[UVA 12633] Super Rooks on Chessboard FFT+计数

如果只有行和列的覆盖,那么可以直接做,但现在有左上到右下的覆盖.

考虑对行和列的覆盖情况做一个卷积,然后就有了x+y的非覆盖格子数.

然后用骑士的左上到右下的覆盖特判掉那些x+y的格子就可以了.

注意题意,Row是从上到下来的,被坑得好惨.

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<ctime>
#include<string>
#include<iomanip>
#include<algorithm>
#include<map>
using namespace std;
#define LL long long
#define FILE "dealing"
#define up(i,j,n) for(LL i=j;i<=n;++i)
#define db double
#define ull unsigned long long
#define eps 1e-10
#define pii pair<LL,LL>
LL read(){
	LL x=0,f=1,ch=getchar();
	while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
	while(ch>=‘0‘&&ch<=‘9‘){x=(x<<1)+(x<<3)+ch-‘0‘;ch=getchar();}
	return f*x;
}
const LL maxn=402000,maxm=20000,mod=(LL)(1e9+7+0.1),limit=(LL)(1e6+1),inf=(LL)(1e9);
bool cmax(LL& a,LL b){return a<b?a=b,true:false;}
bool cmin(LL& a,LL b){return a>b?a=b,true:false;}
namespace FFT{
	db pi=acos(-1.0);
	struct cp{
		db x,y;
		cp(db x=0,db y=0):x(x),y(y){}
		cp operator+(const cp& b){return cp(x+b.x,y+b.y);}
		cp operator-(const cp& b){return cp(x-b.x,y-b.y);}
		cp operator*(const cp& b){return cp(x*b.x-y*b.y,x*b.y+y*b.x);}
	}w[maxn],a[maxn],b[maxn];
	LL R[maxn],H,L;
	void FFT(cp* a,LL f){
		up(i,0,L-1)if(i<R[i])swap(a[i],a[R[i]]);
		for(LL len=2;len<=L;len<<=1){
			LL l=len>>1;
			cp wn(cos(pi/l),f*sin(pi/l));
			up(i,1,l-1)w[i]=w[i-1]*wn;
			for(LL st=0;st<L;st+=len)
				for(LL k=0;k<l;k++){
					cp x=a[st+k],y=w[k]*a[st+k+l];
					a[st+k]=x+y;a[st+k+l]=x-y;
				}
		}
		if(f==-1)up(i,0,L-1)a[i].x/=L;
	}
	void solve(LL* c,LL* d,LL n,LL m,LL* ch){
		n++,m++;
		up(i,0,n-1)a[i].x=c[i],a[i].y=0;
		up(i,0,m-1)b[i].x=d[i],b[i].y=0;
		for(H=0,L=1;L<n+m-1;H++)L<<=1;
		up(i,n,L)a[i].x=a[i].y=0;
		up(i,m,L)b[i].x=b[i].y=0;
		up(i,1,L)R[i]=(R[i>>1]>>1)|((i&1)<<(H-1));
		w[0].x=1;
		FFT(a,1);FFT(b,1);
		up(i,0,L-1)a[i]=a[i]*b[i];
		FFT(a,-1);
		up(i,1,n+m-1)ch[i]=(LL)(a[i].x+0.5);
	}
};
LL n,m,K;
LL a[maxn],b[maxn],c[maxn],d[maxn];
int main(){
	freopen(FILE".in","r",stdin);
	freopen(FILE".out","w",stdout);
	LL T=read();
	up(j,1,T){
		n=read(),m=read(),K=read();
		up(i,1,n)a[i]=1;
		up(i,1,m)b[i]=1;
		up(i,1,n+m)d[i]=0;
		up(i,1,K){
			LL x=n-read()+1,y=read();
			a[x]=0,b[y]=0;
			d[x+y]=1;
		}
		FFT::solve(a,b,n,m,c);
		LL ans=0;
		up(i,1,n+m)if(c[i]&&!d[i])ans+=c[i];
		printf("Case %lld: %lld\n",j,ans);
	}
	return 0;
}