可重复覆盖的DLX

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在精确覆盖的基础上只修改remove() 和resume() 就好了，每次选中一行，只把这一行对应的列以及这一行删除即可。

很显然的，这样降低了矩阵缩小的速度，但是出现A*优化的补救方法，H()是在VJ上扒下来的，233。

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <queue>
#include <cmath>
#include <stack>
#include <map>

#pragma comment(linker, "/STACK:1024000000")
#define EPS (1e-8)
#define LL long long
#define ULL unsigned long long
#define _LL __int64
#define INF 0x3f3f3f3f
#define Mod 100007

using namespace std;

const int MAXCLOUM = 230,MAXROW = 230,MAXPOINT = 230*230;

bool MARKCLOUM[MAXCLOUM];
int sta[MAXROW],sta_anw;//答案栈以及栈内元素个数
int SIZE[MAXCLOUM];//每一列元素个数，不包括虚拟列指针
int PRE[MAXROW];//每一行最后加入的元素位置，初始为-1.
int C[MAXCLOUM];//每一列的头指针的位置
int ROW[MAXPOINT],COL[MAXPOINT];//每个元素的行列坐标
int L[MAXPOINT],R[MAXPOINT],U[MAXPOINT],D[MAXPOINT];//每个元素的四个指针
int Top;//第一个未分配元素的地址

inline int Creat(int row,int col)
{
    COL[Top] = col;
    ROW[Top] = row;

    U[Top] = Top;
    D[Top] = Top;
    L[Top] = Top;
    R[Top] = Top;

    return Top++;
}

void Init(int row,int col)
{
    sta_anw = 0;
    Top = 0;
    C[0] = Creat(0,0);
    for(int i = 1; i <= col; ++i)
    {
        C[i] = Creat(0,i);
        R[C[i]] = R[C[i-1]];
        L[R[C[i-1]]] = C[i];
        R[C[i-1]] = C[i];
        L[C[i]] = C[i-1];
        SIZE[i] = 0;
    }

    memset(PRE,-1,sizeof(PRE));
}

void Insert(int row,int col)
{
    //cout<<"row = "<<row<<" col = "<<col<<endl;
    int now = Creat(row,col);

    D[U[C[col]]] = now;
    U[now] = U[C[col]];
    D[now] = C[col];
    U[C[col]] = now;
    ++SIZE[col];

    if(PRE[row] != -1)
    {
        L[R[PRE[row]]] = now;
        R[now] = R[PRE[row]];
        L[now] = PRE[row];
        R[PRE[row]] = now;
    }

    PRE[row] = now;
}

void Remove(int c)
{
    for(int i = D[c]; i != c; i = D[i])
    {
        L[R[i]] = L[i];
        R[L[i]] = R[i];
    }
}

void Resume(int c)
{
    for(int i = U[c]; i != c; i = U[i])
    {
        L[R[i]] = i;
        R[L[i]] = i;
    }
}

int Min;

int H()
{
     memset(MARKCLOUM,false,sizeof(MARKCLOUM));

     int i,j,k,ans = 0;

     for(i = R[C[0]];i != C[0] ;i = R[i])
     {
         if(MARKCLOUM[COL[i]])
            continue;

         MARKCLOUM[COL[i]] = true;
         ans++;

         for(j = D[i];j != i; j = D[j])
         {
             for(k = R[j];k != j ; k = R[k])
             {
                 MARKCLOUM[COL[k]] = true;
             }
         }
     }
     return ans;
}

bool Dance(int dis)
{
    int now,i,j;

    if(dis+H() >= Min)
        return false;

    if(R[C[0]] == C[0])
        return Min = dis,true;

    for(i = now = R[C[0]];i != C[0];i = R[i])
    {
        if(SIZE[COL[i]] < SIZE[COL[now]])
            now = i;
    }

    for(i = D[now]; i != now; i = D[i])
    {
        Remove(i);
        for(j = R[i]; j != i; j = R[j])
            Remove(j);

        Dance(dis+1);

        for(j = L[i]; j != i; j = L[j])
            Resume(j);
        Resume(i);
    }

    return false;
}

int Map[20][20];

int main()
{
    //freopen("data.in","r",stdin);
    int n,m,i,j,x,y,k,l;

    while(scanf("%d %d",&n,&m) != EOF)
    {
        for(i = 0;i < n; ++i)
        {
            for(j = 0;j < m; ++j)
                scanf("%d",&Map[i][j]);
        }

        scanf("%d %d",&x,&y);

        int sum = 0;

        for(i = 0;i < n; ++i)
            for(j = 0;j < m; ++j)
                if(Map[i][j])
                    Map[i][j] = ++sum;

        Init(n*m,sum);
        Min = min(sum,((n+x-1)/x)*((m+y-1)/y));

        sum = 0;

        for(i = 0;i <= n-x; ++i)
            for(j = 0;j <= m-y; ++j)
            {
                sum++;
                for(l = 0;l < x; ++l)
                {
                    for(k = 0;k < y; ++k)
                    {
                        if(Map[i+l][j+k])
                            Insert(i*m+j,Map[i+l][j+k]);
                    }
                }
            }

        Dance(0);

        printf("%d\n",Min);
    }

    return 0;
}

可重复覆盖的DLX

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原文地址：http://blog.csdn.net/zmx354/article/details/43148383