loj6388 「THUPC2018」赛艇 / Citing

大概就是一个形状要嵌到矩阵里头这样子，看看代码就明白了qwq

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;
int n, m, k, lim=1, limcnt, rev[8388655];
char ss[5000005];
const double PI=acos(-1.0);
struct Complex{
	double x, y;
	Complex(double u=0.0, double v=0.0){
		x = u; y = v;
	}
	Complex operator+(const Complex &u)const{
		return Complex(x+u.x, y+u.y);
	}
	Complex operator-(const Complex &u)const{
		return Complex(x-u.x, y-u.y);
	}
	Complex operator*(const Complex &u)const{
		return Complex(x*u.x-y*u.y, x*u.y+y*u.x);
	}
}a[8388655], b[8388655];
void fft(Complex a[], int opt){
	for(int i=0; i<lim; i++)
		if(i<rev[i])
			swap(a[i], a[rev[i]]);
	for(int i=2; i<=lim; i<<=1){
		int tmp=i>>1;
		Complex wn=Complex(cos(2*PI/i), opt*sin(2*PI/i));
		for(int j=0; j<lim; j+=i){
			Complex w=Complex(1.0, 0.0);
			for(int k=0; k<tmp; k++){
				Complex tmp1=a[j+k], tmp2=w*a[j+k+tmp];
				a[j+k] = tmp1 + tmp2;
				a[j+k+tmp] = tmp1 - tmp2;
				w = w * wn;
			}
		}
	}
	if(opt==-1)
		for(int i=0; i<lim; i++)
			a[i].x /= lim;
}
int main(){
	cin>>n>>m>>k;
	for(int i=0; i<n; i++){
		scanf("%s", ss);
		for(int j=0; j<m; j++)
			a[i*m+j].x = ss[j]=='1';
	}
	int nx=0, ny=0, mxy=0, mxx=0, mnx=0, mny=0;
	scanf("%s", ss+1);
	for(int i=1; i<=k; i++){
		if(ss[i]=='w')	nx--;
		if(ss[i]=='s')	nx++;
		if(ss[i]=='a')	ny--;
		if(ss[i]=='d')	ny++;
		mxy = max(mxy, ny); mxx = max(mxx, nx);
		mny = min(mny, ny); mnx = min(mnx, nx);
	}
	nx = ny = 0;
	mxx -= mnx; mxy -= mny;
	b[(nx-mnx)*m+(ny-mny)].x = 1;
	for(int i=1; i<=k; i++){
		if(ss[i]=='w')	nx--;
		if(ss[i]=='s')	nx++;
		if(ss[i]=='a')	ny--;
		if(ss[i]=='d')	ny++;
		b[(nx-mnx)*m+(ny-mny)].x = 1;
	}
	int u=n*m-1;
	while(lim<=2*u)	lim <<= 1, limcnt++;
	for(int i=0; i<lim; i++)
		rev[i] = (rev[i>>1]>>1) | ((i&1)<<(limcnt-1));
	reverse(b, b+1+u);
	fft(a, 1);
	fft(b, 1);
	for(int i=0; i<lim; i++)
		a[i] = a[i] * b[i];
	fft(a, -1);
	int ans=0;
	for(int i=0; i<n-mxx; i++)
		for(int j=0; j<m-mxy; j++)
			if(a[u+i*m+j].x<0.5)
				ans++;		
	cout<<ans<<endl;
	return 0;
}