题面

题解

我们枚举一下发射源，并把敌人和激光塔按极角排序，那么一组合法解就是两个极角之差不超过\(\pi\)且中间有敌人的三元组数，预处理一下前缀和然后用双指针就行了

//minamoto#include
    
     #define R register#define ll long long#define inline __inline__ __attribute__((always_inline))#define fp(i,a,b) for(R int i=(a),I=(b)+1;i
     
      I;--i)#define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)using namespace std;char buf[1<<21],*p1=buf,*p2=buf;inline char getc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}int read(){    R int res,f=1;R char ch;    while((ch=getc())>'9'||ch<'0')(ch=='-')&&(f=-1);    for(res=ch-'0';(ch=getc())>='0'&&ch<='9';res=res*10+ch-'0');    return res*f;}const int N=3205;struct node{    int x,y,t;    inline node(){}    inline node(R int xx,R int yy,R int tt):x(xx),y(yy),t(tt){}    inline ll operator *(const node &b)const{return 1ll*x*b.y-1ll*y*b.x;}    inline bool sgn()const{return x?x>0:y>0;}    inline bool operator <(const node &b)const{        return sgn()!=b.sgn()?sgn()
      
       <0;    }}a[N],b[N],c[N],st[N];ll res;int s[N],f[N],g[N],D,S,T,top;int main(){//  freopen("testdata.in","r",stdin);    D=read();    fp(i,1,D)a[i].x=read(),a[i].y=read();    S=read();    fp(i,1,S)b[i].x=read(),b[i].y=read();    T=read();    fp(i,1,T)c[i].x=read(),c[i].y=read();    fp(k,1,S){        top=0;        fp(i,1,D)st[++top]=node(a[i].x-b[k].x,a[i].y-b[k].y,0);        fp(i,1,T)st[++top]=node(c[i].x-b[k].x,c[i].y-b[k].y,1);        sort(st+1,st+1+top);        fp(i,1,top)st[top+i]=st[i];        fp(i,1,top<<1){            s[i]=s[i-1],f[i]=f[i-1],g[i]=g[i-1];            st[i].t?++f[i],g[i]+=s[i]:++s[i];        }        for(R int i=1,j=1;i<=top;++i)if(st[i].t){            if(j