SGOI 海上搜索 状压DP

SSL1317 本来以为暴力枚举矩形左上角dfs推右下角的点和DP复杂度一样结果一想发现复杂度整整多了个n = =

#include
#include
#include
#include
#include
#include
#include
#define LL long long
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define down(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
inline LL read()
{
	LL d=0,f=1;char s=getchar();
	while(s<'0'||s>'9'){if(s=='-')f=-1;s=getchar();}
	while(s>='0'&&s<='9'){d=d*10+s-'0';s=getchar();}
	return d*f;
}
#define N 55
#define NM 64
unsigned long a[N][N],f[N][N],ff[N][N];
int pre[5][3];
int n,m,nm,ans=10000;

void floodfill(int x,int y,int color)
{
	if(a[x][y]==0)return;
	if(f[x][y]==color)return;
	f[x][y]=color;
	fo(i,1,4)
	floodfill(x+pre[i][1],y+pre[i][2],color);
}

void prework()
{
	memset(a,0,sizeof(a));
	memset(f,0,sizeof(f));
	n=read(),m=read();
	fo(i,1,n)
	{
		fo(j,1,m)
		{
			a[i][j]=read()^1;
		}
	}
	int cnt=1;
	fo(i,-1,1)
	fo(j,-1,1)
	if(abs(i+j)==1)
	pre[cnt][1]=i,pre[cnt++][2]=j;
	cnt=0;
	fo(i,1,n)
	fo(j,1,m)
	if(a[i][j]==1&&f[i][j]==0)floodfill(i,j,++cnt);
	nm=cnt;
}

void DP()
{
	fo(i,1,n)
	fo(j,1,m)
	{
		memset(ff,0,sizeof(ff));
		fo(p,i,n)
		fo(q,j,m)
		{
			ff[p][q]=ff[p-1][q]|ff[p][q-1];
			if(f[p][q]&&!(ff[p][q]&(1<<(f[p][q]-1))))
			ff[p][q]+=1<<(f[p][q]-1);
			if(ff[p][q]==(1<