牛客挑战赛33 C 艾伦的立体机动装置（几何）

思路：

我们需要枚举展开多少条边 然后把上底面的点放到和下底面一个平面 然后算两点之间的距离 注意判断直线与线段是否有交点

#include <bits/stdc++.h>

using namespace std;
const double eps = 1e-8;
const double inf = 1e20;
const double pi = acos(-1.0);
const int N = 1e5+7;
//Compares a double to zero
int sgn(double x){
    if(fabs(x) < eps)return 0;
    if(x < 0)return-1;
    else return 1;
}
//square of a double
inline double sqr(double x){return x*x;}

struct Point{
    double x,y;
Point(){}
    Point(double _x,double _y){
        x = _x;
        y = _y;
    }
    void input(){
        scanf("%lf%lf",&x,&y);
    }
    void output(){
        printf("%.2f-%.2f\n",x,y);
    }
    bool operator == (Point b)const{
        return sgn(x-b.x) == 0 && sgn(y-b.y) == 0;
    }
    bool operator < (Point b)const{
        return sgn(x-b.x)== 0-sgn(y-b.y)?0:x<b.x;
    }
    Point operator-(const Point &b)const{
        return Point(x-b.x,y-b.y);
    }
    //叉积
    double operator ^(const Point &b)const{
        return x*b.y-y*b.x;
    }
    //点积
    double operator *(const Point &b)const{
        return x*b.x + y*b.y;
    }
    //返回长度
    double len(){
        return hypot(x,y);//库函数
    }
    //返回长度的平方
    double len2(){
        return x*x + y*y;
    }
    //返回两点的距离
    double distance(Point p){
        return hypot(x-p.x,y-p.y);
    }
    Point operator +(const Point &b)const{
        return Point(x+b.x,y+b.y);
    }
    Point operator *(const double &k)const{
        return Point(x*k,y*k);
    }
    Point operator /(const double &k)const{
        return Point(x/k,y/k);
    }
    //计算 pa 和 pb 的夹角
    //就是求这个点看 a,b 所成的夹角
    //测试 LightOJ1203
    double rad(Point a,Point b){
        Point p = *this;
        return fabs(atan2( fabs((a-p)^(b-p)),(a-p)*(b-p) ));
    }
    //化为长度为 r 的向量
    Point trunc(double r){
        double l = len();
        if(!sgn(l))return *this;
        r /= l;
        return Point(x*r,y*r);
    }
    //逆时针旋转 90 度
    Point rotleft(){
        return Point(-y,x);
    }
    //顺时针旋转 90 度
    Point rotright(){
        return Point(y,-x);
    }
    //绕着 p 点逆时针旋转 angle
    Point rotate(Point p,double angle){
        Point v = (*this)-p;
        double c = cos(angle), s = sin(angle);
        return Point(p.x + v.x*c-v.y*s,p.y + v.x*s + v.y*c);
    }
}p[N];
struct Line{
    Point s,e;
Line(){}
    Line(Point _s,Point _e){
        s = _s;
        e = _e;
    }
    bool operator ==(Line v){
        return (s == v.s)&&(e == v.e);
    }
    //根据一个点和倾斜角 angle 确定直线,0<=angle<pi
    Line(Point p,double angle){
        s = p;
        if(sgn(angle-pi/2) == 0){
            e = (s + Point(0,1));
        }
        else{
            e = (s + Point(1,tan(angle)));
        }
    }
    //ax+by+c=0
    Line(double a,double b,double c){
        if(sgn(a) == 0){
            s = Point(0,-c/b);
            e = Point(1,-c/b);
        }
        else if(sgn(b) == 0){
            s = Point(-c/a,0);
            e = Point(-c/a,1);
        }
        else{
            s = Point(0,-c/b);
            e = Point(1,(-c-a)/b);
        }
    }
    void input(){
        s.input();
        e.input();
    }
    void adjust(){
        if(e < s){
            swap(s,e);
        }
    }
    //求线段长度
    double length(){
        return s.distance(e);
    }
    //返回直线倾斜角 0<=angle<pi
    double angle(){
        double k = atan2(e.y-s.y,e.x-s.x);
        if(sgn(k) < 0)k += pi;
        if(sgn(k-pi) == 0)k-= pi;
