UESTC 1332 Beauty Contest(凸包直径)

题目链接：http://acm.uestc.edu.cn/problem.php?pid=1332

题意：给定n个点。求最远点对。

思路：模板题。

 

View Code 

 #include <iostream>

 #include <stdio.h>

 #include <cmath>

 #include <algorithm>

 #define max(x,y) ((x)>(y)?(x):(y))

 using namespace std;

 

 struct point

 {

     double x,y;

 

     point(){}

     point(double _x,double _y)

     {

         x=_x;

         y=_y;

     }

 

 

     void get()

     {

         scanf("%lf%lf",&x,&y);

     }

 };

 

 const double EPS=1e-8;

 const int MAX=100005;

 point p[MAX],q[MAX];

 int n,m;

 

 

 int DB(double x)

 {

     if(x>EPS) return 1;

     if(x<-EPS) return -1;

     return 0;

 }

 

 double Dis(point a,point b)

 {

     return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));

 }

 

 //判断p在向量ab的哪一侧

 //右侧：返回正值

 //左侧：返回负值

 //在向量ab上返回0

 double cross(point a,point b,point p)

 {

     return (b.x-a.x)*(p.y-a.y)-(b.y-a.y)*(p.x-a.x);

 }

 

 int cmp(point a,point b)

 {

     double x=Dis(a,p[0]),y=Dis(b,p[0]);

     int flag=DB(cross(p[0],a,b));

     if(flag) return flag==1;

     return DB(x-y)<=0;

 }

 

 void Graham(point p[],int n,point q[],int &m)

 {

     point temp;

     int i,k=0,a,b;

     for(i=1;i<n;i++)

     {

         a=DB(p[i].y-p[k].y);

         b=DB(p[i].x-p[k].x);

         if(a==-1||!a&&b==-1) k=i;

     }

     if(k!=0) temp=p[0],p[0]=p[k],p[k]=temp;

     sort(p+1,p+n,cmp);

     q[0]=p[0];

     q[1]=p[1];

     p[n]=p[0];

     m=2;

     for(i=2;i<=n;i++)

     {

         while(m>1&&DB(cross(q[m-2],q[m-1],p[i]))<=0) m--;

         q[m++]=p[i];

     }

     m--;

 }

 

 

 double calMaxLen(point q[],int n)

 {

     q[n]=q[0];

     int i,p=1;

     double ans=0,x,y;

     for(i=0;i<n;i++)

     {

         while(1)

         {

             x=cross(q[i],q[i+1],q[p+1]);

             y=cross(q[i],q[i+1],q[p]);

             if(DB(x-y)<=0) break;

             p=(p+1)%n;

         }

         ans=max(ans,max(Dis(q[i],q[p]),Dis(q[i+1],q[p+1])));

     }

     return ans;

 }

 

 int main()

 {

     while(scanf("%d",&n)!=-1)

     {

         int i;

         for(i=0;i<n;i++) p[i].get();

         Graham(p,n,q,m);

         double ans=calMaxLen(q,m);

         printf("%.0lf\n",ans*ans);

     }

     return 0;

 }