POJ 2676 Sudoku (搜索，Dancing Links)

题目：

http://poj.org/problem?id=2676

题意：

数独，每行1-9，每列1-9，每3*3小格1-9，填数，不能重复

方法：Dancing Links(16ms)或者DFS暴搜(400-900ms)

Dancing Links(DLX) 是为了解决矩阵精确覆盖问题的算法，算法效率非常高

使用DLX解决的问题必须转化为矩阵精确覆盖问题:

1、DLX详解：

http://wenku.baidu.com/view/d8f13dc45fbfc77da269b126.html

2、转化方法：

非常详细的讲解：http://www.cnblogs.com/grenet/p/3163550.html

约束1：每个格子只能填一个数：dlx.Link(t, encode(0, i, j));

约束2：每行需1-9：dlx.Link(t, encode(1, i, k - 1));

约束3：每列需1-9：dlx.Link(t, encode(2, j, k - 1));

约束4：每3*3格子需1-9：dlx.Link(t, encode(3, (i / 3) * 3 + j / 3, k - 1));

 1 void build()
 2 {
 3     for (int i = 0; i < 9; i++)
 4         for (int j = 0; j < 9; j++)
 5             for (int k = 1; k <= 9; k++)
 6                 if (mtx[i][j] == '0' || mtx[i][j] == k + '0')
 7                 {
 8                     int t = encode(i, j, k - 1);
 9                     dlx.Link(t, encode(0, i, j));
10                     dlx.Link(t, encode(1, i, k - 1));
11                     dlx.Link(t, encode(2, j, k - 1));
12                     dlx.Link(t, encode(3, (i / 3) * 3 + j / 3, k - 1));
13                 }
14 }

DLX模板(转自kuangbin(http://www.cnblogs.com/kuangbin/p/3752854.html))：

const int maxnode = 100010;
const int MaxM = 1010;
const int MaxN = 1010;
struct DLX
{
    int n, m, size;
    int U[maxnode], D[maxnode], R[maxnode], L[maxnode], Row[maxnode], Col[maxnode];
    int H[MaxN], S[MaxM];
    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;
        size = m;
        for (int i = 1; i <= n; i++)
            H[i] = -1;
    }
    void Link(int r, int c)
    {
        ++S[Col[++size] = c];
        Row[size] = r;
        D[size] = D[c];
        U[D[c]] = size;
        U[size] = c;
        D[c] = size;
        if (H[r] < 0)H[r] = L[size] = R[size] = size;
        else
        {
            R[size] = R[H[r]];
            L[R[H[r]]] = size;
            L[size] = H[r];
            R[H[r]] = size;
        }
    }
    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;
    }
    //d为递归深度
    bool Dance(int d)
    {
        if (R[0] == 0)
        {
            ansd = d;
            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;
    }
};

代码：

/********************************************
*ACM Solutions
*
*@Title:
*@Version: 1.0
*@Time: 2014-xx-xx
*@Solution: http://www.cnblogs.com/xysmlx/p/xxxxxxx.html
*
*@Author: xysmlx(Lingxiao Ma)
*@Blog: http://www.cnblogs.com/xysmlx
*@EMail: xysmlx@163.com
*
*Copyright (C) 2011-2015 xysmlx(Lingxiao Ma)
********************************************/
// #pragma comment(linker, "/STACK:102400000,102400000")
#include <cstdio>
#include <iostream>
#include <cstring>
#include <string>
#include <cmath>
#include <set>
#include <list>
#include <map>
#include <iterator>
#include <cstdlib>
#include <vector>
#include <queue>
#include <stack>
#include <algorithm>
#include <functional>
using namespace std;
typedef long long LL;
#define pb push_back
#define ROUND(x) round(x)
#define FLOOR(x) floor(x)
#define CEIL(x) ceil(x)
const int maxn = 0;
const int maxm = 0;
const int inf = 0x3f3f3f3f;
const LL inf64 = 0x3f3f3f3f3f3f3f3fLL;
const double INF = 1e30;
const double eps = 1e-6;
const int P[4] = {0, 0, -1, 1};
const int Q[4] = {1, -1, 0, 0};
const int PP[8] = { -1, -1, -1, 0, 0, 1, 1, 1};
const int QQ[8] = { -1, 0, 1, -1, 1, -1, 0, 1};

const int maxnode = 100010;
const int MaxM = 1010;
const int MaxN = 1010;
struct DLX
{
    int n, m, size;
    int U[maxnode], D[maxnode], R[maxnode], L[maxnode], Row[maxnode], Col[maxnode];
    int H[MaxN], S[MaxM];
    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;
        size = m;
        memset(H, -1, sizeof(H));
        // for (int i = 1; i <= n; i++)
        //     H[i] = -1;
    }
    void Link(int r, int c)
    {
        ++S[Col[++size] = c];
        Row[size] = r;
        D[size] = D[c];
        U[D[c]] = size;
        U[size] = c;
        D[c] = size;
        if (H[r] < 0)H[r] = L[size] = R[size] = size;
        else
        {
            R[size] = R[H[r]];
            L[R[H[r]]] = size;
            L[size] = H[r];
            R[H[r]] = size;
        }
    }
    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;
    }
    //d为递归深度
    bool Dance(int d)
    {
        if (R[0] == 0)
        {
            ansd = d;
            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;
    }
} dlx;

int kase;
int n, m;
char mtx[20][20];
void init()
{
    n = 9, m = 9;
    kase++;
    dlx.init(1, 324);
}
void input()
{
    for (int i = 0; i < 9; i++)
        scanf("%s", mtx[i]);
}
void debug()
{
    //
}
int encode(int a, int b, int c)
{
    return a * 81 + b * 9 + c + 1;
}
void decode(int code, int &a, int &b, int &c)
{
    code--;
    c = code % 9;
    code /= 9;
    b = code % 9;
    code /= 9;
    a = code;
}
void build()
{
    for (int i = 0; i < 9; i++)
        for (int j = 0; j < 9; j++)
            for (int k = 1; k <= 9; k++)
                if (mtx[i][j] == '0' || mtx[i][j] == k + '0')
                {
                    int t = encode(i, j, k - 1);
                    dlx.Link(t, encode(0, i, j));
                    dlx.Link(t, encode(1, i, k - 1));
                    dlx.Link(t, encode(2, j, k - 1));
                    dlx.Link(t, encode(3, (i / 3) * 3 + j / 3, k - 1));
                }
}
void solve()
{
    build();
    dlx.Dance(0);
    for (int i = 0; i < dlx.ansd; i++)
    {
        int r, c, k;
        decode(dlx.ans[i], r, c, k);
        mtx[r][c] = k + '1';
    }
}
void output()
{
    for (int i = 0; i < 9; i++)
        printf("%s
", mtx[i]);
}
int main()
{
    // int size = 256 << 20; // 256MB
    // char *p = (char *)malloc(size) + size;
    // __asm__("movl %0, %%esp
" :: "r"(p));

    // std::ios_base::sync_with_stdio(false);
#ifdef xysmlx
    freopen("in.cpp", "r", stdin);
#endif

    kase = 0;
    int T;
    scanf("%d", &T);
    while (T--)
    {
        init();
        input();
        solve();
        output();
    }
    return 0;
}