fzu 1686 神龍的難題

 http://acm.fzu.edu.cn/problem.php?pid=1686

DLX重複覆蓋，大小爲n1*m1的矩陣爲行，每個怪物爲列

# include <math.h>
# include <stdio.h>
# include <string.h>
# include <stdlib.h>
# include <algorithm>
# include <iostream>
# include <string>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <vector>
# include <cstring>
# include <list>
# include <ctime>

# define For(i,a)   for((i)=0;i<(a);(i)++)
# define MAX(x,y)   ((x)>(y)? (x):(y))
# define MIN(x,y)   ((x)<(y)? (x):(y))
# define MEM(a)     (memset((a),0,sizeof(a)))
# define MEME(a)    (memset((a),-1,sizeof(a)))
# define MEMX(a)    (memset((a),0x7f,sizeof(a)))

using namespace std;

typedef long long           ll      ;
typedef unsigned long long  ull     ;
typedef unsigned int        uint    ;
typedef unsigned char       uchar   ;

#define N 260
#define M 260
int U[N*M],R[N*M],L[N*M],D[N*M];
int S[M],H[N],C[N*M];
int head,size,ans;
void build(int r,int c)
{
    S[c]++; C[size]=c;
    U[size]=U[c]; D[U[c]]=size;
    D[size]=c; U[c]=size;
    if(H[r]==-1) H[r]=R[size]=L[size]=size;
    else
    {
        L[size]=L[H[r]]; R[L[H[r]]]=size;
        R[size]=H[r]; L[H[r]]=size;
    }
    size++;
}
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 h()
{
    int i,j,k,ret=0;
    char vis[M];
    MEM(vis);
    for(i=R[head];i!=head;i=R[i])
    {
        if(vis[i]==0)
        {
            ret++;vis[i]=1;
            for(j=D[i];j!=i;j=D[j])
                for(k=R[j];k!=j;k=R[k])
                    vis[C[k]]=1;
        }
    }
    return ret;
}
void dfs(int dep)
{
    int i,j,c,min;
    if(R[head]==head)
    {
        if(dep<ans) ans=dep;
        return ;
    }
    if(dep+h()>=ans)
        return ;
    for(i=R[head],min=10000000;i!=head;i=R[i])
        if(S[i]<min)
        {
            min=S[i]; c=i;
        }
    remove(c);
    L[R[c]]=L[c]; R[L[c]]=R[c];
    for(i=D[c];i!=c;i=D[i])
    {
        L[R[i]]=i;R[L[i]]=i;
        for(j=R[i];j!=i;j=R[j])
            remove(j);
        dfs(dep+1);
        for(j=L[i];j!=i;j=L[j])
            resume(j);
        L[R[i]]=L[i];R[L[i]]=R[i];
    }
    L[R[c]]=c; R[L[c]]=c;
    resume(c);
}
int a[20][20];
int main()
{
    int n,m,n1,m1;
    int i,j,k,r,p;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        k=1;MEM(a);
        for(i=1;i<=n;i++)
            for(j=1;j<=m;j++)
            {
                scanf("%d",&p);
                if(p==1) a[i][j]=k++;
            }
        scanf("%d%d",&n1,&m1);
        for(i=0;i<k;i++)
        {
            R[i]=i+1;L[i]=i-1;
            U[i]=i; D[i]=i;
            S[i]=0;
        }
        R[k-1]=0;L[0]=k-1;
        MEME(H);
        head=0;size=k;
        p=0;
        for(i=1;i<=n-n1+1;i++)
            for(j=1;j<=m-m1+1;j++)
            {
                for(k=i;k<=i+n1-1;k++)
                    for(r=j;r<=j+m1-1;r++)
                        if(a[k][r]!=0)
                            build(p,a[k][r]);
                p++;
            }
        ans=1000000;
        dfs(0);
        printf("%d\n",ans);
    }
    return 0;
}