POJ2187 旋转卡壳 求最长直径

给定平面上的一些散点集，求最远两点距离的平方值。

题解：

旋转卡壳求出凸包，然后根据单调性，求出最远两点的最大距离

 #pragma GCC optimize(2)
 #pragma G++ optimize(2)
 #include<cstring>
 #include<cmath>
 #include<iostream>
 #include<algorithm>
 #include<cstdio>

 #define eps 0.00000001
 #define N 50007
 using namespace std;
 inline int read()
 {
     int x=,f=;char ch=getchar();
     while(!isdigit(ch)){if(ch=='-')f=-;ch=getchar();}
     while(isdigit(ch)){x=(x<<)+(x<<)+ch-'';ch=getchar();}
     return x*f;
 }

 int n,top;
 double ans;

 double sqr(double x){return x*x;}
 struct P
 {
     double x,y;
     P(){}
     P(double _x,double _y):x(_x),y(_y){}
     friend P operator+(P a,P b){return P(a.x+b.x,a.y+b.y);}
     friend P operator-(P a,P b){return P(a.x-b.x,a.y-b.y);}
     friend double operator*(P a,P b){return a.x*b.y-a.y*b.x;}
     friend double operator/(P a,P b){return a.x*b.x+a.y*b.y;}
     friend bool operator==(P a,P b){return fabs(a.x-b.x)<eps&&fabs(a.y-b.y)<eps;}
     friend bool operator!=(P a,P b){return !(a==b);}
     friend bool operator<(P a,P b)
     {
         if(fabs(a.y-b.y)<eps)return a.x<b.x;
         return a.y<b.y;
     }
     friend double dis2(P a){return sqr(a.x)+sqr(a.y);}
     friend void print(P a){printf("%.2lf %.2lf\n",a.x,a.y);}
 }p[N],q[N];

 bool cmp(P a,P b)
 {
     if(fabs((b-p[])*(a-p[]))<eps)return dis2(a-p[])<dis2(b-p[]);
     return (a-p[])*(b-p[])>;//叉乘大于0，表示向左转，a的斜率更小。
 }
 void Graham()//选出凸包上的点。
 {
     for (int i=;i<=n;i++)
         if(p[i]<p[])swap(p[i],p[]);
     sort(p+,p+n+,cmp);
     q[++top]=p[],q[++top]=p[];
     for (int i=;i<=n;i++)
     {
         while((q[top]-q[top-])*(p[i]-q[top-])<eps&&top>)top--;//如果当前的点的斜率更小，就替换
         q[++top]=p[i];
     }
 }
 void RC()//求直径。
 {
     q[top+]=q[];//因为凸包是一个圈。
     int now=;
     for (int i=;i<=top;i++)
     {
         while((q[i+]-q[i])*(q[now]-q[i])<(q[i+]-q[i])*(q[now+]-q[i]))
         {
             now++;
             if(now==top+)now=;
         }
         ans=max(ans,dis2(q[now]-q[i]));
     }
 }
 int main()
 {
     n=read();
     for (int i=;i<=n;i++)
         p[i].x=read(),p[i].y=read();
     Graham();
     RC();
     printf("%d",(int)ans);
 }