Poj 2187 旋转卡壳

Poj 2187 旋转卡壳求解

传送门
旋转卡壳，是利用凸包性质来求解凸包最长点对的线性算法，我们逐渐改变每一次方向，然后枚举出这个方向上的踵点对(最远点对),类似于用游标卡尺卡着凸包旋转一周，答案就在这其中的某个方向上。

直接暴力和旋转卡壳速度对比(仅此题)

#include <queue>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define ll long long
#define inf 1000000000LL
#define mod 1000000007
using namespace std;
int read()
{
    int x=0,f=1;
    char ch=getchar();
    while(ch<'0'||ch>'9')
    {
        if(ch=='-')f=-1;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9')
    {
        x=x*10+ch-'0';
        ch=getchar();
    }
    return x*f;
}
const int N = 5e4+10;
const double PI = acos(-1.0);
const double eps = 1e-12;

int dcmp(double x) {
    if(fabs(x)<eps) return 0; else return x<0? -1:1;
}

struct Pt {
    double x,y;
    Pt(double x=0,double y=0) :x(x),y(y) {};
};
typedef Pt vec;

vec operator - (Pt a,Pt b) { return vec(a.x-b.x,a.y-b.y); }
vec operator + (vec a,vec b) { return vec(a.x+b.x,a.y+b.y); }
bool operator == (Pt a,Pt b) {
    return dcmp(a.x-b.x)==0 && dcmp(a.y-b.y)==0;
}
bool operator < (const Pt& a,const Pt& b) {
    return a.x<b.x || (a.x==b.x && a.y<b.y);
}

vec rotate(vec a,double x) {
    return vec(a.x*cos(x)-a.y*sin(x),a.x*sin(x)+a.y*cos(x));
}
double cross(vec a,vec b) { return a.x*b.y-a.y*b.x; }
double dist(Pt a,Pt b) {
    //return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
    return ((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

vector<Pt> ConvexHull(vector<Pt> p) {
    sort(p.begin(),p.end());
    p.erase(unique(p.begin(),p.end()),p.end());
    int n=p.size() , m=0;
    vector<Pt> ch(n+1);
    for(int i=0;i<n;i++) {
        while(m>1 && cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0) m--;
        ch[m++]=p[i];
    }
    int k=m;
    for(int i=n-2;i>=0;i--) {
        while(m>k && cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0) m--;
        ch[m++]=p[i];
    }
    if(n>1) m--;
    ch.resize(m); return ch;
}
vector<Pt>q,con;
double RC(){
    con=ConvexHull(q);
    int n=con.size();
    if(n==2){         //处理特殊情况
        return dist(con[0],con[1]);
    }
    int i=0,j=0;
    for(int k=0;k<n;k++){
       if(!(con[i]<con[k])) i=k;
       if(con[j]<con[k]) j=k;
    }
    double res=0;
    int si=i,sj=j;
    while(i!=sj||j!=si){
        res=max(res,dist(con[i],con[j]));
        if(cross(con[(i+1)%n]-con[i],con[(j+1)%n]-con[j])<0){
            i=(i+1)%n;
        }else{
            j=(j+1)%n;
        }
    }
    return res;
}
int main(){
     int n=read();
     int x,y;
     for(int i=0;i<n;i++){
           x=read();y=read();
           q.push_back(Pt((double)x,(double)y));
     }
     printf("%.0f
",RC());
     return 0;
}