Codeforces Gym - 101635K Blowing Candles [Rotating Candles]

Question meaning: give you some points on the plane, find the shortest diameter of a convex hull

Idea: Rotate the jammed shell, and then just do a little bit of it. The farthest point of the rotating jammed shell is almost the same. The cur band table is the found opposite point, and then with the current straight line p[i], p[i+1], find out distance

Code:

#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-16;
int sgn(double x)
{
    if(fabs(x) < eps)return 0;
    if(x < 0)return -1;
    else return 1;
}
struct point
{
    double x,y;
    point(int _x = 0,int _y = 0)
    {
        x = _x;
        y = _y;
    }
    point operator -(const point &b)const
    {
        return point(x - b.x, y - b.y);
    }
    double operator ^(const point &b)const
    {
        return x*b.y - y*b.x;
    }
    double operator *(const point &b)const
    {
        return x*b.x + y*b.y;
    }
    void sc()
    {
        scanf("%lf%lf",&x,&y);
    }
};

struct Line
{
    point s,e;
    Line() {}
    Line(point _s,point _e)
    {
        s = _s;
        e = _e;
    }
    pair<int,point> operator &(const Line &b)const
    {
        point res = s;
        if(sgn((s-e)^(b.s-b.e)) == 0)
        {
            if(sgn((s-b.e)^(b.s-b.e)) == 0)
                return make_pair(0,res);
            else return make_pair(1,res);
        }
        double t = ((s-b.s)^(b.s-b.e))/((s-e)^(b.s-b.e));
        res.x += (e.x-s.x)*t;
        res.y += (e.y-s.y)*t;
        return make_pair(2,res);
    }
};
double dist(point a,point b)
{
    return (double)sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
point NearestpointToLineSeg(point P,Line L)
{
    point result;
    double t = ((P-L.s)*(L.e-L.s))/((L.e-L.s)*(L.e-L.s));
    if(t >= 0.000 && t <= 1.0000)
    {
        result.x = L.s.x + (L.e.x - L.s.x)*t;
        result.y = L.s.y + (L.e.y - L.s.y)*t;
    }
    else
    {
        if(dist(P,L.s) < dist(P,L.e))
            result = L.s;
        else result = L.e;
    }
    return result;
}
double emmm(point P,Line L)
{
    point zz=NearestpointToLineSeg(P,L);
    return dist(zz,P);
}
const  int MAXN = 2e5 + 7 ;
point List[MAXN];
int Stack[MAXN],top;
bool _cmp(point p1,point p2)
{
    double tmp = (p1-List[0])^(p2-List[0]);
    if(tmp > 0)return true;
    else if(tmp == 0 && dist(p1,List[0]) <= dist(p2,List[0]))
        return true;
    else return false;
}
void Graham(int n)
{
    point p0;
    int k = 0;
    p0 = List[0];
    for(int i = 1; i < n; i++){
        if(p0.y > List[i].y || (p0.y == List[i].y && p0.x > List[i].x))
        {
            p0 = List[i];
            k = i;
        }
    }
    swap(List[k],List[0]);
    sort(List+1,List+n,_cmp);
    if(n == 1)
    {
        top = 1;
        Stack[0] = 0;
        return;
    }
    if(n == 2)
    {
        top = 2;
        Stack[0] = 0;
        Stack[1] = 1;
        return;
    }
    Stack[0] = 0;
    Stack[1] = 1;
    top = 2;
    for(int i = 2; i < n; i++)
    {
        while(top > 1 &&
                ((List[Stack[top-1]]-List[Stack[top-2]])^(List[i]-List[Stack[top-2]])) <= 0)
            top--;
        Stack[top++] = i;
    }
}
double rotating_calipers(point p[],int n)
{
    double years = 999999999999999999999.0 ;
    point v;
    int cur = 1;
    for(int i = 0; i < n; i++)
    {
        v = p[i]-p[(i+1)%n];
        while((v^(p[(cur+1)%n]-p[cur])) < 0){
            cur = (cur+1)%n;s
        }
        Line qw=Line(p[(i)%n],p[(i+1)%n]);
//        point zz=NearestpointToLineSeg(p[cur],qw);
        double as=emmm(p[cur],qw);
        //as=max(as,zx);
        ans=min(ans,as);
//        ans=min(ans,max(dist(p[i],p[cur]),dist(p[(i+1)%n],p[(cur+1)%n])));
    }
    return ans;
}
point p[MAXN];
intmain ()
{
    int n,r;
    while(~scanf("%d%d",&n,&r))
    {

        for(int i = 0; i < n; i++)List[i].sc();
        Graham(n);
        for(int i = 0; i < top; i++)p[i] = List[Stack[i]];
        printf("%.16f\n",rotating_calipers(p,top));
    }
    return 0;
}