UVA 12633 Super Rooks on Chessboard ——FFT

发现对角线上的和是一个定值。

然后就不考虑斜着，可以处理出那些行和列是可以放置的。

然后FFT，统计出每一个可行的项的系数和就可以了。

#include <map>
#include <cmath>
#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define F(i,j,k) for (int i=j;i<=k;++i)
#define D(i,j,k) for (int i=j;i>=k;--i)
#define ll long long
#define mp make_pair

#define maxn 200005

struct Complex{
	double x,y;
	Complex () {}
	Complex (double _x,double _y) {x=_x;y=_y;}
	Complex operator + (Complex a) {return Complex(x+a.x,y+a.y);}
	Complex operator - (Complex a) {return Complex(x-a.x,y-a.y);}
	Complex operator * (Complex a) {return Complex(x*a.x-y*a.y,x*a.y+y*a.x);}
}A[maxn],B[maxn],C[maxn];

int t,r,c,n,k,m,len,line[maxn],row[maxn],ang[maxn],rev[maxn],cntl=0,cntr=0,kas=0;
ll ans=0;

const double pi=acos(-1.0);

void FFT(Complex * x,int n,int flag)
{
	F(i,0,n-1) if (rev[i]>i) swap(x[i],x[rev[i]]);
	for (int m=2;m<=n;m<<=1)
	{
		Complex wn=Complex(cos(2*pi/m),flag*sin(2*pi/m));
		for (int i=0;i<n;i+=m)
		{
			Complex w=Complex(1.0,0);
			for (int j=0;j<(m>>1);++j)
			{
				Complex u=x[i+j],v=x[i+j+(m>>1)]*w;
				x[i+j]=u+v; x[i+j+(m>>1)]=u-v;
				w=w*wn;
			}
		}
	}
	if (flag==-1) F(i,0,n-1) x[i].x=(x[i].x+0.3)/n;
}

int main()
{
	scanf("%d",&t);
	while (t--)
	{
		memset(line,0,sizeof line);
		memset(row,0,sizeof row);
		memset(ang,0,sizeof ang);
		cntl=cntr=0;
		scanf("%d%d%d",&r,&c,&k);
		F(i,1,k)
		{
			int x,y;
			scanf("%d%d",&x,&y);
			x=r-x+1;
			if (!line[x]) cntl++;
			if (!row[y]) cntr++;
			line[x]=1; row[y]=1;
			ang[x+y]=1;
		}
		ans=((ll)r-(ll)cntl)*((ll)c-(ll)cntr);
		n=r+c+10;m=1;len=0;
		while (m<=n) m<<=1,len++;n=m;
		F(i,0,n-1)
		{
			int ret=0,t=i;
			F(j,1,len) ret<<=1,ret|=t&1,t>>=1;
			rev[i]=ret;
		}
		F(i,0,n-1) A[i].x=B[i].x=A[i].y=B[i].y=0;
		F(i,1,r) if (!line[i]) A[i].x=1.0;
		F(i,1,c) if (!row[i]) B[i].x=1.0;
		FFT(A,n,1); FFT(B,n,1);
		F(i,0,n-1) C[i]=A[i]*B[i];
		FFT(C,n,-1);
		F(i,1,r+c) if (ang[i]) ans-=(ll)C[i].x;
		printf("Case %d: ",++kas);printf("%lld\n",ans);
	}
}