POJ2187-Beauty Contest-凸包

平面最远点对

由于点数为1e5，而整数点的情况下，凸包上点的个数为sqrt(M)，M为范围。

这样求出凸包之后n^2枚举维护距离就可以了

否则就用旋转卡壳。

这里用了挑战上的做法，比较简洁。

 #include <cstdio>
 #include <algorithm>
 #define LL long long

 using namespace std;

 const int maxn = 5e4+;
 struct Point{
     int x,y;
     Point(int _x=,int _y=):x(_x),y(_y){}
     bool operator < (const Point &rhs) const{
         if(x == rhs.x) return y < rhs.y;
         else return x < rhs.x;
     }
     LL operator *(const Point &rhs) const{
         return (LL)x*rhs.x+(LL)y*rhs.y;
     }
     LL operator ^(const Point &rhs) const{
         return (LL)x*rhs.y - (LL)y*rhs.x;
     }
     Point operator -(const Point &rhs) const{
         return Point(x-rhs.x,y-rhs.y);
     }
 }p[maxn],ch[maxn];
 typedef Point Vector;

 LL dist(Point a,Point b)
 {
     return (a-b)*(a-b);
 }

 int ConvexHull(Point *pt,int n)
 {
     sort(pt,pt+n);
     int k = ;
     for(int i=;i<n;i++)
     {
         while(k >  && ((ch[k-]-ch[k-])^(pt[i]-ch[k-])) <= ) k--;
         ch[k++] = pt[i];
     }
     for(int i=n-,t = k;i>=;i--)
     {
         while(k > t && ((ch[k-]-ch[k-])^(pt[i]-ch[k-])) <= ) k--;
         ch[k++] = pt[i];
     }
     return k;
 }
 int N;
 int main()
 {
     while(~scanf("%d",&N))
     {
         for(int i=;i<N;i++)
         {
             scanf("%d%d",&p[i].x,&p[i].y);
         }
         int cnt = ConvexHull(p,N);
         LL ans = ;
         for(int i=;i<cnt;i++)
         {
             //printf("%d %d\n",ch[i].x,ch[i].y);
             for(int j=i+;j<cnt;j++)
             {
                 ans = max(ans,dist(ch[i],ch[j]));
             }
         }
         printf("%lld\n",ans);
     }
 }