Codeforces Round #365 (Div. 2) C - Chris and Road 二分找切点

 // Codeforces Round #365 (Div. 2)
 // C - Chris and Road 二分找切点
 // 题意：给你一个凸边行，凸边行有个初始的速度往左走，人有最大速度，可以停下来，竖直走。
 // 问走到终点的最短时间
 // 思路：
 // 1.贪心来做
 // 2.我觉的二分更直观
 // 可以抽象成：一条射线与凸边行相交，判断交点。二分找切点

 #include <bits/stdc++.h>
 using namespace std;
 #define LL long long
 const int inf = 0x3f3f3f3f;
 const int MOD =;
 const int N =1e6+;
 #define clc(a,b) memset(a,b,sizeof(a))
 const double eps = 1e-;
 struct Point{
     double x,y;
     Point(){}
     Point(int x_,int y_):x(x_),y(y_){}
     Point operator + (const Point &t)const{
         return Point(x+t.x,y+t.y);
     }
     Point operator - (const Point &t)const{
         return Point(x-t.x,y-t.y);
     }
     double operator * (const Point &t)const{
         return x*t.y-y*t.x;
     }
     double operator ^ (const Point &t)const{
         return x*t.x+y*t.y;
     }
     bool operator < (const Point & a) const{
         if(x==a.x) return y>a.y;
         return x<a.x;
     }
 }p[];

 int n;
 bool judge(double k,double b){
     int tot=,cnt=;
     for(int i=;i<=n;i++){
         if(k*p[i].x+b-p[i].y<) tot++;
         else cnt++;
         if(cnt&&tot) return false;
     }
     return true;
 }
 int main(){
     int w;
     double v,u;
     scanf("%d%d%lf%lf",&n,&w,&v,&u);
     for(int i=;i<=n;i++){
         scanf("%lf%lf",&p[i].x,&p[i].y);
     }
     double k=u/v;
     if(judge(k,0.0)){
         printf("%.8f\n",(double)(w/u));
     }
     else{
         double l=,r=1e9;
         double ans;
         while(r-l>eps){
              double mid = (l+r)/2.0;
              if(judge(k,-mid/v*u)){
                 ans=(double)(w/u)+mid/v;
                 r=mid;
              }
              else{
                 l=mid;
              }
         }
         printf("%.8f\n",ans);
     }
     return ;
 }