UVA 12633 Super Rooks on Chessboard (生成函数+FFT)

 1 #include <cmath>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 #define N1 (1<<18)
 6 #define M1 (N1<<1)
 7 #define ll long long 
 8 #define dd double
 9 #define idx(X) (X-‘a‘)
10 using namespace std;
11 
12 int gint()
13 {
14     int ret=0,fh=1; char c=getchar();
15     while(c<‘0‘||c>‘9‘){if(c==‘-‘)fh=-1;c=getchar();}
16     while(c>=‘0‘&&c<=‘9‘){ret=ret*10+c-‘0‘;c=getchar();}
17     return ret*fh;
18 }
19 int T,n,m,num,r[N1];
20 int nt[N1],x[N1],y[N1];
21 const dd pi=acos(-1);
22 
23 struct cp{
24 dd x,y;
25 friend cp operator + (const cp &s1,const cp &s2){ return (cp){s1.x+s2.x,s1.y+s2.y}; }
26 friend cp operator - (const cp &s1,const cp &s2){ return (cp){s1.x-s2.x,s1.y-s2.y}; }
27 friend cp operator * (const cp &s1,const cp &s2){ return (cp){s1.x*s2.x-s1.y*s2.y,s1.y*s2.x+s1.x*s2.y}; }
28 }a[N1],b[N1],c[N1];
29 
30 void init()
31 {
32     memset(a,0,sizeof(a));
33     memset(b,0,sizeof(b));
34     memset(c,0,sizeof(c));
35     memset(nt,0,sizeof(nt));
36 }
37 void FFT(cp *s,int len,int type)
38 {
39     int i,j,k; cp wn,w,t;
40     for(i=0;i<len;i++) if(i<r[i]) swap(s[i],s[r[i]]);
41     for(k=2;k<=len;k<<=1)
42     {
43         wn=(cp){cos(2.0*type*pi/k),sin(2.0*type*pi/k)};
44         for(i=0;i<len;i+=k)
45         {
46             w=(cp){1,0};
47             for(j=0;j<(k>>1);j++,w=w*wn)
48             {
49                 t=w*s[i+j+(k>>1)];
50                 s[i+j+(k>>1)]=s[i+j]-t;
51                 s[i+j]=s[i+j]+t;
52             }
53         }
54     }
55 }
56 void FFT_Main(int len)
57 {
58     FFT(a,len,1); FFT(b,len,1);
59     for(int i=0;i<len;i++) c[i]=a[i]*b[i];
60     FFT(c,len,-1);
61     for(int i=0;i<len;i++) c[i].x=c[i].x/len;
62 }
63 
64 int main()
65 {
66     scanf("%d",&T); int t;
67     for(t=1;t<=T;t++){
68     
69     init();
70     scanf("%d%d%d",&n,&m,&num);
71     int i,j,de,len,L; ll ans=0;
72     for(i=1;i<=num;i++) x[i]=n-gint(), y[i]=gint()-1, a[x[i]].x=-1, b[y[i]].x=-1, nt[x[i]+y[i]]=1;
73     for(i=0;i<n;i++) a[i].x++;
74     for(i=0;i<m;i++) b[i].x++;
75     for(len=1,L=0;len<n+m-1;len<<=1,L++);
76     for(i=0;i<len;i++) r[i]=(r[i>>1]>>1)|((i&1)<<(L-1));
77     FFT_Main(len);
78     for(i=0;i<=n+m-2;i++) if(!nt[i]) ans+=(int)(c[i].x+0.1);
79     printf("Case %d: %lld\n",t,ans);
80     
81     }
82     return 0;
83 }