OJ题号
牛客挑战赛33 C 艾伦的立体机动装置
https://ac.nowcoder.com/acm/contest/1115/C
简单题意
中文题
正解思路
- 因为下表面是可以走的，所以不能直接算s点和t点展开的直线距离。
- 枚举经过下表面的每一条边，再在每条边上三分求出最短距离。注意最短距离的初始值要设为不经过下表面任何一条边的最短距离。
slen记录每个顶点到s点的距离，getdis是获得上表面或下表面的两点距离，finddis是获得如果经过下表面某个点的路程距离，tridiv三分。因为我的下标是从0开始的，所以s点是p[s-1]。
详细链接
#include<bits/stdc++.h>
#define ll long long
using namespace std;
int s,t,n;
double ans,slen[100001],h;
pair<double,double>p[100001];
inline double getdis(pair<double,double>p1,pair<double,double>p2)
{
return sqrt((p1.first-p2.first)*(p1.first-p2.first)+(p1.second-p2.second)*(p1.second-p2.second));
}
inline double finddis(pair<double,double>now,int l,int r)
{
double len = min(slen[l]+getdis(p[l],now),slen[r]+getdis(p[r],now));
return sqrt(h*h+len*len)+getdis(now,p[t-1]);
}
inline double tridiv(int l,int r)
{
double minn = 0x3f3f3f3f;
pair<double,double> left = p[l];
pair<double,double> right = p[r];
while(getdis(left,right)>1e-3)
{
pair<double,double> first = make_pair((2*left.first+right.first)/3,(2*left.second+right.second)/3);
pair<double,double> second = make_pair((left.first+2*right.first)/3,(left.second+2*right.second)/3);
double firdis = finddis(first,l,r);
double secdis = finddis(second,l,r);
if(firdis>secdis)
{
left = first;
minn = min(minn,secdis);
}
else
{
right = second;
minn = min(minn,firdis);
}
}
return minn;
}
int main()
{
scanf("%d%lf",&n,&h);
for(int i = 0; i<n; i++)
scanf("%lf%lf",&p[i].first,&p[i].second);
scanf("%d%d",&s,&t);
double s1 = 0,s2 = 0,sum = 0;
int last = s-1;
slen[last] = 0;
for(int i = s%n; i!=s-1; i++,i%=n)
{
double len = getdis(p[last],p[i]);
sum+=len;
slen[i] = slen[last]+len;
last = i;
}
sum+=getdis(p[last],p[s-1]);
for(int i = 0; i<n; i++)
slen[i] = min(slen[i],sum-slen[i]);
ans = sqrt(slen[t-1]*slen[t-1]+h*h);
last = n-1;
for(int i = 0; i<n; i++,last++,last%=n)
ans = min(ans,tridiv(last,i));
printf("%.6lf",ans);
return 0;
}
点与线模板，但我不会用这个模板做这道题。
#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);
}
