POJ3074 Sudoku —— Dancing Links 精确覆盖

题目链接：http://poj.org/problem?id=3074

Sudoku

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10451		Accepted: 3776

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

题解：

Dancing Links博客（来自万仓一黍 ）

Dancing Links的一些特点：

1.矩阵中每个元素的值只能是0或1（在实际操作中只记录1）。

2.行代表着放置情况， 列代表着约束条件。其中矩阵中的行和列的编号从1开始。

3.选择若干行，使得其满足所有约束条件。

对于此题：

1.行：9*9*9，表明有9*9个格子，每个格子有9中情况。

2.列：9*9*4，首先每个格子能且仅能放1个数字，其次每一行的九个数字能且仅能被放一次， 再者列如行者，最后每个九宫格的九个数字能且仅能被放一次。

3.所以构成了(9*9*9) * (9*9*4)的矩阵，然后直接套模板。

代码如下：

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <vector>
 #include <queue>
 #include <stack>
 #include <map>
 #include <string>
 #include <set>
 #define ms(a,b) memset((a),(b),sizeof((a)))
 using namespace std;
 typedef long long LL;
 const int N = ;
 const int MaxN = N*N*N+;
 const int MaxM = N*N*+;
 const int maxnode = MaxN* + MaxM + ;

 char g[MaxN];
 struct DLX      //矩阵的行和列是从1开始的
 {
     int n, m, size; //size为结点数
     int U[maxnode], D[maxnode], L[maxnode], R[maxnode], Row[maxnode], Col[maxnode];
     int H[MaxN], S[MaxM];   //H为每一行的头结点，但不参与循环。S为每一列的结点个数
     int ansd, ans[MaxN];

     void init(int _n, int _m)   //m为列
     {
         n = _n;
         m = _m;
         for(int i = ; i<=m; i++)   //初始化列的头结点
         {
             S[i] = ;
             U[i] = D[i] = i;
             L[i] = i-;
             R[i] = i+;
         }
         R[m] = ; L[] = m;
         size = m;
         for(int i = ; i<=n; i++) H[i] = -;    //初始化行的头结点
     }

     void Link(int r, int c)
     {
         size++; //类似于前向星
         Col[size] = c;
         Row[size] = r;
         S[Col[size]]++;
         D[size] = D[c];
         U[D[c]] = size;
         U[size] = c;
         D[c] = size;
         if(H[r]==-) H[r] = L[size] = R[size] = size; //当前行为空
         else        //当前行不为空： 头插法，无所谓顺序，因为Row、Col已经记录了位置
         {
             R[size] = R[H[r]];
             L[R[H[r]]] = size;
             L[size] = H[r];
             R[H[r]] = size;
         }
     }

     void remove(int c)  //c是列的编号， 不是结点的编号
     {
         L[R[c]] = L[c]; R[L[c]] = R[c];     //在列的头结点的循环队列中， 越过列c
         for(int i = D[c]; i!=c; i = D[i])
         for(int j = R[i]; j!=i; j = R[j])
         {
             //被删除结点的上下结点仍然有记录
             U[D[j]] = U[j];
             D[U[j]] = D[j];
             S[Col[j]]--;
         }
     }

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

     bool Dance(int d)
     {
         if(R[]==)
         {
             for(int i = ; i<d; i++) g[(ans[i]-)/] = (ans[i]-)% + '';
             for(int i = ; i<N*N; i++) printf("%c", g[i]);
             printf("\n");
             return true;
         }

         int c = R[];
         for(int i = R[]; i!=; i = R[i])   //挑结点数最少的那一列，否则会超时，那为什么呢?
             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(Dance(d+)) return true;
             for(int j = L[i]; j!=i; j = L[j]) resume(Col[j]);
         }
         resume(c);
         return false;
     }
 };

 //i、j从0开始，代表着位置； k从1开始，代表着数字
 void place(int &r, int &c1, int &c2,int &c3, int &c4, int i, int j, int k)
 {
     //c1为每个格子一个数， c2为行， c3为列， c4为九宫格
     r = (i*N+j)*N+k; c1 = i*N+j+; c2 = N*N+i*N+k;
     c3 = N*N*+j*N+k; c4 = N*N*+((i/)*+(j/))*N+k;
 }

 DLX dlx;
 int main()
 {
     while(scanf("%s", g) && strcmp(g,"end") )
     {
         dlx.init(N*N*N, N*N*);
         int r, c1, c2, c3,c4;
         for(int i = ; i<N; i++)
             for(int j = ; j<N; j++)
                 for(int k = ; k<=N; k++)
                 if(g[i*N+j]=='.' || g[i*N+j]==''+k)
                 {
                     place(r,c1,c2,c3,c4,i,j,k); //获取位置
                     dlx.Link(r,c1);     //加入到矩阵中， 下同
                     dlx.Link(r,c2);
                     dlx.Link(r,c3);
                     dlx.Link(r,c4);
                 }
         dlx.Dance();      //一起摇摆
     }
     return ;
 }