设$A[i][j]=[a[i][j]=G],B[i][j]=[b[i][j]=L]$,枚举右下角,则对应$(A-B)^2$的和就是匹配成功的格子数。
$(a-b)^2=a^2+b^2-2ab$,将矩阵展开成一维,然后将$B$翻转,用FFT求出$sum a imes b$即可。
时间复杂度$O(BRClog(RC))$。
#include<cstdio> #include<cmath> #include<algorithm> using namespace std; const int N=530000,M=505; int Case,T,r,c,n,m,i,j,k,pos[N],tmp,val,s[M][M],ans,sx,sy,c0,c1;char a[M][M],b[M][M]; struct comp{ double r,i;comp(double _r=0,double _i=0){r=_r;i=_i;} comp operator+(const comp&x){return comp(r+x.r,i+x.i);} comp operator-(const comp&x){return comp(r-x.r,i-x.i);} comp operator*(const comp&x){return comp(r*x.r-i*x.i,r*x.i+i*x.r);} comp conj(){return comp(r,-i);} }A[N],B[N]; const double pi=acos(-1.0); void FFT(comp a[],int n,int t){ for(int i=1;i<n;i++)if(i<pos[i])swap(a[i],a[pos[i]]); for(int d=0;(1<<d)<n;d++){ int m=1<<d,m2=m<<1; double o=pi*2/m2*t;comp _w(cos(o),sin(o)); for(int i=0;i<n;i+=m2){ comp w(1,0); for(int j=0;j<m;j++){ comp&A=a[i+j+m],&B=a[i+j],t=w*A; A=B-t;B=B+t;w=w*_w; } } } if(t==-1)for(int i=0;i<n;i++)a[i].r/=n; } int main(){ scanf("%d%d",&r,&c); for(i=0;i<r;i++)scanf("%s",a[i]); for(i=0;i<r;i++)for(j=0;j<c;j++){ tmp=a[i][j]=='G'; A[i*c+j]=comp(tmp,0); s[i][j]=tmp*tmp; if(i)s[i][j]+=s[i-1][j]; if(j)s[i][j]+=s[i][j-1]; if(i&&j)s[i][j]-=s[i-1][j-1]; } for(k=1;k<r*c;k<<=1);k<<=1; j=__builtin_ctz(k)-1; for(i=0;i<k;i++)pos[i]=pos[i>>1]>>1|((i&1)<<j); FFT(A,k,1); scanf("%d",&Case); for(T=1;T<=Case;T++){ scanf("%d%d",&n,&m); for(i=0;i<n;i++)scanf("%s",b[i]); for(i=0;i<k;i++)B[i]=comp(); for(val=i=0;i<n;i++)for(j=0;j<m;j++){ tmp=b[i][j]=='L'; val+=tmp*tmp; B[(n-i-1)*c+m-j-1]=comp(tmp,0); } FFT(B,k,1); for(i=0;i<k;i++)B[i]=A[i]*B[i]; FFT(B,k,-1); ans=-1; for(i=n-1;i<r;i++)for(j=m-1;j<c;j++){ tmp=s[i][j]+val; if(i>=n)tmp-=s[i-n][j]; if(j>=m)tmp-=s[i][j-m]; if(i>=n&&j>=m)tmp+=s[i-n][j-m]; tmp-=(int(B[i*c+j].r+0.5))*2; if(tmp>ans)ans=tmp,sx=i-n+1,sy=j-m+1; } for(c0=c1=i=0;i<n;i++)for(j=0;j<m;j++)if(a[sx+i][sy+j]==b[i][j])if(b[i][j]=='G')c0++;else c1++; printf("Case #%d: %d %d %d %d ",T,sx+1,sy+1,c0,c1); } return 0; }