【模板】点集直径——p1452

首先要了解对踵点的定义https://blog.csdn.net/qq_34921856/article/details/80689629

然后是旋转卡壳算法的全家桶 https://blog.csdn.net/qq_34921856/article/details/80690822

但是代码实现有点复杂，单单是求点集直径更通俗的博客 https://jvruo.com/archives/79/

模板有问题，测了老半天

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<vector>
#include<algorithm>
using namespace std;

typedef double db;
const db eps=1e-8;
const db pi=acos(-1);
int sign(db k){
    if (k>eps) return 1; else if (k<-eps) return -1; return 0;
}
int cmp(db k1,db k2){return sign(k1-k2);}
int inmid(db k1,db k2,db k3){return sign(k1-k3)*sign(k2-k3)<=0;}// k3 在 [k1,k2] 内 
struct point{
    db x,y;
    point operator + (const point &k1) const{return (point){k1.x+x,k1.y+y};}
    point operator - (const point &k1) const{return (point){x-k1.x,y-k1.y};}
    point operator * (db k1) const{return (point){x*k1,y*k1};}
    point operator / (db k1) const{return (point){x/k1,y/k1};}
    bool operator < (const point k1) const{
        int a=cmp(x,k1.x);
        if (a==-1) return 1; else if (a==1) return 0; else return cmp(y,k1.y)==-1;
    }
    db abs(){return sqrt(x*x+y*y);}
    db abs2(){return x*x+y*y;}
    db dis(point k1){return ((*this)-k1).abs();}
};
int inmid(point k1,point k2,point k3){return inmid(k1.x,k2.x,k3.x)&&inmid(k1.y,k2.y,k3.y);}
db cross(point k1,point k2){return k1.x*k2.y-k1.y*k2.x;}
db dot(point k1,point k2){return k1.x*k2.x+k1.y*k2.y;} 

int onS(point k1,point k2,point q){return inmid(k1,k2,q)&&sign(cross(k1-q,k2-k1))==0;}
int checkconvex(vector<point>A){ //判断是否是凸包
    int n=A.size(); A.push_back(A[0]); A.push_back(A[1]);
    for (int i=0;i<n;i++) if (sign(cross(A[i+1]-A[i],A[i+2]-A[i]))==-1) return 0;
    return 1;
}
vector<point> ConvexHull(vector<point>A,int flag=1){ // 求凸包：flag=0 不严格 flag=1 严格 
    int n=A.size(); vector<point>ans(n*2); 
    sort(A.begin(),A.end()); int now=-1;
    for (int i=0;i<A.size();i++){
        while (now>0&&sign(cross(ans[now]-ans[now-1],A[i]-ans[now-1]))<flag) now--;
        ans[++now]=A[i];
    } int pre=now;
    for (int i=n-2;i>=0;i--){
        while (now>pre&&sign(cross(ans[now]-ans[now-1],A[i]-ans[now-1]))<flag) now--;
        ans[++now]=A[i];
    } ans.resize(now); return ans;
}

db RC(vector<point> q)
{
    if(q.size()==2)return (q[1]-q[0]).abs2();
    q.push_back(q[0]);
    int now=1;
    db ans=0;
    for(int i=0;i<q.size()-1;i++)//逆时针枚举凸包上的边(i,i+1) 
    {
        while(cross((q[i+1]-q[i]),(q[now]-q[i]))<cross((q[i+1]-q[i]),(q[now+1]-q[i])))
        {
            now++;
            if(now==q.size()-1)now=0;
        }
        ans=max(ans,(q[now]-q[i]).abs2());
    }
    return ans;
}

int n;
vector<point> v;

int main(){
    cin>>n;
        v.clear();
        for(int i=1;i<=n;i++){
            point p;
            scanf("%lf%lf",&p.x,&p.y);
            v.push_back(p);
        }
        
        vector<point>res=ConvexHull(v,1);
        db ans=RC(res);
        printf("%lld\n",(long long)(ans));
}