return    true;
         for(int j = L[i];j != i;j = L[j])
             resume(C[j]);
     }
     resume(c);

     return    false;
 }

 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])
         {
             D[U[j]] = D[j];
             U[D[j]] = U[j];
             S[C[j]] --;
         }
 }

 void    resume(int c)
 {
     L[R[c]] = c;
     R[L[c]] = c;
     for(int i = D[c];i != c;i = D[i])
         for(int j = L[i];j != i;j = L[j])
         {
             D[U[j]] = j;
             U[D[j]] = j;
             S[C[j]] ++;
         }
 }