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Dancing Links 小结 (因为之前丢了一次稿,未完待续)

Dancing Links (DLX)是Knuth为了解决精确覆盖问题而提出的算法,很多搜索问题可以转化位精确覆盖问题从而使用Dancing Links解决(效率会比DFS高很多,因为里面常常蕴涵着意想不到的剪枝)

信息学竞赛中的DLX的问题类似网络流,只需建图+贴版即可

 

参考文献:

1、DLX的原理:Knuth的论文:

原版:http://arxiv.org/abs/cs/0011047

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

2、DLX在搜索中的应用(DLX应用于信息学竞赛的论文):http://bbs.whu.edu.cn/wForum/elite.php?file=%2Fgroups%2FGROUP_3%2FACM_ICPC%2Fnx08%2FD.08010000%2FM.1215524645.R0&ap=563

 

首先介绍精确覆盖问题概念:

精确覆盖:在一个全集$X$中若干子集的集合为$S$,精确覆盖是指,$S$的子集$S*$,满足$X$中的每一个元素在$S*$中\textbf{恰好}出现一次。

即:给你一个0-1矩阵,选取若干行,使得选取的行组成的新矩阵的每一列有且仅有一个1

 

关于DLX的原理这里就不详细介绍了,有兴趣了解者可以去阅读Knuth的论文

 

DLX精确覆盖模板(From kuangbin):

 1 const int maxnode = 100010; 2 const int maxr = 1010; 3 const int maxc = 1010; 4 struct DLX 5 { 6     int n, m, size; 7     int U[maxnode], D[maxnode], R[maxnode], L[maxnode], Row[maxnode], Col[maxnode]; 8     int H[maxc], S[maxr]; 9     int ansd, ans[maxc];10     void init(int _n, int _m)11     {12         n = _n;13         m = _m;14         for (int i = 0; i <= m; i++)15         {16             S[i] = 0;17             U[i] = D[i] = i;18             L[i] = i - 1;19             R[i] = i + 1;20         }21         R[m] = 0; L[0] = m;22         size = m;23         for (int i = 1; i <= n; i++)24             H[i] = -1;25     }26     void Link(int r, int c)27     {28         ++S[Col[++size] = c];29         Row[size] = r;30         D[size] = D[c];31         U[D[c]] = size;32         U[size] = c;33         D[c] = size;34         if (H[r] < 0)H[r] = L[size] = R[size] = size;35         else36         {37             R[size] = R[H[r]];38             L[R[H[r]]] = size;39             L[size] = H[r];40             R[H[r]] = size;41         }42     }43     void remove(int c)44     {45         L[R[c]] = L[c]; R[L[c]] = R[c];46         for (int i = D[c]; i != c; i = D[i])47             for (int j = R[i]; j != i; j = R[j])48             {49                 U[D[j]] = U[j];50                 D[U[j]] = D[j];51                 --S[Col[j]];52             }53     }54     void resume(int c)55     {56         for (int i = U[c]; i != c; i = U[i])57             for (int j = L[i]; j != i; j = L[j])58                 ++S[Col[U[D[j]] = D[U[j]] = j]];59         L[R[c]] = R[L[c]] = c;60     }61     //d为递归深度62     bool Dance(int d)63     {64         if (R[0] == 0)65         {66             ansd = d;67             return true;68         }69         int c = R[0];70         for (int i = R[0]; i != 0; i = R[i])71             if (S[i] < S[c])72                 c = i;73         remove(c);74         for (int i = D[c]; i != c; i = D[i])75         {76             ans[d] = Row[i];77             for (int j = R[i]; j != i; j = R[j])remove(Col[j]);78             if (Dance(d + 1))return true;79             for (int j = L[i]; j != i; j = L[j])resume(Col[j]);80         }81         resume(c);82         return false;83     }84 };
View Code

PS:init函数初始化,0-1矩阵中为1的位置(i,j)用Link(i,j)添加约束,Dance函数为解决过程(bool表示能否完成),ansd为选取的数量,and[]为选取的行

 

精确覆盖问题可以延伸为重复覆盖问题,定义如下:

重复覆盖:在一个全集$X$中若干子集的集合为$S$,重复覆盖是指,$S$的子集$S*$,满足$X$中的每一个元素在$S*$中\textbf{至少}出现一次。

即:给你一个0-1矩阵,选取若干行,使得选取的行组成的新矩阵的每一列至少有一个1

 

DLX重复覆盖模板

 1 const int maxnode = 360000; 2 const int maxc = 500; 3 const int maxr = 500; 4 const int inf = 0x3f3f3f3f; 5 struct DLX 6 { 7     int L[maxnode], R[maxnode], D[maxnode], U[maxnode], C[maxnode]; 8     int S[maxc], H[maxr], size; 9     int ans;10     ///不需要S域11     void Link(int r, int c)12     {13         S[c]++; C[size] = c;14         U[size] = U[c]; D[U[c]] = size;15         D[size] = c; U[c] = size;16         if (H[r] == -1) H[r] = L[size] = R[size] = size;17         else18         {19             L[size] = L[H[r]]; R[L[H[r]]] = size;20             R[size] = H[r]; L[H[r]] = size;21         }22         size++;23     }24     void remove(int c)25     {26         for (int i = D[c]; i != c; i = D[i])27             L[R[i]] = L[i], R[L[i]] = R[i];28     }29     void resume(int c)30     {31         for (int i = U[c]; i != c; i = U[i])32             L[R[i]] = R[L[i]] = i;33     }34     int h() ///用精确覆盖去估算剪枝35     {36         int ret = 0;37         bool vis[maxc];38         memset (vis, false, sizeof(vis));39         for (int i = R[0]; i; i = R[i])40         {41             if (vis[i])continue;42             ret++;43             vis[i] = true;44             for (int j = D[i]; j != i; j = D[j])45                 for (int k = R[j]; k != j; k = R[k])46                     vis[C[k]] = true;47         }48         return ret;49     }50     //根据具体问题选择限制搜索深度或直接求解。51     bool Dance(int k)52     {53         if (k + h() >= ans) return 0;54         if (!R[0])55         {56             if (k < ans)ans = k;57             return 1;58         }59         int c = R[0];60         for (int i = R[0]; i; i = R[i])61             if (S[i] < S[c])c = i;62         for (int i = D[c]; i != c; i = D[i])63         {64             remove(i);65             for (int j = R[i]; j != i; j = R[j])66                 remove(j);67             Dance(k + 1);68             for (int j = L[i]; j != i; j = L[j])69                 resume(j);70             resume(i);71         }72         return 0;73     }74     void initL(int x) ///col is 1~x,row start from 175     {76         for (int i = 0; i <= x; ++i)77         {78             S[i] = 0;79             D[i] = U[i] = i;80             L[i + 1] = i; R[i] = i + 1;81         }///对列表头初始化82         R[x] = 0;83         size = x + 1; ///真正的元素从m+1开始84         memset (H, -1, sizeof(H));85         ///mark每个位置的名字86     }87 };
View Code

输入:Link()
输出:ans, bool Dance(int k)

 

重点来了:如何建图?这里选取典型例题说明

 

(因为之前丢了一次稿,未完待续)

Dancing Links 小结 (因为之前丢了一次稿,未完待续)