1773 -- 【模擬試題】最遠距離點對
Description
給定平面上的n個點，找出它們之間最遠的點對。
Input
多組數據，每組第一行n代表點數，接着n行爲點的座標，座標爲整數，不超過Longint範圍。n<=30000。
Output
每組一行，最遠點對的距離，保留2位小數
Sample Input
4 0 0 1 1 0 1 1 0
Sample Output
1.41
第一次旋轉卡殼呢：
#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
struct Vector{
double x,y;
Vector operator +(Vector &a){
Vector v1;
v1.x=this->x+a.x,v1.y=this->y+a.y;
return v1;
};
Vector operator -(Vector &a){
Vector v1;
v1.x=this->x- a.x,v1.y=this->y -a.y;
return v1;
};
double operator * (Vector &a){
return ((this->x)*a.y)-(a.x*(this->y));
};
}P[500055] ;
int n;
int ch[500005];
bool vis[500005];
int top=0;
bool cmp(Vector a,Vector b){
return (a.y<b.y)||((a.y==b.y)&&(a.x<b.x));
}
bool Judge(Vector a,Vector b,Vector c){
Vector q=b-c;
Vector w=a-c;
return q*w>=0;
}
double Get_Area(Vector a,Vector b,Vector c){
a=a-c;
b=b-c;
return a*b;
}
double Get_Long(Vector a,Vector b){
return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
void TUBao(){
sort(P+1,P+1+n,cmp);
ch[++top]=1;
int k=2;
while(k<=n){
while(top>1&&Judge(P[ch[top]],P[k],P[ch[top-1]]))
vis[ch[top--]]=0;
ch[++top]=k;
vis[k]=1;
k++;
}
int Max=top;
k=n-1;
while(k>=1){
while(vis[k])k--;
while(top>Max&&Judge(P[ch[top]],P[k],P[ch[top-1]]))
vis[ch[top--]]=0;
ch[++top]=k;
vis[k]=1;
k--;
}
}
void Rotating_calipers(){
int p=2;
double ans=0.0;
for(int q=1;q<top;q++){
while(Get_Area(P[ch[q]],P[ch[q+1]],P[ch[p+1]])>Get_Area(P[ch[q]],P[ch[q+1]],P[ch[p]]))p=(p+1)%(n+1);
ans=max(ans,max(Get_Long(P[ch[q]],P[ch[p]]),Get_Long(P[ch[q+1]],P[ch[p+1]])));
}
printf("%.2lf",ans);
}
int main(){
while(scanf("%d",&n)==1){
for(int i=1;i<=n;i++){
scanf("%lf%lf",&P[i].x,&P[i].y);
}
TUBao();
Rotating_calipers();
}
return 0;
}