POJ - 3074（DancingLink）（数独）

题目：POJ - 3074 

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

 

分析：行：每个格子可以放9种，共有9 * 9 个格子，所以行的数目为：9 * 9 * 9；

列：每个格子有四种状态：（1）这个格子有x；（2）这一行有 x；（3）这一列有x；（4）这个9宫格有x；

所以共有 N * N* 4列；

#include 
#include 
#include 

using namespace std;

const int N = 9;
const int maxn = N * N * N + 10;
const int maxnode = maxn * 4 + maxn + 10;
char g[maxn];
struct DLX
{
    int n,m,sz;
    int U[maxnode],D[maxnode],R[maxnode],L[maxnode],Row[maxnode],Col[maxnode];
    int H[maxn],S[maxn];
    int ansd,ans[maxn];
    void init(int _n,int _m)
    {
        n = _n;
        m = _m;
        for(int i = 0;i <= m;i ++)
        {
            S[i] = 0;
            U[i] = D[i] = i;
            L[i] = i - 1;
            R[i] = i + 1;
        }
        R[m] = 0;
        L[0] = m;
        sz = m;
        for(int i = 1;i <= n;i ++)
        {
            H[i] = -1;
        }
    }

    void Link(int r,int c)
    {
        ++ S[Col[++ sz] = c];
        Row[sz] = r;
        D[sz] = D[c];
        U[D[c]] = sz;
        U[sz] = c;
        D[c] = sz;
        if(H[r] < 0) H[r] = L[sz] = R[sz] = sz;
        else
        {
            R[sz] = R[H[r]];
            L[R[H[r]]] = sz;
            L[sz] = H[r];
            R[H[r]] = sz;
        }
    }

    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]];
            }
        }
    }

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

    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])
        {
            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;
    }
};

void place(int &r,int &c1,int &c2,int &c3,int &c4,int i,int j,int k)
{
    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) == 1)
    {
        if(strcmp(g,"end") == 0) break;
        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);
                        dlx.Link(r,c1);
                        dlx.Link(r,c2);
                        dlx.Link(r,c3);
                        dlx.Link(r,c4);
                    }
                }
            }
        }
        dlx.Dance(0);
    }
    return 0;
}
/*
.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
*/