LOJ#2307. 「NOI2017」分身术

$n \leq 100000$个点，$m \leq 100000$次询问，每次问删掉一些点后的凸包面积。

不会啦写个20暴力，其实是可以写到50的。当个计算几何板子练习。

 //#include<iostream>
 #include<cstring>
 #include<cstdio>
 //#include<math.h>
 //#include<set>
 //#include<queue>
 //#include<bitset>
 //#include<vector>
 #include<algorithm>
 #include<stdlib.h>
 using namespace std;

 #define LL long long
 int qread()
 {
     char c; int s=,f=; while ((c=getchar())<'' || c>'') (c=='-') && (f=-);
     do s=s*+c-''; while ((c=getchar())>='' && c<=''); return s*f;
 }

 //Pay attention to '-' , LL and double of qread!!!!

 int n,m;
 #define maxn 100011
 int Abs(int x) {return x>?x:-x;}
 struct Poi
 {
     int x,y;
     LL operator * (const Poi &b) const {return 1ll*x*b.y-1ll*y*b.x;}
     Poi operator - (const Poi &b) const {return (Poi){x-b.x,y-b.y};}
     bool operator < (const Poi &b)
     {
         if (x== && y==) return ;
         if (b.x== && b.y==) return ;
         Poi c=(Poi){,};
         LL tmp=(c-(*this))*(c-b);
         return tmp> || (tmp== && Abs(x)<Abs(b.x));
     }
 }p[maxn],q[maxn],sta[maxn]; int lq,top;
 bool vis[maxn];

 int main()
 {
     n=qread(); m=qread();
     for (int i=;i<=n;i++) {p[i].x=qread(); p[i].y=qread();}

     LL s=-;
     int K,op;
     while (m--)
     {
         memset(vis,,sizeof(vis));
         K=qread();
         while (K--) vis[(qread()+s+n)%n+]=;
         int my=0x3f3f3f3f,id=; lq=top=;
         for (int i=;i<=n;i++) if (!vis[i])
         {
             q[++lq]=p[i];
             if (p[i].y<my) {my=p[i].y; id=lq;}
         }
         for (int i=;i<=lq;i++) if (i!=id) q[i].x-=q[id].x,q[i].y-=q[id].y; q[id].x=q[id].y=;
         sort(q+,q++lq);
 //        for (int i=1;i<=lq;i++) cout<<q[i].x<<' '<<q[i].y<<endl;cout<<endl;
         sta[++top]=q[lq];
         for (int i=;i<=lq;i++)
         {
             while (top> && (sta[top]-sta[top-])*(q[i]-sta[top])<=) top--;
             sta[++top]=q[i];
         }
 //        for (int i=1;i<=top;i++) cout<<sta[i].x<<' '<<sta[i].y<<endl; cout<<endl;
         s=;
         for (int i=;i<top;i++) s+=sta[i]*sta[i+];
         printf("%lld\n",s);
     }
     return ;
 }