        return k;
    }
    //点和直线关系
    //1 在左侧
    //2 在右侧
    //3 在直线上
    int relation(Point p){
        int c = sgn((p-s)^(e-s));
        if(c < 0)return 1;
        else if(c > 0)return 2;
        else return 3;
    }
    // 点在线段上的判断
    bool pointonseg(Point p){
        return sgn((p-s)^(e-s)) == 0 && sgn((p-s)*(p-e)) <= 0;
    }
    //两向量平行 (对应直线平行或重合)
    bool parallel(Line v){
        return sgn((e-s)^(v.e-v.s)) == 0;
    }
    //两线段相交判断
    //2 规范相交
    //1 非规范相交
    //0 不相交
    int segcrossseg(Line v){
        int d1 = sgn((e-s)^(v.s-s));
        int d2 = sgn((e-s)^(v.e-s));
        int d3 = sgn((v.e-v.s)^(s-v.s));
        int d4 = sgn((v.e-v.s)^(e-v.s));
        if( (d1^d2)==-2 && (d3^d4)==-2 )return 2;
        return (d1==0 && sgn((v.s-s)*(v.s-e))<=0) ||
        (d2==0 && sgn((v.e-s)*(v.e-e))<=0) ||
        (d3==0 && sgn((s-v.s)*(s-v.e))<=0) ||
        (d4==0 && sgn((e-v.s)*(e-v.e))<=0);
    }
    //直线和线段相交判断
    //-*this line -v seg
    //2 规范相交
    //1 非规范相交
    //0 不相交
    int linecrossseg(Line v){
        int d1 = sgn((e-s)^(v.s-s));
        int d2 = sgn((e-s)^(v.e-s));
        if((d1^d2)==-2) return 2;
        return (d1==0||d2==0);
    }
    //两直线关系
    //0 平行
    //1 重合
    //2 相交
    int linecrossline(Line v){
        if((*this).parallel(v))
        return v.relation(s)==3;
        return 2;
    }
    //求两直线的交点
    //要保证两直线不平行或重合
    Point crosspoint(Line v){
        double a1 = (v.e-v.s)^(s-v.s);
        double a2 = (v.e-v.s)^(e-v.s);
        return Point((s.x*a2-e.x*a1)/(a2-a1),(s.y*a2-e.y*a1)/(a2-a1
        ));
    }
    //点到直线的距离
    double dispointtoline(Point p){
        return fabs((p-s)^(e-s))/length();
    }
    //点到线段的距离
    double dispointtoseg(Point p){
        if(sgn((p-s)*(e-s))<0 || sgn((p-e)*(s-e))<0)
        return min(p.distance(s),p.distance(e));
        return dispointtoline(p);
    }
    //返回线段到线段的距离
    //前提是两线段不相交，相交距离就是 0 了
    double dissegtoseg(Line v){
        return min(min(dispointtoseg(v.s),dispointtoseg(v.e)),min(v
        .dispointtoseg(s),v.dispointtoseg(e)));
    }
    //返回点 p 在直线上的投影
    Point lineprog(Point p){
        return s + ( ((e-s)*((e-s)*(p-s)))/((e-s).len2()) );
    }
    //返回点 p 关于直线的对称点
    Point symmetrypoint(Point p){
        Point q = lineprog(p);
        return Point(2*q.x-p.x,2*q.y-p.y);
    }
};
double sum[N],ans=1e18;
int s,t;
void work(Point x,Line tmp,int i){
    Line res=Line(x,p[t]);
    //printf("%lf\n",res.length());
    if(res.linecrossseg(tmp)!=0){
        ans=min(ans,res.length());
    }
}
int main(){
    int n,h;
    scanf("%d%d",&n,&h);
    for(int i=0;i<n;i++){
        p[i].input();
    }
    scanf("%d%d",&s,&t);
    --s; --t;
    for(int i=1;i<n;i++){
        sum[i]=sum[i-1]+p[i-1].distance(p[i]);
    }
    double tot=sum[n]=sum[n-1]+p[n-1].distance(p[0]);
    for(int i=0;i<n;i++){
        Point x=p[i],y=p[(i+1)%n];
        Point z=(x-y).rotleft();
        z=z*(h*1.0/z.len());

        Point t=y+z;

        Line tmp=Line(t,t+z.rotleft());
        //printf("%lf\n",tot);
        double len;
        if(s>=i+1){
            len=sum[s]-sum[i+1];
        }else{
            len=tot-sum[i+1]+sum[s];
        }
        Point l,r;
        l=t+(tmp.e-tmp.s)/tmp.length()*len;
        r=t+(tmp.e-tmp.s)/tmp.length()*(len-tot);
//        cout<<i<<" "<<r.x<<" "<<r.y<<endl;
        work(l,Line(x,y),i);
        work(r,Line(x,y),i);
    }
    printf("%.6lf\n",ans);
}