Dancing Links用来解决如下精确匹配的问题:
选择若干行使得每一列恰好有一个1。Dancing Links通过对非零元素建立双向十字循环链表。上面的例子建立的链表如下所示:
计算的时候使用搜索的策略。每次选出1最少的一列,比如c,然后选择这一列中的某一行,比如r,(r,c)=1,然后r中所有1所在的列,那些其他行这些列有1的都删掉(这些行不会在r算入答案后也在答案里,否则就有某些列多于一个1出现)。然后这就变成一个规模更小的问题,继续搜索。无解时要回溯。
class CDancingLinks
{
protected:
struct DancingLinksNode
{
DancingLinksNode* left;
DancingLinksNode* right;
DancingLinksNode* down;
DancingLinksNode* up;
int col;
int row;
}; typedef DancingLinksNode Node; int *m_columnEleNumbers;
int m_colNumber;
int m_rowNumber;
Node* m_pool;
Node** m_head;
int m_curUsePoolIndex; void _Remove(Node* cur)
{
--m_columnEleNumbers[cur->col];
for(Node* p=cur->down;p!=cur;p=p->down)
{
p->left->right=p->right;
p->right->left=p->left;
}
} void _Resume(Node* cur)
{
++m_columnEleNumbers[cur->col];
for(Node* p=cur->up;p!=cur;p=p->up)
{
p->left->right=p;
p->right->left=p;
}
} bool _SearchSolution(const int depth,std::vector<int> &solution)
{
Node* p=_GetNode();
if(p->left==p) return true; int Min=m_rowNumber+;
int MinColumnIndex=;
for(Node* q=p->left;q!=p;q=q->left)
{
if(m_columnEleNumbers[q->col]<Min)
{
Min=m_columnEleNumbers[q->col];
MinColumnIndex=q->col;
}
} for(Node* q=_GetNode(MinColumnIndex)->down;q!=_GetNode(MinColumnIndex);q=q->down)
{
_Remove(q);
solution.push_back(q->row);
for(Node* rr=q->right;rr!=q;rr=rr->right) _Remove(rr);
if(_SearchSolution(depth+,solution)) return true;
for(Node* rr=q->left;rr!=q;rr=rr->left) _Resume(rr);
solution.pop_back();
_Resume(q);
} return false;
} Node* _GetNode(int id) { return m_pool+id; } void _ReleaseMemory()
{
if(m_columnEleNumbers)
{
delete[] m_columnEleNumbers;
m_columnEleNumbers=nullptr;
} if(m_pool)
{
delete[] m_pool;
m_pool=nullptr;
}
if(m_head)
{
delete[] m_head;
m_head=nullptr;
}
} public: CDancingLinks():m_colNumber(-),m_rowNumber(-),
m_columnEleNumbers(nullptr),m_pool(nullptr),m_head(nullptr) {} /***
列下标为[1,Column]
***/
CDancingLinks(const int Column,const int Row):
m_columnEleNumbers(nullptr),m_pool(nullptr),m_head(nullptr)
{
SetSize(Column,Row);
} /***
列下标为[1,Column]
***/
void SetSize(const int Column,const int Row)
{
m_colNumber=Column;
m_rowNumber=Row; _ReleaseMemory(); m_columnEleNumbers=new int[m_colNumber+];
m_pool=new Node[m_colNumber*(m_rowNumber+)+];
m_head=new Node*[m_rowNumber+];
Clear();
} void Clear()
{
for(int i=;i<=m_colNumber;++i)
{
Node* cur=_GetNode(i);
cur->left=((i==m_colNumber)?_GetNode():_GetNode(i+));
cur->right=((==i)?_GetNode(m_colNumber):_GetNode(i-));
m_columnEleNumbers[i]=; cur->up=cur->down=_GetNode(i);
cur->col=i;
cur->row=;
}
for(int i=;i<=m_rowNumber;++i) m_head[i]=NULL;
m_curUsePoolIndex=m_colNumber+;
} ~CDancingLinks()
{
_ReleaseMemory();
} void AddElement(const int row,const int col)
{ Node* cur=m_pool+(m_curUsePoolIndex++); cur->up=_GetNode(col);
cur->down=_GetNode(col)->down;
m_pool[col].down->up=cur;
m_pool[col].down=cur; if(m_head[row]==NULL)
{
m_head[row]=cur->left=cur->right=cur;
}
else
{
cur->left=m_head[row]->left;
cur->right=m_head[row];
m_head[row]->left->right=cur;
m_head[row]->left=cur;
}
++m_columnEleNumbers[col];
cur->col=col;
cur->row=row;
} bool GetSolution(std::vector<int> &Solution)
{
return _SearchSolution(,Solution);
}
};