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 = 9;
const int MaxN = N*N*N+10;
const int MaxM = N*N*4+10;
const int maxnode = MaxN*4 + MaxM + 10;

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 = 0; i<=m; i++)   //初始化列的头结点
        {
            S[i] = 0;
            U[i] = D[i] = i;
            L[i] = i-1;
            R[i] = i+1;
        }
        R[m] = 0; L[0] = m;
        size = m;
        for(int i = 1; i<=n; i++) H[i] = -1;    //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]==-1) 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[0]==0)
        {
            for(int i = 0; i<d; i++) g[(ans[i]-1)/9] = (ans[i]-1)%9 + '1';
            for(int i = 0; i<N*N; i++) printf("%c", g[i]);
            printf("\n");
            return true;
        }

        int c = R[0];
        for(int i = R[0]; i!=0; 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+1)) 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+1; c2 = N*N+i*N+k;
    c3 = N*N*2+j*N+k; c4 = N*N*3+((i/3)*3+(j/3))*N+k;
}

DLX dlx;
int main()
{
    while(scanf("%s", g) && strcmp(g,"end") )
    {
        dlx.init(N*N*N, N*N*4);
        int r, c1, c2, c3,c4;
        for(int i = 0; i<N; i++)
            for(int j = 0; j<N; j++)
                for(int k = 1; k<=N; k++)
                if(g[i*N+j]=='.' || g[i*N+j]=='0'+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(0);      //一起摇摆
    }
    return 0;
}