Codeforces Round #365 (Div. 2) C

// 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 =998244353; 
const int N =1e6+10; 
#define clc(a,b) memset(a,b,sizeof(a))
const double eps = 1e-7;
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[10010];

int n;
bool judge(double k,double b){
    int tot=0,cnt=0;
    for(int i=1;i<=n;i++){
        if(k*p[i].x+b-p[i].y<0) 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=1;i<=n;i++){
        scanf("%lf%lf",&p[i].x,&p[i].y);
    }
    double k=u/v;
    if(judge(k,0.0)){
        printf("%.8f
",(double)(w/u));
    }
    else{
        double l=0,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
",ans);
    }
    return 0;
}