求凸包周长，用旋转卡壳，具体可见yyb大佬的博客
顺便一提这题暴力+随机化也能过
暴力代码
//minamoto
#include<bits/stdc++.h>
#define rint register int
#define int long long
using namespace std;
#define getc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
char buf[1<<21],*p1=buf,*p2=buf;
int read(){
int res,f=1;char ch;
while((ch=getc())>'9'||ch<'0')(ch=='-')&&(f=-1);
for(res=ch-'0';(ch=getc())>='0'&&ch<='9';res=res*10+ch-'0');
return res*f;
}
const int N=100005;
int x[N],y[N],ans=-1;int n;
inline int dis(int i,int j){return (x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]);}
signed main(){
// freopen("testdata.in","r",stdin);
srand(time(0));
n=read();
for(rint i=1;i<=n;++i)x[i]=read(),y[i]=read();
while(clock()<CLOCKS_PER_SEC*0.8)
for(rint i=rand()%n+1,j=1;j<=n;++j)ans=max(ans,dis(i,j));
printf("%d
",ans);return 0;
}
旋转卡壳代码
//minamoto
#include<bits/stdc++.h>
#define rint register int
#define inf 0x3f3f3f3f
using namespace std;
#define getc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
char buf[1<<21],*p1=buf,*p2=buf;
int read(){
int res,f=1;char ch;
while((ch=getc())>'9'||ch<'0')(ch=='-')&&(f=-1);
for(res=ch-'0';(ch=getc())>='0'&&ch<='9';res=res*10+ch-'0');
return res*f;
}
const int N=1e5+5;
struct node{int x,y;}p[N],st[N];
int n,k,top,ans;
inline bool cmp(node a,node b){
double A=atan2(a.y-p[1].y,a.x-p[1].x);
double B=atan2(b.y-p[1].y,b.x-p[1].x);
return A!=B?A<B:a.x<b.x;
}
inline int cross(node a,node b,node c){return (b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x);}
inline int dis(node a,node b){return (b.y-a.y)*(b.y-a.y)+(b.x-a.x)*(b.x-a.x);}
void get(){
if(top==1)return (void)(ans=dis(st[0],st[1]));
st[++top]=st[0];
for(rint i=0,j=2;i<top;++i){
while(cross(st[i],st[i+1],st[j])<cross(st[i],st[i+1],st[j+1]))j=(j+1)%top;
ans=max(ans,max(dis(st[i],st[j]),dis(st[i+1],st[j])));
}
}
int main(){
// freopen("testdata.in","r",stdin);
n=read(),p[0]={inf,inf};
for(rint i=1;i<=n;++i){
p[i].x=read(),p[i].y=read();
if(p[0].y>p[i].y||(p[0].y==p[i].y&&p[0].x>p[i].x))p[0]=p[i],k=i;
}
swap(p[k],p[1]),sort(p+2,p+1+n,cmp);
st[0]=p[1],st[1]=p[2],top=1;
for(rint i=3;i<=n;++i){
while(top&&cross(st[top-1],p[i],st[top])>=0)--top;
st[++top]=p[i];
}
get();printf("%d
",ans);return 0;
}