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
1 #pragma GCC optimize(2) 2 #pragma G++ optimize(2) 3 #include<cstring> 4 #include<cmath> 5 #include<iostream> 6 #include<algorithm> 7 #include<cstdio> 8 9 #define eps 0.00000001 10 #define N 50007 11 using namespace std; 12 inline int read() 13 { 14 int x=0,f=1;char ch=getchar(); 15 while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();} 16 while(isdigit(ch)){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();} 17 return x*f; 18 } 19 20 int n,top; 21 double ans; 22 23 double sqr(double x){return x*x;} 24 struct P 25 { 26 double x,y; 27 P(){} 28 P(double _x,double _y):x(_x),y(_y){} 29 friend P operator+(P a,P b){return P(a.x+b.x,a.y+b.y);} 30 friend P operator-(P a,P b){return P(a.x-b.x,a.y-b.y);} 31 friend double operator*(P a,P b){return a.x*b.y-a.y*b.x;} 32 friend double operator/(P a,P b){return a.x*b.x+a.y*b.y;} 33 friend bool operator==(P a,P b){return fabs(a.x-b.x)<eps&&fabs(a.y-b.y)<eps;} 34 friend bool operator!=(P a,P b){return !(a==b);} 35 friend bool operator<(P a,P b) 36 { 37 if(fabs(a.y-b.y)<eps)return a.x<b.x; 38 return a.y<b.y; 39 } 40 friend double dis2(P a){return sqr(a.x)+sqr(a.y);} 41 friend void print(P a){printf("%.2lf %.2lf ",a.x,a.y);} 42 }p[N],q[N]; 43 44 bool cmp(P a,P b) 45 { 46 if(fabs((b-p[1])*(a-p[1]))<eps)return dis2(a-p[1])<dis2(b-p[1]); 47 return (a-p[1])*(b-p[1])>0;//叉乘大于0，表示向左转，a的斜率更小。 48 } 49 void Graham()//选出凸包上的点。 50 { 51 for (int i=1;i<=n;i++) 52 if(p[i]<p[1])swap(p[i],p[1]); 53 sort(p+2,p+n+1,cmp); 54 q[++top]=p[1],q[++top]=p[2]; 55 for (int i=3;i<=n;i++) 56 { 57 while((q[top]-q[top-1])*(p[i]-q[top-1])<eps&&top>1)top--;//如果当前的点的斜率更小，就替换 58 q[++top]=p[i]; 59 } 60 } 61 void RC()//求直径。 62 { 63 q[top+1]=q[1];//因为凸包是一个圈。 64 int now=2; 65 for (int i=1;i<=top;i++) 66 { 67 while((q[i+1]-q[i])*(q[now]-q[i])<(q[i+1]-q[i])*(q[now+1]-q[i])) 68 { 69 now++; 70 if(now==top+1)now=1; 71 } 72 ans=max(ans,dis2(q[now]-q[i])); 73 } 74 } 75 int main() 76 { 77 n=read(); 78 for (int i=1;i<=n;i++) 79 p[i].x=read(),p[i].y=read(); 80 Graham(); 81 RC(); 82 printf("%d",(int)ans); 83 }