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); }